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Punjab State Board PSEB 5th Class Hindi Book Solutions Chapter 11 साहसी दीपा Textbook Exercise Questions and Answers.

PSEB Solutions for Class 5 Hindi Chapter 11 साहसी दीपा (2nd Language)

साहसी दीपा अभ्यास

नीचे गुरुमुखी और देवनागरी लिपि में दिये गये शब्दों को पढ़ो और हिन्दी शब्दों को लिखने का अभ्यास करो :

• ਲੜਕੀ = लड़की
• ਨੌਕਰੀ = नौकरी
• ਸਹੇਲੀ = सहेली
• ਨਦੀ = नदी
• ਬਹਾਦੁਰ = बहादुर
• ਮਗਰਮੱਛ = मगरमच्छ

नीचे एक ही अर्थ के लिए पंजाबी और हिन्दी भाषा में शब्द दिये गये हैं। इन्हें ध्यान से पढ़ो और हिन्दी शब्दों को लिखो :

• ਬਾਰ੍ਹਾਂ = बारह
• ਜੀਉਂਦੇ = जीवित
• ਪ੍ਰਸੰਸਾ = सराहना।
• ਇਨਾਮ = पुरस्कार

पढ़ो, समझो और लिखो

(क) च् + छ = च्छ = अच्छा, मगरमच्छ
न् + य = न्य = धन्यवाद
ल् + द = ल्द = जल्दी
स् + क = स्क = पुरस्कार

बताओ

प्रश्न 1.
दीपा कैसी लड़की थी?
उत्तर :
दीपा साहसी लड़की थी।

प्रश्न 2.
दीपा नदी के किनारे क्या कर रही थी?
उत्तर :
दीपा नदी के किनारे कपड़े धो रही थी।

प्रश्न 3.
‘बचाओ-बचाओ’ की आवाजें किसने लगाईं?
उत्तर :
बचाओ-बचाओ’ की आवाजें हरीश ने लगाईं।

प्रश्न 4.
दीपा की सहेली ने उसे नदी में कूदने से मना क्यों किया?
उत्तर :
पानी में तेजी से बढे आ रहे मगरमच्छ को देखकर दीपा की सहेली ने दीपा को पानी में कूदने से मना किया।

प्रश्न 5.
दीपा ने रवि को डूबने से कैसे बचाया?
उत्तर :
दीपा झट से पानी में कूद गई और रवि को पानी से निकाल कर बाहर ले आई। इस प्रकार उसने रवि को डूबने से और मगरमच्छ से बचाया।

प्रश्न 6.
दीपा को कौन-सा पुरस्कार, कब मिला?
उत्तर :
दीपा को वीरता का पुरस्कार 26 जनवरी को मिला।

वाक्य पूरे करो

होशियार किनारे मगरमच्छ बहादुर धन्यवाद वीरता पुरस्कार।

1. दीपा ………………………. और ………………………. लड़की थी।
2. नदी में ………………………. था।
3. दीपा रवि को ………………………. पर ले आई।
4. रवि और हरीश के माता-पिता ने दीपा का ………………………. किया।
5. दीपा को 26 जनवरी के दिन ………………………. मिला।

उत्तर :

1. दीपा बहादुर और होशियार लड़की थी।
2. नदी में मगरमच्छ था।
3. दीपा रवि को किनारे पर ले आई।
4. रवि और हरीश के माता-पिता ने दीपा का धन्यवाद किया।
5. दीपा को 26 जनवरी के दिन वीरता पुरस्कार मिला।

वाक्य बनाओ

1. हिम्मत,
2. धन्यवाद,
3. बहादुरी,
4. पुरस्कार,
5. ठान लेना,
6. हिम्मत बंधाना,
7. एक न मानना,
8. खाली हाथ लौटना,
9. अपना-सा मुँह लेकर रह जाना।

उत्तर :

1. हिम्मत-वह बहुत हिम्मत करके मेरे पास आया।
2. धन्यवाद-मुख्याध्यापक ने अतिथि का धन्यवाद किया।
3. मगरमच्छ-मगरमच्छ पानी में रहता है।
4. पुरस्कार-दीपा को उसकी बहादुरी पर वीरता पुरस्कार मिला।
5. ठान लेना-मैंने इस प्रतियोगिता को जीतने की ठान ली है।
6. एक न मानना-वह अपने माता-पिता की एक नहीं मानता, अपने मन की करता है।
7. हिम्मत बँधाना-दीपा ने हरीश को हिम्मत बँधाई।
8. खाली हाथ लौटना-मगरमच्छ को खाली हाथ लौटना पड़ा।
9. अपना सा मुँह लेकर रह जाना-मगरमच्छ शिकार हाथ से निकलते देखकर अपना सा मुँह लेकर रह गया।

पढ़ो, समझो और लिखो

साल = वर्ष
लड़की = बालिका
किनारा = तट
जीवित – ज़िन्दा
उत्तर :
उपरोक्त रेखांकित शब्दों के पर्यायवाची शब्द सामने दर्शाये गए हैं। विद्यार्थी इन पर्याय-शब्दों को स्मरण रखें।

लिंग बदलो

1. माता = ……………………….
2. लड़की = ……………………….

उत्तर :

1. माता = पिता।
2. लड़की = लड़का।

समझो

गाँव = शहर
घर = बाहर
डूबना = तैरना
वीर = कायर
उत्तर :
उपरोक्त रेखांकित शब्दों के विपरीतार्थक शब्द उनके सामने लिखे गए हैं। विद्यार्थी इन्हें समझकर कण्ठस्थ कर लें।

इनसे नये शब्द बनाओ

1. च्छ
2. न्य
3. स्क
4. ल्द

उत्तर :

1. च्छ = स्वच्छ।
2. न्य = अन्य।
3. स्क= पुरस्कार।
4. ल्द = जल्द।

रचनात्मक अभिव्यक्ति

चित्र देखकर नीचे दिए गए शब्दों की सहायता से वाक्य पूरे करो

साहसी दुकान दीवाली पटाखों आग बच्चा बचा

…………………………………. का त्योहार है। बाज़ार में …………………………………. की दुकानें सजी हुई हैं। एक दुकान में अचानक …………………………………. लग गई है। एक …………………………………. आग में फँस गया है। वह ज़ोर-ज़ोर से चिल्ला रहा है। एक बालिका उसे जान पर खेलकर …………………………………. रही है। वह …………………………………. बालिका है।
उत्तर :
दीवाली का त्योहार है। बाज़ार में पटाखों की दुकानें सजी हुई हैं। एक दुकान में अचानक आग लग गई है। एक बच्चा आग में फँस गया है। वह ज़ोर-ज़ोर से चिल्ला रहा है। एक बालिका उसे जान पर खेलकर बचा रही है। वह साहसी बालिका है।

साहसी दीपा बहुवैकल्पिक प्रश्न

प्रश्न 1.
दीपा कैसी लड़की थी?
(i) साहसी
(i) निडर
(iii) योगी
(iv) सुंदर।
उत्तर :
(i) साहसी

प्रश्न 2.
दीपा नदी किनारे क्या कर रही थी?
(i) कपड़े धो रही थी
(ii) कपड़े सिल रह थे
(iii) खेल रही थी
(iv) सो रही थी।
उत्तर :
(i) कपड़े धो रही थी

प्रश्न 3.
‘ठान लेना’ का अर्थ है
(i) पक्का इरादा
(ii) पक्की बात
(iii) पक्के लोग
(iv) पक्की रस्सी।
उत्तर :
(i) पक्का इरादा

प्रश्न 4.
अगर ‘गाँव’-शहर है तो ‘घर ………………………………. है।
(i) बाहर
(ii) बाह्य
(iii) बाहरी
(iv) बाहु।
उत्तर :
(i) बाहर।

साहसी दीपा Summary in Hindi

पाठ का सार माधोपुर गाँव में दीपा नामक लड़की रहती थी। उसकी आयु बारह वर्ष की थी। उसके पिता शहर में नौकरी करते थे और माँ घर में बीमार रहती थी। दीपा बड़ी बहादुर लड़की थी। सारे घर के कामकाज करती और नदी में कपड़े धोने भी जाती थी।

एक दिन दीपा अपनी सहेली के साथ नदी के किनारे कपडे धो रही थी और थोड़ी ही दूरी पर रवि और हरीश नदी में नहा रहे थे। नहाते-नहाते रवि नदी में डूबने लगा तो हरीश ने उसे बचाने की कोशिश की और सहायता के लिए चिल्लाने लगा।

उसकी आवाज़ सुनकर दीपा उसकी सहायता के लिए दौड़ी तो उसकी सहेली ने उसे मना कर दिया क्योंकि उसने देखा कि पानी में मगरमच्छ रवि की ओर चला आ रहा था। लेकिन दीपा ने इसकी परवाह नहीं की और पानी में कूद गई। उसने जल्दी से रवि को पकड़ा और किनारे पर ले आई।

इस प्रकार उसने रवि को मगरमच्छ से बचा लिया। इतने में दीपा की सहेली रवि और हरीश के माता-पिता को ले आई। अपने बच्चों को सुरक्षित देखकर वे बहुत खुश हुए और दीपा का धन्यवाद किया। गाँव के सभी लोगों ने भी दीपा के साहस की सराहना की।

दीपा की बहादुरी पर उसे 26 जनवरी को पुरस्कार दिया गया।

साहसी दीपा शब्दार्थ Meanings

• होशियार = सावधान, चौकस
• कोशिश = प्रयास, यत्न
• मगरमच्छ = गहरे जल में पाया जाने वाला बहुत बड़ा, भीषण एवं हिंसक जंतु, घड़ियाल
• चंगुल = हाथ की उँगलियों के मोड़ने से बनी मुद्रा, पकड़, काबू
• धन्यवाद = कृतज्ञता
• सराहना = प्रशंसा प्रकट करने का शब्द
• पुरस्कार = इनाम

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 2 Structure of Atom Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 2 Structure of Atom

PSEB 11th Class Chemistry Guide Structure of Atom InText Questions and Answers

Question 1.
(i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
Answer:
(i) Mass of one electron = 9.1 × 10^-31 g
Number of electrons in 1.0 g = \(\frac{1}{9.1 \times 10^{-31}}\) = 1.098 x 10³⁰ electrons

(ii) One mole of electron = 6.022 × 10²³ electrons
Mass of 1 electron = 9.1 × 10^-31 kg
Mass of 6.022 × 1023 electrons – (9.1 x 10^-31 kg) × (6.022 x 10²³)
= 5.48 × 10^-7 kg
Charge on one electron = 1.602 × 10^-19 coulomb
Charge on one mole of electrons = 1.602 × 10^-19 × 6.022 × 10²³
= 9.65 × 10⁴ coulomb.

Question 2.
(i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of ¹⁴C.
(Assume that mass of a neutron = 1.675 × 10^-27 kg)
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH₃ at STP.
Will the answer change if the temperature and pressure are changed?
Answer:
(i) Number of electrons present in 1 molecule of methane (CH₄)
= {1(6) + 4(1)} = 10 electrons
Number of electrons present in 1 mole i.e., 6.022 × 10²³ molecules of
methane = 6.022 × 10²³ × 10 = 6.022 × 10²⁴ electrons
(ii) (a) Mass of a neutron = 1.675 × 10^-27 kg
14.0 g of ¹⁴C contains 1 mole of atoms of C-14
∴ 14.0 g of 14C contains 8 × 6.022 × 10²³ neutrons
[∵ each atom of C-14 contains 8 neutrons]
7 mg (= 7 × 10^-3 g) of ¹⁴C contains
= \(\frac{6.022 \times 10^{23} \times 8 \times 7 \times 10^{-3}}{14}\) = 2.4092 × 10²¹ neutrons

(b) Mass of one neutron = 1.675 × 10²⁷ kg
Mass of 2.4092 × 10²¹ neutrons
= (2.4092 × 10²¹)(1.675 × 10^-27 kg) = 4.035 × 10^-6 kg

(iii) 1 mole of NH₃ contains protons = 7 + 3 = 10 moles of protons
= 6.022 × 10²³ × 10 protons = 6.022 × 10²⁴ protons

(a) 17 gNH₃ contains 6.022 × 10²⁴ protons.
Number of protons in 34 mg NH₃ will contain
= \(6.022 \times 10^{24} \times 34 \times 10^{-3}\) = 1.2046 x 10²² Protons

(b) Mass of one proton = 1.675 x 10^-27 kg
Total mass of protons in 34 mg of NH₃
– (1.675 × 10^-27 kg) (1.2046 × 10²²)
= 2.0177 × 10^-5 kg
The number of protons, electrons and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

Question 3.
How many neutrons and protons are there in the following nuclei?
\({ }_{6}^{13} \mathrm{C},{ }_{8}^{16} \mathrm{O},{ }_{12}^{24} \mathrm{Mg},{ }_{26}^{56} \mathrm{Fe},{ }_{38}^{88} \mathrm{Sr}\)
Answer:
\({ }_{6}^{13} \mathrm{C}\)C : Atomic mass =13
Atomic number = Number of protons = 6
Number of neutrons = (Atomic mass) – (Atomic number)
= 13 – 6 = 7
\({ }_{8}^{16} \mathrm{O}\) :
Atomic mass =16
Atomic number = 8
Number of protons = 8
Number of neutrons = (Atomic mass) – (Atomic number)
= 16 – 8 = 8
\({ }_{12}^{24} \mathrm{Mg}\) : Atomic mass = 24
Atomic number = Number of protons = 12
Number of neutrons = (Atomic mass) – (Atomic number)
= 24-12 = 12
\({ }_{26}^{56} \mathrm{Fe}\) : Atomic mass = 56
Atomic number = Number of protons = 26
Number of neutrons = (Atomic mass) – (Atomic number)
= 56 – 26 = 30
\({ }_{38}^{88} \mathrm{Sr}\) : Atomic mass = 88
Atomic number = Number of protons = 38
Number of neutrons = (Atomic mass) – (Atomic number)
= 88 – 38 = 50

Question 4.
Write the complete symbol for the atom with the given atomic number (Z) and atomic mass (A)
(i) Z = 17, A = 35
(ii) Z = 92, A = 233
(iii) Z = 4, A = 9
Answer:
(i) Z = 17, A = 35
Since the no. of protons = 17 – no. of electrons
∴ The atom is chlorine, Cl: \({ }_{17}^{35} \mathrm{Cl}\)

(ii) Z = 92, A = 233
No. of protons = 92
∴ The atom is uranium, U : \({ }_{92}^{233} U\)

(iii) Z = 4, A = 9
No. of protons = 4
∴ The atom is Beryllium, Be : \({ }_{4}^{9} \mathrm{Be}\)

Question 5.
Yellow light emitted from a sodium lamp has a wavelength (A.) of 580 nm. Calculate the frequency (υ) and wave number (\(\overline{\mathbf{v}}\)) of the ‘ yellow light.
Answer:
We know that,
Frequency, v = \(\frac{c}{\lambda}\)
1 nm = 10^-9 m
580 nm = 580 × 10^-9 m = 580 x 10^-7 cm
v = \(\) = 5.17 × 10¹⁴s^-1
(velocity of light = 3 × 10⁸ m/s)
Wave number, \(\bar{v}=\frac{1}{\lambda}\)
= \(\frac{1}{580 \times 10^{-7}}\) = 1.72 × 10⁴ cm^-1

Question 6.
Find energy of each of the photons which
(i) correspond to light of frequency 3 × 10¹⁵ Hz.
(ii) have wavelength of 0.50 Å
Answer:
(i) Energy E = hv
where, h = Planck’s constant = 6.626 × 10^-34 Js
and v = frequency of light = 3 × 10¹⁵ Hz
E= (6.626 × 10^-34)(3 × 10¹⁵)
E = 1.988 × 10^-18 J

(ii) Energy E = \(\frac{h c}{\lambda}\)
where, h = Planck’s constant = 6.626 × 10^-34 Js
c = velocity of light in vacuum = 3 × 10⁸ m/s
= wavelength = 0.50Å = 0.50 × 10^-10
E = \(\frac{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{0.50 \times 10^{-10}}\) = 3.976 × 10^-15 J

Question 7.
Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10^-10 s.
Answer:
Frequency of light (v) = \(\frac{1}{\text { Time period }}=\frac{1}{2.0 \times 10^{-10} \mathrm{~s}}\) = 5.0 × 10⁹s^-1
Wavelength of light (λ) = \(\frac{c}{v}\)
where, c = velocity of light in vacuum = 3 × 10⁸ m/s
λ = \(\frac{3 \times 10^{8}}{5.0 \times 10^{9}}\) = 6.0 × 10^-2 m
Wave number(\(\bar{v}\)) of light
= \(\frac{1}{\lambda}=\frac{1}{6.0 \times 10^{-2}}\) = 1.66 × 10¹m^-1 = 16.66m^-1

Question 8.
What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?
Answer:
Energy (E) of a photon = hv = \(\frac{h c}{\lambda}\)
where, = wavelength of light = 4000 pm = 4000 × 10^-12 m
(∵ 1 pm = 10^-12 m)
c = velocity of light in vacuum = 3 × 10⁸ m/s
h = Planck’s constant = 6.626 × 10^-34 Js
E = \(\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{4000 \times 10^{-12}}\) = 4.9695 × 10^-17J
Number of protons, N = \(\frac{1}{4.9695 \times 10^{-17}}\) = 2.0122 × 10¹⁶ protons.

Question 9.
A photon of wavelength 4 × 10^-7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV = 1.6020 × 10^-19 J).
Answer:
(i) Energy (E) of a photon = hv = \(\frac{h c}{\lambda}\)
where,
h = Planck’s constant = 6.626 × 10^-34 Js
c = velocity of light in vacuum = 3 × 10⁸ m/s
λ = wavelength of photon =4 × 10^-7 m
E = \(\frac{\left(6.626 \times 10^{-34}\right)\left(3 \times 10^{8}\right)}{4 \times 10^{-7}}\)
= 4.9695 × 10^-19j = \(\frac{4.9695 \times 10^{-19}}{1.602 \times 10^{-19}}\) [∵ 1 eV = 1.602 x 10^-19J]
= 3.10 eV
Hence, the energy of the photon is 3.10 eV.

(ii) The kinetic energy of emission E_k is given by
= hv – hv₀ = (E – W) eV
= 3.10 – 2.13 eV = 0.97 eV
Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron is given by
KE = \(\frac{1}{2}\) mv² – hv – hv₀ = 0.97 eV
\(\frac{1}{2}\)mv² = 0.97 × 1.602 × 10^-19 J
(∵ 1 eV = 1.602 x 10^-19 J)
\(\frac{1}{2}\) × 9.11 × 10^-31 kg × v² = 0.97 × 1.602 × 10^-19 J
(∵ mass of electron = 9.11 × 10^-31 kg)
v² = \(\frac{0.97 \times 1.602 \times 10^{-19} \times 2}{9.11 \times 10^{-31}}\) = 0.341 × 10¹²
v = 0.584 × 10⁶ = 5.84 × 10⁵ m/s
Hence, the velocity of the photoelectron is 5.84 × 10⁵ ms^-1.

Question 10.
Electromagnetic radiation of wavelength 242 run is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol^-1.
Answer:
Energy (E) = \(\frac{h c}{\lambda}\)
= \(\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{242 \times 10^{-9} \mathrm{~m}}\)
= 0.0821 × 10^-17 J
This energy is sufficient for ionisation of one Na atom.
E = 6.02 × 10²³ × 0.0821 × 10^-17 J/mol
E = 4.945 × 10⁵ J/mol = 4.945 × 10² kJ/mol

Question 11.
A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 (am. Calculate the rate of emission of quanta per second.
Answer:
Power of bulb, P = 25 watt = 25 Js^-1
Wavelength,
λ = 0.57 µm = 0.57×10^-6 m
Energy of one photon,

Question 12.
Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800Å. Calculate threshold frequency (v₀) and work function (W₀) of the metal.
Answer:
Threshold wavelength of radiation
λ₀ = 6800Å = 6800 × 10^-10 m
Threshold frequency (v₀) of the metal
= \(\frac{c}{\lambda_{0}}=\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{6800 \times 10^{-10} \mathrm{~m}}\) = 4.41 × 10¹⁴s^-1
Thus, the threshold frequency (v₀) of the metal is 4.41 × 10¹⁴s^-1.
Hence, work function (W₀) of the metal = hv₀
= (6.626 × 10^-34Js)(4.41 × 10¹⁴s^-1)= 2.922 × 10^-19 J

Question 13.
What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2 ?
Answer:
The n_i = 4 to n_f= 2 transition will give rise to a spectral line of the
Balmer series. The energy involved in the transition is given by the relation,
E = 2.18 × 10^-18 \(\left[\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}\right]\)

Question 14.
How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n – 1 orbit).
Answre:
Energy change, ΔE = E_f – E_i

Hence, far higher energy (25 times) is required to remove an electron from first orbit.

Question 15.
What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?
Answer:
The maximum number of emission lines
= \(\frac{n(n-1)}{2}=\frac{6(6-1)}{2}\) = 3 × 5 = 15

Question 16.
(i) The energy associated with the first orbit in the hydrogen atom is -2.18 × 10^-18 J atom^-1. What is the energy associated with the fifth orbit?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.
Answer:
(i) Energy associated with the fifth orbit of hydrogen atom is calculated as :
E₅ = \(\frac{-\left(2.18 \times 10^{-18}\right)}{(5)^{2}}=\frac{-2.18 \times 10^{-18}}{25}\)
E₅ = -8.72 × 10^-20 J
(ii) Radius of Bohr’s nth orbit for hydrogen atom is given by,
r_n = (0.0529 nm) n²
For, n= 5
r₅ = (0.0529 nm) (5)²
r₅ = 1.3225 nm

Question 17.
Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Answer:
For the Balmer series, a transition from n₁= 2 to n₂ = 3 is allowed.

Question 18.
What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is -2.18 × 10^-11 ergs.
Answer:
ΔE = E₅ – E₁ = 2.18 × 10^-11 (\(\left(\frac{1}{n_{i}^{2}}-\frac{1}{n_{f}^{2}}\right)\))erg
(n_i = 1st orbit and n_f = 5th orbit)

Question 19.
The electron energy in hydrogen atom is given by E_n = (-218 × 10^-18)/n²J. Calculate the energy required to remove an electron completely from n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?
Answer:

Question 20.
Calculate the wavelength of an electron moving with a velocity of 2.05 × 10⁷ ms^-1.
Answer:
According to de Broglie’s equation,
λ = \(\frac{h}{m v}\)
where, λ = wavelength of moving electron
m = mass of electron = 9.1 × 10^-31 kg
h= Planck’s constant – 6.626 × 10^-34 Js
v = velocity of electron = 2.05 × 10⁷ ms^-1
λ = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(2.05 \times 10^{7} \mathrm{~ms}^{-1}\right)}\)
λ = 3.548 × 10^-11 m
Hence, the wavelength of the electron moving with a velocity of
2.5× 10⁷ ms^-1 is 3.548 x 10^-11m.

Question 21.
The mass of an electron is 9.1 x 10^-31 kg. If its K.E. is 3.0 x 10^-25 J, calculate its wavelength.
Answer:
Kinetic energy, K.E. = \(\frac{1}{2}\) mv²
∴ Velocity(v) = \(\sqrt{\frac{2 \mathrm{KE}}{m}}=\sqrt{\frac{2\left(3.0 \times 10^{-25}\right)}{9.1 \times 10^{-31} \mathrm{~kg}}}=\sqrt{0.6593 \times 10^{6}}\)
_ 6.626 x 10~34 Js
~ (9.1 x 10-31 kg)(811.579 ms-1)
λ = 8.968 x 10^-7 m
Hence, the wavelength of the eliectron is 8.968 x 10^-7 m or 8968 Å.

Question 22.
Which of the following are isoelectronic species i.e., those having the same number of electrons?
Na⁺, K⁺, Mg²⁺, Ca²⁺, S^2-, Ar
Ans. No. of electrons in Na⁺ = 10 [11-1]
No. of electrons in K⁺ = 18 [19-1]
No. of electrons in Mg²⁺ = 10 [12-2]
No. of electrons in Ca²⁺ =18 [20 – 2]
No. of electrons in S^2- = 18 [16 + 2]
No. of electrons in Ar = 18
∴ Na⁺ , Mg²⁺ are isoelectronic (10 e^– each)
∴ Ca²⁺, K⁺, S^2-, Ar are isoelectronic (18 e^– each).

Question 23.
(i) Write the electronic configurations of the following ions:
(a) H^– (b) Na⁺ (c)0^2- (d)F^–
(ii) What are the atomic numbers of elements whose outermost electrons are represented by
(a) 3s¹ (b) 2p³ (c) 3p⁵?
(iii) Which atoms are indicated by the following configurations?
(a) [He] 2s¹ (b) [Ne] 3s² 3p³ (c) [Ar] 4s²3d¹.
Answer:
(i) (a) H^– ion
The electronic configuration of H atom is 1s¹.
A negative charge on the species indicates the gain of an electron by it.
∴ Electronic configuration of H^– = 1s²

(b) Na⁺ ion
The electronic configuration of Na atom is 1s² 2s² 2p⁶ 3s¹.
A positive charge on the species indicates the loss of an electron by it.
∴ Electronic configuration of Na⁺ = 1s² 2s² 2p⁶ 3s° or 1s² 2s² 2p⁶

(c) O^2- ion
The electronic configuration of O atom is 1s²2s² 2p⁴
A dinegative charge on the species indicates that two electrons are gained by it.
∴ Electronic configuration of 0^2- ion is 1s² 2s² 2p⁶

(d) F^– ion
The electronic configuration of F atom is 1s²2s²2p⁵.
A negative charge on the species indicates the gain of an electron by it.
∴ Electronic configuration of F“ ion is 1s² 2s² 2p⁶.

(ii) (a) 3s¹
Completing the electronic configuration of the element as 1s² 2s² 2p⁶ 3s1. .-. Number of electrons present in the atom of the element = 2 +2 + 6 +1=11
∴ Atomic number of the element = 11

(b) 2p³
Completing the electronic configuration of the element as 1s² 2s² 2p³
∴ Number of electrons present in the atom of the element = 2+ 2+ 3 = 7
∴ Atomic number of the element = 7

(c) 3p⁵
Completing the electronic configuration of the element as 1s² 2s² 2p⁶ 3s² 3 P⁵
∴ Number of electrons present in the atom of the element
2 + 2 + 6 + 2 + 5 = 17
∴ Atomic number of the element = 17

(iii) (a) [He] 2s¹
The electronic configuration of the element is [He]2s¹ = 1s²2s¹.
∴ Atomic number of the element = 3
Hence, the element with the electronic configuration [He] 2s¹ is lithium (Li).

(b) [Ne] 3s² 3p³
The electronic configuration of the element is
[Ne] 3s² 3p³ = 1s² 2s² 2p⁶ 3s² 3p³
∴ Atomic number of the element = 15
Hence, the element with the electronic configuration [Ne] 3s² 3p³ is phosphorus (P).

(c) [Ar] 4s² 3d¹
The electronic configuration of the element is [Ar] 4s² 3d¹ = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹.
∴ Atomic number of the element = 21
Hence, the element with the electronic configuration [Ar] 4s² 3d² is scandium (Sc).

Question 24.
What is the lowest value of n that allows g orbitals to exist?
Answer:
For g-orbitals, l = 4.
As for any value ‘n’ of principal quantum number, the Azimuthal quantum number (l) can have a value from zero to (n —1).
∴ For l = 4, minimum value of n = 5

Question 25.
An electron is in one of the 3d orbitais. Give the possible values of n, l and m_l for this electron.
Answer:
For the 3d orbital:
Principal quantum number (n) = 3
Azimuthal quantum number (L) = 2
Magnetic quantum number (m_l) = – l to + l including 0 = -2 -1, 0, 1, 2

Question 26.
An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.
Answer:
No. of protons in a neutral atom = No. of electrons = 29
Electronic configuration = 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹.

Question 27.
Give the number of electrons in the species : H₂⁺, H₂ and 0₂⁺.
Answer:
H₂⁺ = one ; H₂ = two ; 0₂⁺ = 15

Question 28.
(i) An atomic orbital has n = 3. What are the possible values of l and m_l ?
(ii) List the quantum numbers m_l and l of electron in 3rd orbital.
(iii) Which of the following orbitals are possible ?
1p, 2s, 2p and 3f.
Answer:
(i) For n = 3; l = 0, 1 and 2.
For l = 0 ; m_l = 0
For l = 1; m_l = +1, 0, -1
For l = 2 ; m_l = +2, +1,0, +1, + 2
(ii) For an electron in 3rd orbital ; n = 3; l = 2 ; m_l can have any of the values -2, -1, 0,
+ 1, +2.
(iii) 1p and 3f orbitals are not possible.

Question 29.
Using s, p and d notations, describe the orbitals with follow ing quantum numbers :
(a) n = 1, l = 0
(b) n = 4, l = 3
(c) n = 3, l = 1
(d) n = 4, l = 2
Answer:
(a) 1s orbital
(b) 4f orbital
(c) 3p orbital
(d) 4d orbital

Question 30.
From the following sets of quantum numbers, state which are possible. Explain why the others are not possible.
(i) n = 0, l = 0, m_l = 0, m_s = +1/2
(ii) n = 1, l = 0, m_l = 0, m_s – – 1/2
(iii) n = 1, l = 1, m_l = 0, m_s= +1/2
(iv) n = 1, l = 0, m_l = +1, m_s= +1/2
(v) n = 3, l = 3, m_l = -3, m_s = +1/2
(vi) n = 3, l = 1, m_l = 0, m_s= +1/2
Answer:
(i) The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
(ii) The set of quantum numbers is possible.
(iii) The set of quantum numbers is not possible because, for n = 1, l can not be equal to 1. It can have 0 value.
(iv) The set of quantum numbers is not possible because for l = 0. mt cannot be + 1. It must be zero.
(v) The set of quantum numbers is not possible because, for n = 3, l ≠ 3.
(vi) The set of quantum numbers is possible.

Question 31.
How many electrons in an atom may have the following quantum numbers ?
(a) n = 4 ; m_s = -1/2
(b) n = 3, l = 0.
Answer:
(a) For n = 4
Total number of electrons = 2n² = 2 × 16 = 32
Half out of these will have ms = —1/2
∴ Total electrons with ms (-1/2) = 16
(b) For n = 3
l= 0 ; m_l = 0, m_s +1/2, -1/2 (two e^–)

Question 32.
Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.
Answer:

Thus, the circumference (2πr) of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength.

Question 33.
Calculate the number of atoms present in :
(i) 52 moles of He
(ii) 52 u of He
(iii) 52 g of He.
Answer:

Question 34.
Calculate the energy required for the process :
He⁺fe) → He²⁺(g) + e^–
The ionisation energy’ for the H atom in the ground state is 2.18 × 10^-18  J atom^-1
Answer:

For H atom (Z = 1), E_n =2.18 × 10^-18 × (l)² J atom^-1 (given)
For He⁺ ion (Z = 2), E_n =2.18 × 10^-18 × (2)² = 8.72 × 10^-18 J atom^-1 (one electron species)

Question .35.
If the diameter of carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across a length of a scale of length 20 cm long.
Answer:

Question 36.
2 × 10⁸ atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.
Answer:
The length of the arrangement = 2.4 cm
Total number of carbon atoms present = 2 ×10⁸

Radius of each carbon atom = 12(1.2 × 10^-8) = 6.0 × 10^-9cm = 0.06 nm

Question 37.
The diameter of zinc atom is 2.6 Å. Calculate :
(a) the radius of zinc atom in pm
(b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side length wise.
Answer:
(a) Radius of zinc atom =Undefined control sequence = 1.3 Å = 1.3 × 10^-10m = 130 × 10^-12m = 130 pm
(b) Length of the scale = 1.6 cm = 1.6 × 10¹⁰ pm
Diameter of zinc atom = 260 pm

Question 38.
A certain particle carries 2.5 x 10^-16 C of static electric charge. Calculate the number of electrons present in it.
Answer:

Question 39.
In Millikan’s experiment, the charge on the oil droplets was found to be – 1.282 x 10^-18C. Calculate the number of electrons present in it.
Answer:

Question 40.
In Rutherford experiment, generally the thin foil of heavy atoms like gold, platinum etc. have been used to be bombarded by the a-particles. If a thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results ?
Answer:
We have studied that in the Rutherford’s experiment by using heavy metals like gold and platinum, a large number of a-particles sufferred deflection while a very few had to retrace their path.

If a thin foil of lighter atoms like aluminium etc. be used in the Rutherford experiment, this means that the obstruction offered to the path of the fast moving a-particles will be comparatively quite less.

As a result, the number of a-particles deflected will be quite less and the particles which are deflected back will be negligible.

Question 41.
Symbols \(_{ 35 }^{ 79 }{ Br }\) and ⁷⁹Br can be written whereas symbols \(_{ 79 }^{ 35 }{ Br }\) and ₃₅Br are not accepted. Answer in brief.
Answer:
In the symbol \(_{ A }^{ B }{ X }\) of an element :
A denotes the atomic number of the element
B denotes the mass number of the element.
The atomic number of the element can be identified from its symbol because no two elements can have the atomic number. However, the mass numbers have to be mentioned in order to identify the elements. Thus,
Symbols \(_{ 35 }^{ 79 }{ Br }\) and ⁷⁹Br are accepted because atomic number of Br will remain 35 even if not mentioned. Symbol \(_{ 79 }^{ 35 }{ Br }\) is not accepted because atomic number of Br cannot be 79 (more than the mass number = 35). Similarly, symbol 35Br cannot be accepted because mass number has to be mentioned. This is needed to differentiate the isotopes of an element.

Question 42.
An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the symbol to the element.
Answer:
An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
Number of neutrons = x + [\(\frac { x\times 31.7 }{ 100 } \) = (x × 0.317x)
Now, Mass no. of element = no. of protons =no. neutrons
81 = x + x + 0-317 x = 2.317 x or x = \(\frac { 81 }{ 2.317 } \) = 35
∴ No. of protons = 35, No. of neutrons = 81 – 35 =46
Atomic number of element (Z) = No. of protons = 35
The element with atomic number (Z) 35 is bromine \(_{ 35 }^{ 81 }{ Br }\).

Question 43.
An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11 -1% more neutrons than the electrons, find the symbol of the ion.
Answer:
Let the no. of electron in the ion = x
∴ the no. of protons = x – 1 (as the ion has one unit negative charge)
and the no. of neutrons = x + \(\frac { x\times 11.1 }{ 100 } \) = 1.111 x
Mass no. or mass of the ion = No. of protons + No. of neutrons
(x – 1 + 1.111 x)
Given mass of the ion = 37
∴ x- 1 + 1.111 x = 37 or 2.111 x = 37 + 1 = 38
x = \(\frac { 38 }{ 2.111 } \) = 18
No. of electrons = 18 ; No. of protons = 18 – 1 = 17
Atomic no. of the ion = 17 ; Atom corresponding to ion = Cl
Symbol of the ion = \(_{ 17 }^{ 37 }{ Cl }\)^–

Question 44.
An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign symbol to the ion.
Answer:
Let the no. of electrons in the ion = x
∴ the no. of the protons = x + 3 (as the ion has three units positive charge)
and the no. of neutrons = x + \(\frac { x\times 31.7 }{ 100 } \) = xc + 0.304 x
Now, mass no. of ion = No. of protons + No. of neutrons
= (x + 3) + (x + 0.304x)
∴ 56 = (x + 3) + (x + 0.304x) or 2.304x = 56 – 3 = 53
x = \(\frac { 53 }{ 2.304 } \) = 23
Atomic no. of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe³⁺.

Question 45.
Arrange the following type of radiations in increasing order of frequency:
(a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.
Answer:
The increasing order of frequency is as follows :
Radiation from FM radio < Radiation from microwave oven < Amber light from traffic signal < X-rays < Cosmic rays from outer space.

Question 46.
Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 x 10²⁴ , calculate the power of this laser.
Answer:
Power of laser (E) = \(\frac{n h c}{\lambda}\)
= \(\frac{\left(5.6 \times 10^{24}\right)\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{\left(337.1 \times 10^{-9} \mathrm{~m}\right)}\)
= 0.3302 x 10⁷ J = 3.33 x 10⁶ J
Hence, the power of the laser is 3.33 x 10⁶ J.

Question 47.
Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance travelled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.
Answer:
(a) Frequency (v) of emission can be calculated as: c = v x λ
where, λ – wavelength = 616 nm = 616 x 10^-9 ms^-1
c = velocity of light = 3.0 x 10⁸ ms^-1
v = \(\frac{c}{\lambda}=\frac{3.0 \times 10^{8}}{616 \times 10^{-9}}\) = 4.87 x 10¹⁴s^-1

(b) Distance travelled in 1 second = 3.0 x 10⁸ m
Distance travelled in 30 seconds = 30 x 3.0 x 10⁸ m = 9 x 10⁹ m
(c) Energy of the photon (quantum)
E= hv = 6.626 x 10^-34 x 4.87 x 10¹⁴J
= 32.26 x 10^-20 J

(d) Number of quanta = \(\frac{Total energy produced}
{Energy of one quantum}\)
= \(\frac{2}{32.26 \times 10^{-20}}\) = 6.2 x 10¹⁸

Question 48.
In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 x 10^-18 J from the radiations of 600 rnn, calculate the number of photons received by the detector.
Answer:
Energy of one photon, E = \(\frac{h c}{\lambda}\)
= \(\frac{\left(6.626 \times 10^{-34} \mathrm{Js}\right)\left(3 \times 10^{8} \mathrm{~ms}^{-1}\right)}{\left(600 \times 10^{-9} \mathrm{~m}\right)}\) = 3.313 x 10^-19 J
Number of photons = \(\frac{Total energy received}{Energy of one photon}\)
= \(\frac{3.15 \times 10^{-18} \mathrm{~J}}{3.313 \times 10^{-19} \mathrm{~J}}\)

Question 49.
Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 x 10¹⁵, calculate the energy of the source.
Answer:
Duration of radiation source = 2 ns = 2 x 10^-9 s
No. of photons (n) = 2.5 x 10¹⁵
Frequency of radiation
v = \(\frac{1}{2.0 \times 10^{-9} \mathrm{~s}}\) = 5.0 x 10⁸s^-1
Energy (E) of source = n x hv
E= (2.5 x 10¹⁵)(6.626 x 10^-34Js) (5.0 x 10⁸)
E = 8.282 x 10^-10J
Hence, the energy of the source (E) is 8.282 x 10^-10 J.

Question 50.
The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.
Answer:
For λ₁ = 589 nm = 589 x 10^-9
Frequency of transition (v₁) = \(\) = 5.093 x 10¹⁴s^-1
Similarly, for λ₂ = 589.6 nm = 589.6 x 10^-9m
Frequency of transition (v₂) = \(\frac{c}{\lambda_{2}}=\frac{3.0 \times 10^{8} \mathrm{~ms}^{-1}}{589.6 \times 10^{-9} \mathrm{~m}}\) = 5.088 x 10¹⁴s^-1
Energy difference (ΔE) between excited states = E₁ – E₂
ΔE = h(v₁ – v₂)
= (6.626 × 10^-34Js) (5.093 × 10¹⁴ – 5.088 × 10¹⁴)s^-1
= (6.626 × 10^-34Js) (0.005 × 10^-14) s^-1
ΔE = 3.31 × 10^-22

Question 51.
The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.
Answer:
It is given that the work function (W₀) for caesium atom is 1.9 eV.
(a) Work function,

λ₀ = 6.53 × 10^-7 m
= 653 × 10^-9m = 653nm.
Hence the threshold wave length is 653nm

(b) threshold frequency
v₀ = \(\frac{W_{0}}{h}\)
v₀ = \(\frac{1.9 \times 1.6 \times 10^{-19} \mathrm{~J}}{6.626 \times 10^{-34} \mathrm{Js}}\) = 4.59 x 10¹⁴s^-1
Hence the threshold frequency of radiation is 4.59 x 10¹⁴s^-1
According to the question :
wavelength used in irradiation = 500nm
Kinetic Energy = h(v – v₀)
= hc( \(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\) )

Question 52.
Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.

Answer:

Mass of electron, m = 9.1 × 10^-31 kg
Here, λ = 500 × 10^-9 m
v_max = 2.55 ×10⁵ms^-1
λ₀ = ?

Question 53.
The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.
Answer:
h = 6.626 × 10^-34 Js
V₀ = 0.35 volt
e = 1.6 × 10^-19 C
λ = 256.7 nm = 256.7 × 10^-9 m
W₀ = ? (eV)

Question 54.
If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 x 10⁷ms^-1, calculate the energy with which it is bound to the nucleus.
Answer:
λ = wavelength = 150 pm = 150 × 10^-12 m
v = velocity = 1.5 × 10⁷ ms^-1
Kinetic energy of the ejected electron
= \(\frac{1}{2}\) mv⁷ = \(\frac{1}{2}\) x 9.1 x 10^-31 x (1.5 x 10⁷ )² J = 0.102 x 10^-15 J
Energy of the incident radiation
hv= hv₀ +\(\frac{1}{2}\)mv²
E = hv = \(\frac{h c}{\lambda}=\frac{6.626 \times 10^{-34} \times 3.0 \times 10^{8}}{150 \times 10^{-12}}\)
= 1.325 × 10^-15 J
Minimum energy required to eject the electron
E₀ = E – K.E.
= (1.325 × 10^-15 – 0.102 × ^-15 ) J
= 1.223 × 10^-15J
= \(\frac{1.223 \times 10^{-15}}{1.602 \times 10^{-19}} \mathrm{eV}\)
[ 1J = 1.602 × 10^-19 eV]
= 7.6 × 10³ eV

Question 55.
Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 x 10¹⁵ (Hz) \(\left[\frac{1}{3^{2}}-\frac{1}{n^{2}}\right]\)
Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.
Answer:

Question 56.
Calculate the wavelength for the emis8ion transition if it starts from the orbit having radius 1.3225 mn and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.
Answer:
The radius of the nth orbit of hydrogen like particles is given by

Thus, the transition ocurs from the 5th orbit to the 2nd orbit. It belongs to the Balmer series.
Wave number ( \(\bar{V}\) ) for the transition is given by,
\(\bar{v}=\frac{1}{\lambda}=1.097 \times 10^{7}\left(\frac{1}{2^{2}}-\frac{1}{5^{2}}\right)=1.097 \times 10^{7}\left(\frac{21}{100}\right)\)
= 2.303 × 10⁶ m^-1
∴ Wavelength ) associated with the emission transition is given by,
λ = \(\frac{1}{\bar{v}}=\frac{1}{2.303 \times 10^{6} \mathrm{~m}^{-1}}\)
= 0.434 × 10^-6 = 434 nm
∴ This transition belongs to Balmer series and comes in the visible region of the spectrum.

Question 57.
Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 x 10⁶ ms^-1, calculate de Broglie wavelength associated with this electron.
Answer:
From Broglie’s equation,
λ = \(\frac{h}{m v}\)
λ = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(1.6 \times 10^{6} \mathrm{~ms}^{-1}\right)}\) (1J = 1 kgm²s^-2)
= 4.55 × 10^-10 m = 455 pm
∴ de Broglie’s wavelength associated with the electron is 455 pm.

Question 58.
Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.
Answer:
From de Broglie’s equation,
λ = \(\frac{h}{m v}\)
v = \(\frac{h}{m \lambda}\)
mass of neutron, m = 1.675 × 10^-27 kg
λ = 800pm = 800 × 10^-12 m
v = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(1.675 \times 10^{-27} \mathrm{~kg}\right)\left(800 \times 10^{-12} \mathrm{~m}\right)}\) = 4.94 × 10²ms^-1
= 494 ms^-1
∴ Velocity associated with the neutron is 494 ms^-1.

Question 59.
If the velocity of the electron in Bohr’s first orbit is 2.19 × 10⁶ms^-1, calculate the de Broglie wavelength associated with it.
Answer:
According to de Broglie’s equation,
λ = \(\frac{h}{m v}\)
Mass of electron = 9.1 × 10^-31 kg, h = 6.626 × 10^-34 Js,
velocity = 2.19 × 10⁶ ms^-1
= \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.1 \times 10^{-31} \mathrm{~kg}\right)\left(2.19 \times 10^{6} \mathrm{~ms}^{-1}\right)}\) = 3.32 × 10^-10m
= 3.32 × 10^-10m × \(\frac{100}{100}\) = 332 × 10^-12 m
λ = 332pm
∴ Wavelength associated with the electron is 332 pm.

Question 60.
The velocity associated with a proton moving in a potential difference of 1000 V is 437 × 10⁵ ms^-1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.
Answer:
Wavelength associated with the velocity of hockey ball,
λ = \(\frac{h}{m v}\)
λ = \(\frac{6.626 \times 10^{-34} \mathrm{Js}}{(0.1 \mathrm{~kg})\left(4.37 \times 10^{5} \mathrm{~ms}^{-1}\right)}\)
λ = 1.516 × 10^-38 m

Question 61.
If the position of the electron is measured within an accuracy of ± 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is \(\frac{h}{4 \pi_{m}}\) × 0.05 nm, is there any problem in defining this value.
Answer:
From Heisenberg’s uncertainty principle,

Since the magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.

Question 62.
The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these comhination(s) has/have the same energy lists.
1. n = 4, l = 2, m_l = -2, m_s = – \(\frac{1}{2}\)
2. n = 3, l = 2, m_l= 1, m_s = + \(\frac{1}{2}\)
3. n = 4, l = 1, m_l = 0, m_s = +\(\frac{1}{2}\)
4. n = 3, l – 2, m_l = -2, m_s = –\(\frac{1}{2}\)
5. n = 3, l = 1, m_l = -1, m_s = + \(\frac{1}{2}\)
6. n = 4, l = 1, m_l = 0, m_s = + \(\frac{1}{2}\)
Answer:
For n = 4 and l = 2, the orbital occupied is 4d.
For n = 3 and l = 2, the orbital occupied is 3d.
For n = 4 and l = 1, the orbital occupied is 4p.
Hence, the six electrons i.e., 1, 2, 3, 4, 5 and 6 are present in the 4d, 3d, 4p, 3d, 3p and 4p orbitals respectively.
Therefore, the increasing order of energies are
5(3p) < 2(3d) = 4(3d) < 6(4p) = 3(4p) < 1(4d).

Question 63.
The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital,6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge.
Answer:
Effective nuclear charge decreases as the distance of the orbitals increases from the nucleus. Hence, 4p electrons experience the lowest effective nuclear charge.

Question 64.
Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p.
Answer:
Nuclear charge is defined as the net positive charge experienced by an electron in the orbital of a multi-electron atom. The closer the orbital, the greater is the nuclear charge experienced by the electron (s) in it.
(i) The electron(s) present in the 2s orbital will experience greater nuclear charge (being closer to the nucleus) than the electron(s) in the 3s orbital.
(ii) 4d will experience greater nuclear charge than 4/ since 4d is closer to the nucleus.
(iii) 3p will experience greater nuclear charge since it is closer to the nucleus than 3/.

Question 65.
The unpaired electrons in A1 and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?
Answer:
₁₃Al = 1s²2s²2p⁶3s²3p¹
₁₄Si = 1s²2s²2p⁶3s²3p²
Si (+4) has greater nuclear charge than aluminium (+3). Hence, 3p unpaired electrons of Si experience greater effective nuclear charge than Al.

Question 66.
Indicate the number of unpaired electrons in
(a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.
Answer:
(a) Phosphorus (P) : Atomic number = 15
The electronic configuration of P is : 1s² 2s² 2p⁶ 3s² 3p³

The orbital picture of P can be represented as :

From the orbital picture, phosphorus has three unpaired electrons
(b) Silicon (Si) : Atomic number = 14
The electronic configuration of Si is : 1s² 2s² 2p⁶ 3s² 3p²
The orbitai picture of Si can be represented as :

From the orbital picture, silicon has two unpaired electrons.
(c) Chromium (Cr) : Atomic number = 24
The electronic configuration of Cr is : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹ 3d⁵
The orbital picture of chromium is :

From the orbital picture, chromium has six unpaired electrons.
(d) Iron (Fe) : Atomic number = 26

The electronic configuration of Fe is : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d⁶
The orbital picture of Fe is:

From the orbital picture, iron has four unpaired electrons.
(e) Krypton (Kr) : Atomic number = 36
The electronic configuration of Kr is : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶
The orbital picture of krypton is :

Since all orbitals are fully occupied, there are no unpaired electrons in krypton.

Question 67.
(a) How many sub-shells are associated with n = 4? (b) How many electrons will be present in the sub-shells having m_s value of \(\) 1/2 for n – 4?
Answer:
(a) n = 4 (Given)
For a given value of ‘rt, V can have values from zero to n -1.
l = 0, 1, 2, 3
Thus, four sub-shells are associated with n = 4, which are s, p, d and f
(b) Number of orbitals in the nth shell = n²
For n = 4
Number of orbitals = 16
If each orbital is taken fully, then it will have 1 electron with m_s value of \(-\frac{1}{2}\)
Number of electrons with ms value of (\(-\frac{1}{2}\)) is 16.

Punjab State Board PSEB 11th Class Chemistry Important Questions Chapter 4 Chemical Bonding and Molecular Structure Important Questions and Answers.

PSEB 11th Class Chemistry Important Questions Chapter 4 Chemical Bonding and Molecular Structure

Very Short Answer Type Questions

Question 1.
In \(\mathrm{PO}_{4}^{3-}\) ion formal charge on the oxygen atom of P—O bond is
Answer:
In \(\mathrm{PO}_{4}^{3-}\) ion, formal charge on each O-atom of P—O bond
= \(\frac{\text { total charge }}{\text { Number of O-atoms }}=-\frac{3}{4}\) = -0.75

Question 2.
Which of the following molecules show super octet?
CO₂, CIF₃, SO₂, IF₅
Answer:
ClF₃ and IF₅ are super octet molecules.

Question 3.
Which of the following has highest lattice energy and why?
CsF, CsCl, CsBr, Csl
Answer:
CsF has highest lattice energy because ‘F’ is smallest in size and is more electronegative, therefore, it has maximum ionic character and maximum force of attraction, hence, highest lattice energy.

Question 4.
Account for the following:
The experimentally determined N—F bond length in NF₃ is greater than the sum of the single covalent radii of N and F.
Answer:
This is because both N and F are small and hence, have high electron density. So, they repel the bond pairs thereby making the N—F bond length larger.

Question 5.
What is valence bond approach for the formation of covalent * bond?
Answer:
A covalent bond is formed by the overlapping of half-filled atomic orbitals.

Question 6.
Why axial bonds of PCI5 are longer than equatorial bonds?
Answer:
This is due to greater repulsion on the axial bond pairs by the equatorial bond pairs of electrons.

Question 7.
Which type of atomic orbitals can overlap to form molecular orbitals?
Answer:
Atomic orbitals with comparable energies and proper orientation can overlap to form molecular orbitals.

Question 8.
Why KHF₂ exists but KHCl₂ does not?
Answer:
Due to H-bonding in HF, we have

This can dissociate to give \(\mathrm{HF}_{2}^{-}\) ion and hence, KHF₂ exists but there is no H-bonding in H-Cl. So, \(\mathrm{HCl}_{2}^{-}\) ion does not exist and hence, KHCl₂ also does not exist.

Question 9.
How many nodal planes are present in n(2px) and n(2px) molecular orbitals?
Answer:
One and two respectively.

Question 10.
What is the magnetic character of the anion of K02?
Answer:
Anion of KO₂ is \(\mathrm{O}_{2}^{-}\) (superoxide ion) which has one unpaired electron and hence is paramagnetic.

Short Answer Type Questions

Question 1.
Describe the change in hybridisation (if any) of the Al-atom in
the following reaction:
AlCl₃ + Cl^– → \(\mathrm{AlCl}_{\mathbf{4}}^{-}\)
Answer:
Electronic configuration of Al in ground state,
₁₃Al = 1s², 2s²,2p⁶,3s²,3p¹_x
In excited state = 1s², 2s², 2p⁶, 3s¹, 3p¹_x, 3p¹_y
In the formation of AlCl₃, Al undergoes sp² hybridisation and it is trigonal
planar in shape. While in the formation of AlCl₄^–, Al undergoes sp3
hybridisation.
It means empty 3P_z orbital also involved in hybridisation.
Thus, the shape of AlCl₄^– ion is tetrahedral.

Question 2.
Arrange the following in order of decreasing bond angle, with appropriate reason
\(\mathrm{NO}_{2}, \mathrm{NO}_{2}^{+}, \mathrm{NO}_{2}^{-}\)
Answer:
\(\mathrm{NO}_{2}, \mathrm{NO}_{2}^{+}, \mathrm{NO}_{2}^{-}\). This is because \(\mathrm{NO}_{2}^{+}\) has no lone pair of electrons
(i.e., has only bond pairs on two sides) and hence it is linear.

NO₂ has one unshared electron while \(\mathrm{NO}_{2}^{-}\) has one unshared electron pair. There are greater repulsion on N—O bonds in case of \(\mathrm{NO}_{2}^{-}\) than in case of NO₂

Question 3.
Among the molecules, \(\mathbf{O}_{2}^{-}, \mathbf{N}_{2}^{+}, \mathrm{CN}^{-}\) and \(\mathbf{O}_{2}^{+}\) identify the species which is isoelectronic with CO.
Answer:
Isoelectronics species are those species which have the same number of electrons. CO in total has 14 electrons (6 from carbon and 8 from oxygen). Out of the given ions CN is the ion which has 14 electrons (6 from carbon 7 from Nitrogen and 1 from the negative charge). Thus CN^– ion is isoelectronic with CO.

Question 4.
Which is more polar : COa or N20? Give reason.
Answer:
N₂Ois more polar than CO₂. This is because CO₂ is linear and symmetrical.
Its net dipole moment is zero im
on the other hand, is linear but unsymmetrical. It is considered as a resonance hybrid of the following two structures

It.has a net dipole moment of 0.116 D.

Question 5.
Aluminium forms the ion Al3+, but not Al4+ why?
Answer:
Aluminium [Ne]3s² 3p¹ can achieve the electronic configuration of the nearest noble gas (Ne) by losing only three electrons. : Al³⁺ = 1s² 2s² 2p⁶.
Aluminium will not form Al⁴⁺ ion because an extremely high amount of energy would be required to remove an electron from the stable noble gas configuration.

Long Answer Type Questions

Question 1.
On the basis of VSEPR theory, predict the shapes of the following
(i) \(\mathbf{N H}_{2}^{-}\) (ii) O₃
Answer:
(i) Shape of \(\mathbf{N H}_{2}^{-}\)
Number of valence electrons on central N atom = 5 + 1 (due to one unit negative charge) = 6
Number of atoms linked to it = 2
∴ Total number of electron pairs around N
= \(\frac{6+2}{2}\) = 4 and number of bond pairs = 2
∴ Number of lone pairs = 4 —2 = 2. Thus, the ion is of the type AB₂E₂
Hence, it has bent shape (V-shape).

(ii) Shape of O₃
While predicting geometry of molecules containing the double (or multiple) bond is considered as one electron pair, e.g., in case of ozone, its two resonating structures are

Thus, the central O-atom is considered to have two bond pairs and one lone pair, i.e., it is of the type AB₂E. Hence, it is a bent molecule. Thus, the two resonating structures will be

Question 2.
In each of the following pairs of compounds, which one is more covalent and why?
(i) AgCl, Agl
(ii) BeCl₂,MgCl₂
(iii) SnCl₂, SnCl₄
(iv) CuO, CuS
Answer:
Applying Fajans’ rules, the result can be obtained in each case as follows :
(i) Agl is more covalent than AgCl. This is because I^– ion is larger in size than Cl^– ion and hence, is more polarised than Cl^– ion.
(ii) BeCl₂ is more covalent thanMgCl₂. This is because Be²⁺ ion is smaller in size than Mg₂ ion and hence has the greater polarising power.
(iii) SnCl₄ is more covalent than SnCl₂. This is because Sn⁴⁺ ion has greater charge and smaller size than Sn²⁺ ion and hence has greater polarising power.
(iv) CuS is more covalent than CuO. This is because S^2- ion has larger size
than O^2- ion and hence is more polarised than O^2- ion.

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 5 States of Matter Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 5 States of Matter

PSEB 11th Class Chemistry Guide States of Matter InText Questions and Answers

Question 1.
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm³ at 30° C?
Answer:
Given,
Initial pressure, P₁ =1 bar
Initial volume, V₁ = 500 dm³
Final volume, V₂ = 200 dm³
Final pressure, P₂ = ?
Sipce the temperature remains constant at30°C.
According to Boyle’s law,
P₁V₁ = P₂V₂
⇒ P₂ = \(\frac{p_{1} V_{1}}{V_{2}}=\frac{1 \times 500}{200}\) bar = 2.5 bar
Therefore, the minimum pressure required is 2.5 bar.

Question 2.
A vessel of 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C. What would be its pressure?
Answer:
Given,
Initial pressure, P₁ =1.2 bar
Initial volume, V₁ = 120 mL
Final volume, V₂ = 180 mL
Final pressure, P₂ = ?
Since the temperature remains constant at 35°C.
According to Boyle’s law,
⇒ P₂ = \(\frac{p_{1} V_{1}}{V_{2}}=\frac{1.2 \times 120}{180}\) bar = 0.8 bar
Therefore, the pressure would be 0.8 bar.

Question 3.
Using the equation of state pV = nRT; show that at a given temperature density of a gas is proportional to gas pressure p.
Answer:
The equation of state is given by
pV = nRT

or d = \(\frac{p M}{R T}\)
If T constant, then d ∝ p.

Question 4.
At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
Answer:
Density (d) of the substance at temperature (T)
d = \(\frac{p M}{R T}\)
When T and d are same and R is constant then
p₁M₁ = p₂M₂
Given, P₁ = 2 bar
p₂ = 5 bar
Molecular mass of nitrogen, M₂ = 28g/mol
Now, M₁ = \(\frac{M_{2} p_{2}^{2}}{p_{1}}=\frac{28 \times 5}{2}\) = 70 g/mol
Hence, the molecular mass of the oxide is 70 g/ mol.

Question 5.
Pressure of 1 g of an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship between their molecular masses.
Answer:
For ideal gas A, the ideal gas equation is given by
p_AV – n_ART …(i)
Where, p_A and n_A represent the pressure and number of moles of gas A. For ideal gas B, the ideal gas equation is given by
p_BV = n_BRT …(ii)
Where, p_B and n_B represent the pressure and number of moles of gas B.
Number of moles of A gas, n_A = \(\frac{1}{M_{A}}\) [M_A = molar mass of gas A]
Number of moles of B gas, n_B = \(\frac{2}{M_{B}}\) [M_B = molar mass of gas B]
Pressure of gas A, p_A = 2 bar
Total pressure P_total = p_A + p_B = 3 bar
Pressure of gas B, p_B = p_total – p_A = 3 – 2 = 1 bar
V, R and T are same for both the gases.
Hence from eqs. (i) and (ii)

Question 6.
The drain cleaner, Drainex contains small hits of aluminum which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20°C and one bar will be released when 0.15g of aluminum reacts?
Answer:

Question 7.
What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4g of carbon dioxide contained in a 9 dm³ flask at 27°C?
Answer:
Mass of CH₄ = \(\frac{\text { Mass of } \mathrm{CH}_{4}}{\text { Molar mass of } \mathrm{CH}_{4}}\)
[Molar mass of CH₄ = 12 + 4 x 1 = 16]
= \(\frac{3.2}{16}\) = 0.2 mol
Mass of CO₂ = \(\frac{4.4}{44}\) = 0.1 mol
[Molar mass of CO₂ = 12 + 2 x 16 = 44]
Total moles = 0.2 + 0.1 = 0.3 mol
Pressure P = \(\frac{n R T}{V}=\frac{0.3 \times 8.314 \times 300}{9 \times 10^{-3}}\) = 8.314 x 10⁴ Pa

Question 8.
What will be the pressure of the gaseous mixture when o.5L of H₂ at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C?
Answer:
To calculate the partial pressure of H₂, Le., pH₂
V₁ = 0.5L, V₂ = 1L, p₁ = 0.8 bar, p₂ = ?
Temperature remaining constant, applying Boyl&s Law
p₁V₁=p₂V₂
0.8 x 0.5= p₂ x 1
or P₂ = 0.4 bar
To calculate the partial pressure of O₂ i.e., PO₂
V₁ = 2.0L, V₂ = 1L, p₁ = 0.7 bar, P₂ = ?
Applying Boyle’s Law
p₁V₁ = p₂V₂
0.7 x 2.0 = 1 x p₂
or p₂ = 1.4 bar.
If p is final pressure of the gas mixture, then according to Dalton’s Law of partial pressures
p = pH₂ + pO₂ = (0.4 +1.4)
= 1.8 bar

Question 9.
Density of a gas is found to be 5.46 g/dm³ at 27°C at 2 bar pressure. What will be its density at STP?
Answer:
Given,
d₁ = 5.46 g/dm³
p₁ = 2 bar
T₁ = 27°C = (27 + 273)K = 300 K
P₂ = 1 bar
T₂ = °C = 273 K
d₂ = ?
Density a = \(\frac{M p}{R T}\)
For same gas at different temperatures and pressures

Question 10.
34.05 ml. of phosphorus vapour weighs 0.0625g at 546°C and 0.1 bar pressure. What is the molar mass of phosphorus?
Answer:
Given,
p= 0.1 bar
V= 34.05 mL = 34.05 x 10^-3L
R = 0.083 bar dm³K^-1 mol^-1
T = 546°C = (546 + 273) K = 819 K
Mass, M = 0.0625 g
Now, pV = nRT
n = \(\frac{p V}{R T}=\frac{0.1 \times 34.05 \times 10^{-3}}{0.083 \times 819}\)= 5.0 x 10^-5 mol^-1
∴ Molar mass of phosphorus = \(\frac{0.0625}{5.0 \times 10^{-5}}\) = 1250 g mol^-1
Hence, the molar mass of phosphorus is = 1250 g mol^-1

Question 11.
A student forgot to add the reaction nixture to the round bottomed flask at 27°C but instead he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer he found the temperature of the flask was 477°C. What fraction of air would have been expelled out?
Answer:
Let the volume of the round bottomed flask = V cm3 at 27°C = 300 K
V₁ = V, T₁ = (27 + 273)K = 300K, V₂ =?,
T₂ = 477° C = (477 + 273)K
According to Charles’s law,
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)
V₂ = \(\frac{V_{1} T_{2}}{T_{1}}=\frac{750 \mathrm{~V}}{300}\) = 2.5V
Therefore, volume of air expelled out = 2.5 V – V = 1.5 V
Hence, fraction of air expelled out = \(\frac{1.5 \mathrm{~V}}{2.5 \mathrm{~V}}\) = 0.6

Question 12.
Calculate the temperature of 4.0 mol of a gas occupying 5 dm³ at 3.32 bar. (R = 0.083 bar dm³K^-1 mol^-1).
Answer:
Given,
n = 4.0 mol, V = 5 dm³, p = 3.32 bar, R = 0.083 bar dm³K^-1mol^-1
Applying ideal gas equation ‘
pV = nRT
T = \(\frac{p V}{n R}=\frac{3.32 \times 5}{4 \times 0.083}\) = 50K
Hence, the required temperature is 50 K.

Question 13.
Calculate the total number of electrons present in 1.4g of dinitrogen gas.
Answer:
Molar mass of dinitrogen (N₂) = 28 g mol^-1 Mass
Moles = \(\frac{Mass}{Molar mass}\)
ⁿN₂ = \(\frac{1.4}{28}\) = 0.05 mol
1 mol = 6.022 x 10²³ molecules
0.05 mol = 0.05 x 6.022 x 10²³ molecules
= 0.3011 x 10²³ molecules
Now, 1 molecule of N₂ contains = 14 electrons.
Therefore, 0.3011 x 10²³ molecules will contain = 14 x 0.3011 x 10²³
= 4.214 x 10²³ electrons

Question 14.
How much time would it take to distribute one Avogadro number of wheat grains, if 10¹⁰ grains are distributed each second?
Answer:
Avogadro number, N_A = 6.022 x 10²³

Hence, the time taken would be 1.909 x 10⁶ years.

Question 15.
Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen confined in a vessel of 1 dm³ at 27°C.
R 0.083 bar dm³ K^-1 mol^-1
Answer:
Moles of O₂ = \(\frac{\text { Mass }}{\text { Molar weight }}=\frac{8}{32}\) = 0.25 mol
Moles of H₂ = \(\frac{4}{2}\) = 2 mol
Therefore, total number of moles in the mixture = 0.25 + 2- 2.25 mol
Given, V = 1 dm³
n = 2.25 mol
R = 0.083 bar dm³K^-1mol^-1
T = 27°C = 300K
pV= nRT
⇒ Pressure, p = \(\frac{n R T}{V}=\frac{2.25 \times 0.083 \times 300}{1}\) = 56.025 bar.
Hence, the total pressure of the mixture is 56.025 bar.

Question 16.
Pay load is defined as the difference between the mass of displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10m, mass 100kg is filled with helium at 1.66 bar at 27°C.(Density of air = 1.2 kg m^-3 and R = 0.083 bar dm³K^-1 mol^-1)
Answer:
Given,
Radius of the balloon, r = 10m
∴ Volume of the balloon = \(\frac{4}{3} \pi r^{3}=\frac{4}{3} \times \frac{22}{7}\) x 10³ = 4190.5 m3(approx)
∴ Mass of displaced air = V_{displaced air} x density of air
= 4190.5 x 1.2 kg = 5028.6 kg
Now, mass of helium (m) filled in balloon
\(m_{\mathrm{He}}=\frac{M p V}{R T}\)
Here, M = 4 x 10^-3 kg mol^-1
p= 1.66 bar
V = Volume of the balloon = 4190.5 x 10³ dm³
R = 0.083 bar dm³K^-1mol^-1
T = 27°C = 300 K
Then m_He = \(\frac{4 \times 10^{-3} \times 1.66 \times 4190.5 \times 10^{3}}{0.083 \times 300}\) = 1117.5kg(approx)
Now, total mass of the balloon filled with helium
= (100 +1117.5)kg = 1217.5kg
Hence, pay load = mass of displaced air – mass of balloon
= (5028.6 -1217.5) kg = 3811.1 kg
Hence, the pay load of the balloon is 3811.1 kg

Question 17.
Calculate the volume occupied by 8.8g of CO2 at 31.1°C and 1 bar pressure. (R = 0.083 bar LK^-1 mol^-1).
Answer:
We know that,
pV = nRT m
pV = \(\frac{m}{M} R T\)
V = \(\frac{m R T}{M p}\)
Here, m = 8.8 g
R = 0.083 bar LK^-1mol^-1
T = 31.1°C = 304.1 K
M = 44 g mol^-1
p = 1 bar
Then, volume V = \(\frac{8.8 \times 0.083 \times 304.1}{44 \times 1}\) = 5.048 L
Hence, the volume occupied is 5.048 L.

Question 18.
2.9 g of a gas at 95°C occupied the same volume as 0.184 g of dihydrogen at 17°C, at the same pressure. What is the molar mass of the gas?
Answer:
Applying the ideal gas equation pV = nRT

Hence, the molar mass of the gas is 40 g mol^-1 .

Question 19.
A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Answer:
Let the weight of dihydrogen be 20 g and the weight of dioxygen be 80g.
Moles of dihydrogen,
ⁿH₂ = 20/2 = 10 mol
Moles of dioxygen,
ⁿO₂ = 80/32 = 2.5 mol
Given,
Total pressure of the mixture, p_total = 1 bar
Then, partial pressure of dihydrogen,
^pH₂ = \(\frac{n_{\mathrm{H}_{2}}}{n_{\mathrm{H}_{2}}+n_{\mathrm{O}_{2}}} \times p_{\text {total }}=\frac{10}{10+2.5} \times 1\) = 0.8 bar
Hence, the partial pressure of dihydrogen is 0.8 bar.

Question 20.
What would be the SI unit for the quantity pV² T² / n?
Answer:
The SI unit of \(\frac{p V^{2} T^{2}}{n}\) is given by
= \(\frac{\left(\mathrm{Nm}^{-2}\right)\left(\mathrm{m}^{3}\right)^{2}(\mathrm{~K})^{2}}{\mathrm{~mol}}\) = Nm⁴K²mol^-1

Question 21.
In terms of Charles’ law explain why -273°C is the lowest possible temperature.
Answer:
According to Charles’ law,
V_t = V₀ [1 + \(\frac{t}{273}\) ]
At t = -273° C
V_t = V₀ = [1 – \(\frac{273}{273}\) ] = 0
Thus, at -273° C, volume of a gas becomes zero and below this temperature the volume becomes negative, which is meaningless.

Question 22.
Critical temperature for carbon dioxide and methane are 31.1° C and -81.9°C respectively. Which of these has stronger intermolecular forces and why?
Answer:
Higher is the critical temperature of a gas, easier is its liquefaction. This means that the intermolecular forces of attraction between the molecules of a gas are directly proportional to its critical temperature. Hence, intermolecular forces of attraction are stronger in the case of C02.

Question 23.
Explain the physical significance of van der Waals parameters.
Answer:
Significance of constant ‘a’ : The value of constant ‘a’ is a measure of the magnitude of intermolecular forces between the molecules of the gas. Its units are atm L mol^-2 . Larger the value of ‘a’ larger will be the intermolecular forces among the gas molecules.
Significance of constant ‘b’ : The constant ‘b’ is called co-volume or excluded volume per mol of a gas. Its units are litre mol^-1.The volume of V is four times the actual volume of the molecules. It is measure of effective size of the gas molecules.

Punjab State Board PSEB 8th Class Social Science Book Solutions History Chapter 14 हस्तशिल्प तथा उद्योग Textbook Exercise Questions and Answers.

PSEB Solutions for Class 8 Social Science History Chapter 14 हस्तशिल्प तथा उद्योग

SST Guide for Class 8 PSEB हस्तशिल्प तथा उद्योग Textbook Questions and Answers

I. नीचे लिखे प्रश्नों के उत्तर लिखें :

प्रश्न 1.
भारत के लघु (छोटे) उद्योगों के पतन के कोई दो कारण लिखें।
उत्तर-

1. इन उद्योगों के मुख्य संरक्षक देशी रियासतों के राजा, उनके परिवार के सदस्य तथा उनके अधिकारी और कर्मचारी थे। जब देशी रियासतों की समाप्ति शुरू हो गई तो पुराने उद्योगों को स्वाभाविक रूप से धक्का लगा।
2. भारत के लघु उद्योगों में बनी वस्तुएं नई श्रेणी के लोगों को पसन्द नहीं थीं। वे अंग्रेज़ों के प्रभाव में थे। अत: इन्हें यूरोप की वस्तुएं भारत की वस्तुओं की अपेक्षा अधिक अच्छी लगती थीं।

प्रश्न 2.
भारत के लघु उद्योगों के द्वारा तैयार की गई वस्तुओं का मूल्य अधिक क्यों होता था ? .
उत्तर-
भारत के लघु उद्योगों द्वारा तैयार की गई वस्तुओं का मूल्य इसलिए अधिक होता था, क्योंकि इन्हें तैयार करने के लिए अधिक परिश्रम करना पड़ता था।

प्रश्न 3.
भारत में सूती कपड़े का प्रथम उद्योग कब तथा कहां स्थापित हुआ ?
उत्तर-
भारत में सूती कपड़े का प्रथम उद्योग (कारखाना) 1853 ई० में मुम्बई में स्थापित हुआ।

प्रश्न 4.
भारत में पहला पटसन उद्योग कब तथा कहां लगाया गया ?
उत्तर-
भारत में पहला पटसन उद्योग 1854 ई० में सीरमपुर (बंगाल) में लगाया गया।

प्रश्न 5.
भारत में कॉफी का पहला बाग कब तथा कहां लगाया गया ?
उत्तर-
भारत में कॉफी का पहला बाग 1840 ई० में दक्षिण भारत में लगाया गया।

प्रश्न 6.
चाय का पहला बाग कब तथा कहां लगाया गया ?
उत्तर-
चाय का पहला बाग़ 1852 ई० में असम में लगाया गया।

प्रश्न 7.
19वीं सदी में लघु उद्योगों के पतन के बारे में लिखें।
उत्तर-
भारत में अंग्रेजी शासन की स्थापना से पहले भारत के गांव आत्म-निर्भर थे। गांवों के लोग जैसे कि लोहार, जुलाहे, किसान, बढ़ई, चर्मकार, कुम्हार आदि मिलकर गांव की आवश्यकताओं को पूरा करने के लिए वस्तुएं तैयार कर लेते थे। उनकी दस्तकारियां या लघु उद्योग उनकी आय के साधन होते थे। परन्तु अंग्रेज़ी शासन की स्थापना होने के कारण गांवों के लोग भी अंग्रेज़ी कारखानों में तैयार की गई वस्तुओं का उपयोग करने लगे क्योंकि वे बढ़िया एवं सस्ती होती थीं। अतः भारत के नगरों एवं गांवों के लघु उद्योगों का पतन होने लगा और कारीगर (शिल्पकार) बेकार हो गये।

प्रश्न 8.
आधुनिक भारतीय उद्योगों के महत्त्व के बारे में लिखें।
उत्तर-
भारत में आधुनिक उद्योगों के विकास से आर्थिक एवं सामाजिक जीवन पर महत्त्वपूर्ण प्रभाव पड़े। इसके परिणामस्वरूप समाज में दो श्रेणियों का जन्म हुआ-पूंजीपति तथा मज़दूर। पूंजीपति मजदूरों का अधिक-से-अधिक शोषण करने लगे। वे मजदूरों से अधिक-से-अधिक काम ले कर कम-से-कम पैसे देते थे। अत: सरकार ने मजदूरों की दशा सुधारने के लिए फैक्टरी-एक्ट पास किये। औद्योगिक विकास होने के कारण कई नये नगरों का निर्माण भी हुआ। ये नगर आधुनिक जीवन एवं संस्कृति के केन्द्र बने।

प्रश्न 9.
नील उद्योग पर नोट लिखें।
उत्तर-
अंग्रेजों को इंग्लैंड में अपने कपड़ा उद्योग के लिए नील की आवश्यकता थी। अतः उन्होंने भारत में नील की खेती को बढ़ावा दिया। इसका आरम्भ 18वीं शताब्दी के अन्त में बिहार तथा बंगाल में हुआ। नील के अधिकतर बड़े-बड़े बाग यूरोप वालों ने लगाए, जहां भारतीयों को काम पर लगाया गया। 1825 में नील की खेती के अधीन 35 लाख बीघा भूमि थी। परन्तु 1879 ई० में नकली नील तैयार होने के कारण नील की खेती में कमी आने लगी। परिणामस्वरूप 1915 ई० तक नील की खेती के अधीन केवल 3-4 लाख बीघा जमीन ही रह गई।

प्रश्न 10.
कोयले की खानों पर नोट लिखें।
उत्तर-
भारत में अंग्रेजों द्वारा स्थापित सभी नये कारखाने कोयले से चलते थे। रेलों के लिए भी कोयला चाहिए था। इसलिए कोयला खानों में से कोयला निकालने की ओर विशेष ध्यान दिया गया। 1854 ई० में बंगाल के रानीगंज जिले में कोयले की केवल 2 खानें थीं। परन्तु 1880 तक इनकी संख्या 56 तथा 1885 तक 123 हो गई।

II. रिक्त स्थानों की पूर्ति करें:

1. देशी रियासतों के राजा महाराजा …………. उद्योगों द्वारा तैयार की गई वस्तुओं का प्रयोग करते थे।
2. नई पीढ़ी के लोग लघु उद्योगों द्वारा तैयार किए गए माल को ………… नहीं करते थे।
3. सभी नए कारखाने ……………. से चलते थे।
उत्तर-

1. लघु
2. पसंद
3. कोयले।।

III. प्रत्येक वाक्य के सामने ‘सही’ (✓) या ‘गलत’ (✗) का चिन्ह लगाएं:

1. भारतीय नगरों एवं गांवों के लघु उद्योगों के पतन से कारीगर बेकार हो गए। – (✓)
2. इंग्लैंड में औद्योगिक क्रांति 19वीं सदी में आई। – (✗)
3. मशीनों द्वारा तैयार की गई वस्तुओं का दाम अधिक होता था। – (✗)
4. 18वीं सदी में भारत का कच्चा माल इंग्लैंड जाने लगा। – (✓)

IV. सही जोड़े बनाएं :

उत्तर-

1. टी (चाय) कंपनी
2. सेरमपुर (बंगाल)
3. रानीगंज।

PSEB 8th Class Social Science Guide हस्तशिल्प तथा उद्योग Important Questions and Answers

वस्तुनिष्ठ प्रश्न (Multiple Choice Questions)

सही विकल्प चुनिए :

प्रश्न 1.
कॉफी का पहला बाग़ कब लगाया गया ?
(i) 1834 ई०
(ii) 1839 ई०
(iii) 1840 ई०
(iv) 1854 ई०।
उत्तर-
1840 ई०

प्रश्न 2.
भारत में नील उद्योग कहां से शुरू हुआ ?
(i) बिहार तथा बंगाल
(ii) कर्नाटक तथा तमिलनाडु
(iii) पंजाब तथा हरियाणा
(iv) मध्य भारत।
उत्तर-
बिहार तथा बंगाल

प्रश्न 3.
भारत में पटसन उद्योग का पहला कारखाना कब लगाया गया ?
(i) 1820 ई०
(ii) 1824 ई०
(iii) 1834 ई०
(iv) 1854 ई०।
उत्तर-
1854 ई० ।

अति छोटे उत्तर वाले प्रश्न

प्रश्न 1.
अंग्रेजी शासन की स्थापना से पहले आर्थिक दृष्टि से गांवों की स्थिति कैसी थी ?
उत्तर-
अंग्रेजी शासन की स्थापना से पहले गांव आर्थिक दृष्टि से आत्म-निर्भर थे।

प्रश्न 2.
भारतीय दस्तकारों द्वारा बनी वस्तुएं मशीनों द्वारा बनी वस्तुओं का मुकाबला क्यों न कर सकीं ?
उत्तर-
भारतीय दस्तकारों द्वारा बनी वस्तुएं मशीनों द्वारा बनी वस्तुओं का मुकाबला इसलिए न कर सकी क्योंकि मशीनी वस्तुएं साफ़ तथा सुन्दर होने के साथ-साथ सस्ती भी थीं।

प्रश्न 3.
नई श्रेणी के लोगों को भारत के लघु उद्योगों द्वारा बनी वस्तुएं क्यों पसन्द नहीं थीं ?
उत्तर-
क्योंकि वे पश्चिमी-सभ्यता के प्रभाव में थे।

प्रश्न 4.
पटसन उद्योग में क्या-क्या वस्तुएं बनाई जाती थीं ?
उत्तर-
टाट तथा बोरियां।

प्रश्न 5.
भारत के कॉफी उद्योग को हानि क्यों पहंची ?
उत्तर-
भारत की कॉफी का मुकाबला ब्राज़ील की कॉफी से था जो बहुत अच्छी थी। इसलिए भारत के कॉफी उद्योग को हानि पहुंची।

प्रश्न 6.
भारत में अंग्रेज़ी काल में विकसित किन्हीं छः आधुनिक उद्योगों के नाम बताओ।
उत्तर-

• सूती कपड़ा उद्योग
• पटसन उद्योग
• कोयला उद्योग
• नील उद्योग
• चाय
• कॉफी।

प्रश्न 7.
फैक्टरी एक्ट क्यों पास किए गए ?
उत्तर-
फैक्टरी एक्ट मज़दूरों की दशा सुधारने के लिए पास किए गए।

प्रश्न 8.
आदि मानव अपने आप को गर्म रखने के लिए किस चीज़ से बने वस्त्र पहनता था ?
उत्तर-
पशुओं की खाल से बने वस्त्र।

छोटे उत्तर वाले प्रश्न

प्रश्न 1.
भारत में कपड़ा बुनने का विकास कैसे हुआ ? खुदाइयों से कपड़े की बुनाई के बारे में क्या प्रमाण मिले हैं ?
उत्तर-
आदि मानव अपने आप को गर्म रखने के लिए पशुओं की खाल से बने कपड़े पहनता था। कताई एवं बुनाई . की खोज इसके बहुत समय पश्चात् हुई थी। ऐसा माना जाता है कि भारत में जुलाहों ने कपड़ा तैयार करने के लिए सबसे पहले घास के रेशों का उपयोग किया था। तत्पश्चात् उन्होंने इस पर नमूने बनाने और करघे पर धागों का उपयोग करना सीखा। समय बीतने पर रेशों और नमूनों में और अधिक सुधार हुआ।

प्रमाण-प्राचीन वस्तुओं का अध्ययन करने वाले वैज्ञानिकों को मोहनजोदड़ो तथा हड़प्पा की खुदाइयों से सूत की कताई तथा रंगदार सूती कपड़े के अवशेष मिले हैं। इस के अतिरिक्त उन्हें कश्मीर में कई स्थानों की खुदाई से चरखे, दरियां आदि प्राप्त हुए हैं। इनसे संकेत मिलता है कि लगभग 4,000 साल.पूर्व लोग कपड़ा बुनना जानते थे।

प्रश्न 2.
भारत में कपड़ा उद्योग का पतन क्यों हुआ ? इसे नया जीवन कैसे मिला ?
उत्तर-
भारतीय कपड़े संसार-भर में प्रसिद्ध थे। यूरोप के व्यापारी कपड़े एवं मसालों का व्यापार करने के लिए ही भारत में आये थे। उन्होंने भारत में सूती कपड़े के कारखाने लगाए थे। इन उद्योगों में साधारण हथकरघों की अपेक्षा अधिक कपड़े का उत्पादन किया जाता था। परन्तु इंग्लैंड में औद्योगिक क्रान्ति आने के कारण भारत में कपड़ा-व्यापार का पतन आरम्भ हो गया। परन्तु 20वीं शताब्दी में महात्मा गांधी के मार्गदर्शन में भारत में फिर से हाथ से बुने सूती एवं – रेशमी कपड़े तैयार किये जाने लगे, जिससे भारतीय कपड़ा-उद्योग पुनः अस्तित्व में आया।
सरकार की नई आर्थिक नीति से भी कपड़ा-उद्योग ने पहले से कहीं अधिक उन्नति की। सरकार ने कपड़े का आयात-निर्यात करने के लिए बहुत-सी सुविधाएं प्रदान की।

प्रश्न 3.
अंग्रेज़ी काल में भारत में पटसन उद्योग पर एक नोट लिखो।
उत्तर-
पटसन उद्योग में मुख्य रूप से टाट तथा बोरियां बनाई जाती थीं। इस उद्योग पर यूरोप के लोगों का अधिकार था। इस उद्योग का पहला कारखाना 1854 ई० में सीरमपुर (बंगाल) में लगाया गया। इसके बाद भी पटसन उद्योग के अधिकतर कारखाने बंगाल में ही लगाये गए। 20वीं शताब्दी के आरम्भ तक इन कारखानों की संख्या 36 हो गई थी।

निबन्धात्मक प्रश्न

प्रश्न 1.
भारत में लघु उद्योगों के पतन के कारणों का वर्णन करें।
उत्तर-
भारत में लघु-उद्योगों के पतन के मुख्य कारण निम्नलिखित थे

1. भारत की देशी रियासतों की समाप्ति-अंग्रेजों ने बहुत-सी भारतीय रियासतों को समाप्त कर दिया था। इस कारण लघु उद्योगों को बहुत हानि पहुंची क्योंकि इन रियासतों के राजा-महाराजा तथा उनके परिवार के सदस्य लघु उद्योगों द्वारा तैयार की गई वस्तुओं का उपयोग करते थे।

2. भारतीय लघु उद्योगों द्वारा तैयार वस्तुओं का महंगा होना-भारतीय लघु उद्योगों द्वारा तैयार की गई वस्तुओं का मूल्य अधिक होता था, क्योंकि उन्हें तैयार करने के लिए अधिक परिश्रम करना पड़ता था। दूसरी ओर मशीनों द्वारा तैयार की गई वस्तुओं का मूल्य कम होता था। अत: लोग लघु उद्योगों द्वारा तैयार की गई वस्तुओं को नहीं खरीदते थे। परिणामस्वरूप भारतीय लघु उद्योगों का पतन आरम्भ हो गया।

3. मशीनी वस्तुओं की सुन्दरता-इंग्लैंड के कारखानों में मशीनों द्वारा तैयार वस्तुएं भारत के लघु उद्योगों द्वारा तैयार वस्तुओं की अपेक्षा अधिक साफ तथा सुन्दर होती थीं। अतः भारतीय लोग मशीनों द्वारा तैयार वस्तुओं को अधिक पसन्द करते थे। यह बात भारत के लघु उद्योगों के पतन का कारण बनी।

4. नई श्रेणी के लोगों की रुचि-नई श्रेणी के लोगों पर पश्चिमी संस्कृति का प्रभाव था। दूसरे, मशीनों द्वारा तैयार वस्तुएं बहुत साफ और सुन्दर होती थीं। अतः नई पीढ़ी के लोग लघु उद्योगों द्वारा तैयार वस्तुओं को पसन्द नहीं करते थे।

5. भारत से कच्चा माल इंग्लैंड भेजना-18वीं शताब्दी में यूरोप में औद्योगिक क्रान्ति आई। इसके कारण वहां बहुत बड़े-बड़े कारखाने स्थापित किये गये। इन कारखानों में माल तैयार करने के लिए कच्चे माल की बहुत आवश्यकता थी, जिसे इंग्लैंड का कच्चा माल पूरा न कर सका। इस कारण भारत का कच्चा माल इंग्लैंड भेजा जाने लगा। इससे भारतीय कारीगरों के पास कच्चे माल की कमी हो गई। फलस्वरूप देश के लघु उद्योग पिछड़ गये।

प्रश्न 2.
प्रमुख आधुनिक भारतीय उद्योगों का वर्णन करें।
उत्तर-
अंग्रेजी शासन के समय भारत में बहुत-से नये उद्योगों की स्थापना हुई जिनमें प्रमुख उद्योग निम्नलिखित थे
1. सूती कपड़ा उद्योग-भारत में सूती कपड़े का पहला उद्योग (कारखाना) 1853 ई० में मुम्बई में लगाया गया। इसके पश्चात् 1877 ई० में कपास उगाने वाले बहुत से क्षेत्रों जैसे कि अहमदाबाद, नागपुर आदि में कपड़ा मिलें स्थापित की गईं। 1879 ई० तक भारत में लगभग 59 सूती कपड़ा मिलें स्थापित की जा चुकी थीं। जिनमें लगभग 43,000 लोग काम करते थे। 1905 ई० में कपड़ा मिलों की संख्या 206 हो गई थी। इनमें लगभग 1,96,000 मजदूर काम करते थे।

2. पटसन का उद्योग-पटसन का उद्योग बोरियां तथा टाट बनाने का काम करता था। इस उद्योग पर यूरोप के लोगों का अधिकार था। इस उद्योग का पहला कारखाना 1854 ई० में सैरमपुर अथवा सीरमपुर (बंगाल) में खोला गया। इसके बाद भी पटसन उद्योग के सबसे अधिक कारखाने बंगाल प्रान्त में ही खोले गये। 20वीं शताब्दी के आरम्भ तक इन कारखानों की संख्या 36 हो गई थी।

3. कोयले की खानें-भारत में अंग्रेजों द्वारा स्थापित सभी नये कारखाने कोयले से चलते थे। रेलों के लिए भी कोयला चाहिए था। इसलिए कोयला खानों में से कोयला निकालने की ओर विशेष ध्यान दिया गया। 1854 ई० में बंगाल के रानीगंज जिले में कोयले की केवल 2 खानें थीं। परन्तु 1880 तक इनकी संख्या 56 तथा 1885 तक 123 हो गई।

4. नील उद्योग-अंग्रेजों को इंग्लैंड में अपने कपड़ा उद्योग के लिए नील की आवश्यकता थी। अत: उन्होंने भारत में नील की खेती को बढ़ावा दिया। इसका आरम्भ 18वीं शताब्दी के अन्त में बिहार तथा बंगाल में हुआ। नील के अधिकतर बड़े-बड़े बाग यूरोप वालों ने लगाए, जहां भारतीयों को काम पर लगाया गया। 1825 में नील की खेती के अधीन 35 लाख बीघा भूमि थी। परन्तु 1879 ई० में नकली नील तैयार होने के कारण नील की खेती में कमी आने लगी। परिणामस्वरूप 1915 ई० तक नील की खेती के अधीन केवल 3-4 लाख बीघा जमीन ही रह गई।

5. चाय-1834 ई० में असम में एक कम्पनी की स्थापना की गई। 1852 ई० में अंग्रेजों ने असम में चाय का पहला बाग लगाया। 1920 ई० तक चाय की खेती लगभग 7 लाख एकड़ भूमि में होने लगी। उस समय 34 करोड़ पाऊंड मूल्य की चाय भारत से बाहर के देशों में भेजी जाती थी। तत्पश्चात् कांगड़ा तथा नीलगिरि की पहाड़ियों में भी चाय के बाग लगाए गए।

6. कॉफी-कॉफी का पहला बाग़ 1840 ई० में दक्षिण भारत में लगाया गया। तत्पश्चात् मैसूर, कुर्ग, नीलगिरि और मालाबार क्षेत्रों में भी कॉफी के बाग लगाये गये। ब्राज़ील की कॉफी के साथ इसका मुकाबला होने के कारण इस उद्योग को बहुत हानि पहुंची।

7. अन्य उद्योग-19वीं शताब्दी के अन्त से लेकर 20वीं शताब्दी के आरम्भ तक बहुत से नये कारखाने स्थापित किये गये। इनमें लोहा-इस्पात, चीनी, कागज, दियासलाई बनाने और चमड़ा रंगने के कारखाने प्रमुख थे।

Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 6 Thermodynamics Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 6 Thermodynamics

PSEB 11th Class Chemistry Guide Thermodynamics InText Questions and Answers

Question 1.
Choose the correct answer. A thermodynamic state function is a quantity
(i) used to determine heat changes
(ii) whose value is independent of path
(iii) used to determine pressure volume work
(iv) whose value depends on temperature only.
Answer:
(ii) A thermodynamic state function is a quantity whose value is independent of path. Functions like p, V, T etc., depend only on the state of a system and not on the path.

Question 2.
For the process to occur under adiabatic conditions, the correct condition is:
(i) ΔT = 0 (ii) Δp = 0
(iii) q = 0 (iv) w = 0
Answer:
(iii) A system is said to be under adiabatic conditions if there is no exchange of heat between the system and its surroundings. Hence, under adiabatic conditions, q = 0.

Question 3.
The enthalpies of all elements in their standard states are:
(i) unity
(ii) zero
(iii) < 0
(iv) different for each element
Answer:
(ii) The enthalpies of all elements in their standard states are zero.

Question 4.
\(\Delta \boldsymbol{U}^{\ominus}\) of combustion of methane is – X kJ mol^-1. The value of \(\Delta \boldsymbol{H}^{\ominus}\) is
(i) = ΔU
(ii) > ΔU
(iii) \(\Delta \boldsymbol{U}^{\ominus}\)
(iv) = 0
Answer:
(iii) CH₄(g) + 2O₂(g) > CO₂(g) + 2H₂O(l)
Δ_ng = n_p-n_r = 1 – 3 =-2
Hence,\(\Delta \boldsymbol{H}^{\ominus}\) = \(\Delta \boldsymbol{U}^{\ominus}\) + Δ n_gRT
\(\Delta \boldsymbol{H}^{\ominus}\) = – X – 2RT
Hence, \(\Delta \boldsymbol{H}^{\ominus}\) < \(\Delta \boldsymbol{U}^{\ominus}\)

Question 5.
The enthalpy of combustion of methane, graphite and dihydrogen at 298 K are, -890.3 kJ mol^-1 – 393.5 kJ mol^-1 and -285.8 kJ mol^-1 respectively. Enthalpy of formation of CH₄(g) will be
(i) -748kJmol^-1
(ii) -52.27kJ mol^-1
(iii) +748kJ mol^-1
(iv) +52.26kJ mol^-1
Answer:
According to the equation
(i) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); ΔH = -890.3 kJ mol^-1

(ii) C(s) + O₂(g) → CO₂(g); ΔH = – 393.5 kJ mol^-1

(iii) H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂O(l); ΔH = -285.8 kJ mol^-1
Multiplying equation (iii) by 2, we get equation (iv).

(iv) 2H₂(g) + O₂(g) → 2H₂O(l); ΔH = – 571.6 kJ mol^-1
Adding eqs. (ii) and (iv), we get

(v) (C(s) +2H₂(g) + 2O₂(g) → CO₂(g) + 2H₂O(l); ΔH = – 965.1 kJ mol^-1
Reversing eqs. (i)

(vi) CO₂(g) + 2H₂O(l) → CH₄(g) + 2O₂(g); ΔH = +890.3 kJ mol^-1
Adding eqs. (v) and (vi), we get
C(s) + 2H₂(g) → CH₄(g); ΔH = – 74.8 kJ mol^-1
Hence, option (i) is correct.

Question 6.
A reaction, A + B → C + D +q is found to have a positive entropy change. The reaction will be
(i) possible at high temperature
(ii) possible only at low temperature
(iii) not possible at any temperature
(iv) possible at any temperature
Answer:
(iv) For a reaction to be spontaneous, AG should be negative.
ΔG = ΔH – TΔS
According to the question, for the given reaction,
ΔS = positive
ΔH = negative (since heat is evolved)
=> ΔG = negative
Therefore, the reaction is spontaneous at any temperature.
Hence, option (iv) is correct.

Question 7.
In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?
Answer:
According to the first law of thermodynamics,
ΔU = q + W …(i)
Given,
q = +701 J (heat is absorbed here, q is positive)
W=- 394 J
(work is done by the system hence W is negative)
Substituting the values in expression (i), we get
ΔU= +701 J + (-394 J)
ΔH = 307 J
Hence, the change in internal energy for the given process is 307 J.

Question 8.
The reaction of cyanamide, NH₂CN(s) with dioxygen was carried out in a bomb calorimeter, and ΔU was found to be – 742.7 kJ mol^-1 at 298K. Calculate enthalpy change for the reaction at 298 K.
NH₂CN(s) + \(\frac{3}{2}\)O₂(g) → N₂(g) + CO₂(g) + H₂O(l)
Answer:
The given reaction is
NH₂CN(s) + \(\frac{3}{2}\)O₂(g) → N₂(g) + CO₂(g) + H₂O(l)
Difference of moles of gaseous products and reactants,
Δ_ng = n_p – n_r = 2 – \(\frac{3}{2}=\frac{1}{2}\) = 0.5 mol
Given, ΔU=- 742.7kJ mol^-1
Enthalpy change ΔH = ΔU + Δ_ngRT
= – 742.7 + (0.5 x 8.314 x 10^-3 x 298)
= – 742.7 +1238.786 x 10^-3
= – 741.46 kJ mol^-1

Question 9.
Calculate the number of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35°C to 55°C. Molar heat capacity of Al is 24 J mol^-1K^-1.
Answer:
Given, mass of Al = 60.0 g
ΔT= 55 – 35 = 20°C
No.of moles of Al = \(\frac{60.0}{27}\) mol
Molar heat capacity of Al
= 24 J mol^-1K^-1
Heat, q = n.C . ΔT
= [ \(\frac{60}{27}\) mol )(24 J mol^-1K^-1)(20 K)
q= 1066.7 J
q = 1.07 kJ

Question 10.
Calculate the enthalpy change on freezing of 1.0 mol of water at 10.0°C to ice at – 10.0°C. Δ_fusH = 6.03 kJ mol^-1 at 0°C.
C_p[H₂O(l)] – 75.3 J mol^-1K^-1
C_p[H₂O(s)] = 36.8J mol^-1K^-1
Answer:
(i) Heat change required to lower the temperature of water from 10.0°Cto0°C
ΔH₁= n x C_p x ΔT= 1.0 x 75.3 x (-10) = – 753 J mol^-1
(ii) Heat change required to convert 1 mol of H₂O(l) at 0°C to H₂O(s) at 0°C
ΔH₂ = ΔH_fusion = – 6.03 kJ mol^-1 as heat is given out
(iii) Heat change required to change 1 mole of ice from 0°C to -10.0° C
ΔH₃ = – 36.8 x 10 x 1 = – 368 J mol^-1
Total heat change
= ΔH₁ + ΔH₂ + ΔH₃ = (- 0.753 – 6.03 – 0.368) kJ mol^-1
∴ Total enthalpy change = – 7.151 kJ mol^-1
As in each step, heat is evolved, each step will have a negative sign with Δ_H

Question 11.
Enthalpy of combustion of carbon to CO₂ is -393,5 kJ mol^-1. Calculate the heat released upon formation of 35.2 g of CO₂ from carbon and dioxygen gas.
Answer:
The reaction for the combustion of carbon into CO₂ is
C(s) + O₂(g) → CO₂(g); ΔH = – 393.5 kJ mol^-1 (1 mole CO₂ – 44g)
Heat released in the formation of 44 g CO₂ = 393.5 kJ
∴ Heat released in the formation of 35.2 g CO₂
\(\frac{393.5 \mathrm{~kJ}}{44 \mathrm{~g}}\) x 5.2g = 314.8 kJ mol^-1

Question 12.
Enthalpies of formation of CO(g), CO₍₂₎(g), N₂O(g) and N₂O₄(g) are -110, -393, 81and 9.7 kJ mol^-1 respectively. Find the value of Δ_rH for the reaction:
N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g)
Answer:
Δ_fH (CO) = -110 kJ mol^-1
ΔH_f(CO₂) = -393 kJ mol^-1
Δ_fH (N₂O)= 81 kJ mol^-1
Δ_fH (N₂O₄) = 9.7 kJ mol^-1
The given reaction is
N₂O₄(g) + 3CO(g) → N₂O(g) + 3CO₂(g); Δ_rH = ?

= [81 + 3 (-393)] – [9.7 + 3 (-110)] kJ = [81 -1179]-[9.7-330] kJ
Δ_rH = – 777.8 kJ.

Question 13.
Given : N₂(g) + 3H₂(g) → 2NH₃(g); \(\Delta_{\boldsymbol{r}} \boldsymbol{H}^{\ominus}\) = -92.4kJ mol^-1.
What is the standard enthalpy of formation of NH₃ gas?
Answer:
Given, N₂(g) + 3H₂(g) → 2NH₃(g); Δ_rH = – 92.4 kJ mol^-1
Chemical reaction for the enthalpy of formation of NH3(g) is as follows :
\(\frac{1}{2}\)N₂(g) + \(\frac{3}{2}\)H₂(g) → NH₃(g)
∴ Standard enthalpy of formation of NH₃(g)
= \(\frac{1}{2}\)\(\Delta H^{\theta}\) = \(\frac{1}{2}\) (-92.4 kJ mol^-1) = -46.2 kJ mol^-1

Question 14.
Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:
CH₃OH(l) + \(\frac{3}{2}\)O₂(g) → CO₂(g) + 2H₂O(l); \(\Delta_{\boldsymbol{r}} \boldsymbol{H}^{\ominus}\) = -726 kJ mol^-1
C_(graphite) + O₂(g) → CO₂(g); \(\Delta_{\boldsymbol{c}} \boldsymbol{H}^{\ominus}\) = – 393 kJ mol^-1
H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂O(l); \(\Delta_{\boldsymbol{f}} \boldsymbol{H}^{\ominus}\) = – 286 kJ mol^-1
Answer:

Question 15.
Calculate the enthalpy change for the process
CCl₄(g) → C(g) + 4Cl(g)
and calculate bond enthalpy of C—Cl in CCl₄(g).
\(\Delta_{\mathbf{v a p}} \boldsymbol{H}^{\ominus}\)(CCl₄) = 30.5kJ mol^-1. \(\Delta_{\boldsymbol{f}} \boldsymbol{H}^{\ominus}\)(CCl₄) = – 135.5kJ mol^-1.
\(\Delta_{a} \boldsymbol{H}^{\ominus}(\mathbf{C})\) = 715.0kJ mol^-1, \(\Delta_{\boldsymbol{a}} \boldsymbol{H}^{\ominus}\left(\mathbf{C l}_{\mathbf{2}}\right)\)(Cl2) = 242 kJ mol^-1; where \(\Delta_{\boldsymbol{a}} \boldsymbol{H}^{\ominus}\) is enthalpy of atomisation.
Answer:
Given,
(i) CCl₄(Z) → CCl₄(g); \(\Delta_{\mathrm{vap}} H^{\ominus}\) = +30.5 kJ mol^-1
(ii) C(s) → C(g); \(\Delta_{a} H^{\ominus}\) = 715.0 kJ mol^-1
(iii) Cl₂(g) → 2Cl(g); \(\Delta_{a} H^{\ominus}\) = 242 kJ mol^-1
(iv) C(s) + 2Cl₂(g) → CCl₄ a); \(\Delta_{f} H^{\ominus}\) = – 135.5KJ mol^-1

Enthalpy change for the given process
CCl₄(g) → C(g) + 4Cl(g); \(\Delta_{r} H^{\ominus}\) = ?

Add (i) and (iv) and subtract (ii) and (iii) x 2
CCl₄(l) + C(s) + 2Cl₂(g) – C(s) – 2Cl₂ (g) → CCl₄(g) + CCl₄(g) – C(g) – 4Cl(g)
or \(\Delta_{r} H^{\ominus}\) = 30.5 -135.5 – 715 – 484 = -1304 kJ
0 = CCl₄(g) – C(g) + 4Cl(g); \(\Delta_{r} H^{\ominus}\) = -1304kJ
CCl₄(g) → C (g) + 4Cl(g); \(\Delta_{r} H^{\ominus}\) = 1304 kJ
There are four bonds of C—Cl in CCl₄.
Bond enthalpy of C—Cl bond
= \(\frac{1304}{4}\) mol^-1
= 326 kJmol^-1

Question 16.
For an isolated system, ΔU = 0, what will be ΔS ?
Ans. For an isolated system, ΔU = 0 and for a spontaneous process, total entropy change must be positive. For example, consider the diffusion of two gases A and B into each other in a closed container which is isolated from the surroundings.

The two gases A and B are separated by a movable partition. When partition is removed, the gases begin to diffuse into each other and the system becomes more disordered. It shows that ΔS > 0 and ΔU = 0 for this process.
Moreover, ΔS =\(=\frac{q_{\mathrm{rev}}}{T}=\frac{\Delta H}{T}=\frac{\Delta U+p \Delta V}{T}=\frac{p \Delta V}{T}\) (∴ ΔU = 0)
i.e., TΔS or ΔS > 0

Question 17.
For the reaction at 298 K, 2A + B > C
ΔH = 400kJ mol^-1 and ΔS = 0.2kJK^-1mol^-1 At what temperature will the reaction become spontaneous considering ΔH and ΔS to be constant over the temperature range?
Answer:
Given, ΔH = 400 kJ mol^-1, ΔS = 0.2 kJ K^-1mol^-1
Gibbs free energy, ΔG = ΔH – TΔS
0 = 400 kJ mol^-1 – T x 0.2 kJ K^-1mol^-1
(ΔG = 0 at equilibrium)
T = \(\frac{400 \mathrm{~kJ} \mathrm{~mol}^{-1}}{0.2 \mathrm{~kJ} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}\)
T = 2000 K
For the reaction to be spontaneous, ΔG must be negative. Hence, for the given reaction to be spontaneous, T should be greater than 2000 K.

Question 18.
For the reaction,
2Cl(g) → Cl₂(g), what are the signs of ΔH and ΔS?
Answer:
ΔH and ΔS are negative.
The given reaction represents the formation of chlorine molecule from chlorine atoms. Here, bond formation is taking place. Therefore, energy is being released. Hence, ΔH is negative.
Also, two moles of atoms have more randomness than one mole of a molecule. Since spontaneity is decreased, ΔS is negative for the given reaction.

Question 19.
For the reaction : 2A(g) + B(g) → 2D(g)
\(\Delta \boldsymbol{U}^{\ominus}\) = – 10.5 kJ and ΔS = – 441 JK^-1.
Calculate \(\Delta \boldsymbol{G}^{\ominus}\) for the reaction, and predict whether the reaction may occur spontaneously.
Answer:
For the given reaction,
2A(g) + B(g) → 2D(g)
Δn_g = 2 – 3 = -1 mol
Substituting the value of \(\Delta \boldsymbol{U}^{\ominus}\) in the expression ofΔH
\(\Delta H^{\ominus}=\Delta U^{\ominus}+\Delta n_{g} R T\)
= (-10.5 kJ) + (-1) x (8.314 x 10^-3kJK^-1 mol^-1) x (298 K)
= -10.5 kJ-2.48 kJ
\(\Delta H^{\ominus}\) = -12.98 kJ
We know that,
\(\Delta G^{\ominus}=\Delta H^{\ominus}-T \Delta S^{\ominus}\)
= -12.98 kJ – (298K) x ( – 44.1 JK^-1)
= -12.98 kJ + 13.14kJ
\(\Delta G^{\ominus}\) = +0.16 kJ
Since, \(\Delta G^{\ominus}\) is positive, the reaction will not occur spontaneously.

Question 20.
The equilibrium constant for a reaction is 10. What will be the value of \(\Delta G^{\ominus}\)? R = 8.314 JK^-1mol^-1, T = 300 K.
Answer:
\(\Delta G^{\ominus}\) = – 2303RT log K_c
Given, K_c = 10, T = 300 K, R = 8.314 J K^-1 mol^-1
\(\Delta G^{\ominus}\) = (- 2.303) (8.314 JK^-1mol^-1) (300K) (loglO) (∵ log 10 = 1)
= – 5744.14 Jmol^-1 = – 5.744 kJmol^-1

Question 21.
Comment on the thermodynamic stability of NO(g), given
\(\frac{1}{2}\)N₂(g) + \(\frac{1}{2}\)O₂(g) → NO(g); \(\Delta_{\boldsymbol{r}} \boldsymbol{H}^{\ominus}\) = 90 kJ mol^-1
NO(g) + \(\frac{1}{2}\)O₂(g) → NO₂(g); \(\Delta_{\boldsymbol{r}} \boldsymbol{H}^{\ominus}\) = – 74 kJ mol^-1
Answer:
NO(g) is unstable because formation of NO is endothermic (energy is absorbed) but NO₂(g) is formed because its formation is exothermic (energy is released).
Hence, unstable NO(g) changes to stable NO₂(g).

Question 22.
Calculate the entropy change in surroundings when 1.00 mol of H₂O(Z) is formed under standard conditions. \(\Delta_{\boldsymbol{f}} \boldsymbol{H}^{\ominus}\) = – 286
kJ mol^-1.
Answer:
Enthalpy change for the formation of 1 mol of H₂O(Z)
H₂(g) + \(\frac{1}{2}\)O₂(g) → H₂O(l); \(\Delta_{f} H^{\mathrm{s}}\) = -286 kJ mol^-1
Energy released in the above reaction is absorbed by the surroundings. * It means,
q_surr = + 286 kJ mol^-1
Entropy change ΔS_surr = \(=\frac{q_{\mathrm{surr}}}{T}=\frac{286 \mathrm{~kJ} \mathrm{~mol}^{-1}}{298 \mathrm{~K}}\)
ΔS_surr = 0.95973 kJ K^-1 mol^-1 = 959.73 J mol^-1K^-1

Punjab State Board PSEB 9th Class Social Science Book Solutions Geography Chapter 3b पंजाब : जलतन्त्र Exercise Questions and Answers.

PSEB Solutions for Class 9 Social Science Geography Chapter 3b पंजाब : जलतन्त्र

SST Guide for Class 9 PSEB पंजाब : जलतन्त्र Textbook Questions and Answers

(क) नक्शा कार्य (Map Work):

प्रश्न 1.
पंजाब के रेखाचित्र में दिखाएं :
(i) रावी, ब्यास, सतलुज तथा घग्गर
(ii) कोई चार नहरें
(iii) कोई चार चो।
उत्तर-
यह प्रश्न विद्यार्थी MBD Map Master की सहायता से स्वयं करें।

प्रश्न 2.
नदियों का प्रदूषण कैसे रोका जाए, इस विषय पर कक्षा में अध्यापक से चर्चा करें।
उत्तर-
यह प्रश्न विद्यार्थी स्वयं करें।

प्रश्न 3.
अपने समीप नदी या नहर में हो रहे प्रदूषण के लिए अध्यापक व अधिकारियों को सूचना दें।
उत्तर-
यह प्रश्न विद्यार्थी स्वयं करें।

PSEB 9th Class SST Solutions Geography Chapter 3a भारत : जलप्रवाह

(ख) निम्न वस्तुनिष्ठ प्रश्नों के उत्तर दें:

प्रश्न 1.
कौन-सी नदी का उद्गम स्थान मान सरोवर के पास रक्षताल झील है ?
(i) घग्गर
(ii) ब्यास
(iii) सतलुज
(iv) ब्रह्मपुत्र।
उत्तर-
(iii) सतलुज।

प्रश्न 2.
पंजाब में कितनी नदियां हैं-
(i) तीन
(ii) चार
(iii) पाँच
(iv) आठ।
उत्तर-(i) तीन।

प्रश्न 3.
रणजीत सागर अथवा थीन डैम का निर्माण कौन-सी नदी पर हुआ है ?
(i) ब्यास
(ii) रावी
(iii) सतलुज
(iv) इनमें से कोई नहीं।
उत्तर-
(ii) रावी।

प्रश्न 4.
भंगी और बाशा चोअ कौन-से जिले में पड़ते हैं ?
(i) फिरोज़पुर
(ii) गुरदासपुर
(iii) होशियारपुर
(iv) कोई भी नहीं।
उत्तर-
(iii) होशियारपुर।

प्रश्न 5.
कौन-सा कथन गलत है और कौन-सा सही है
(i) रावी, ब्यास व सतलुज बारामारसी नदियां हैं।
(ii) काली बेई व पावर्ती, ब्यास की सहायक नदियां हैं।
(iii) प्राकृतिक जल का शुद्धतम रूप वर्षा का जल है। .
(iv) पंजाब में 10 हैडवर्क्स तथा 20,786 किलोमीटर नहरें हैं।
उत्तर-

1. सही,
2. सही,
3. सही,
4. गलत।

प्रश्न 6.
बिस्त दोआब में बिस्त से क्या अभिप्राय है ?
उत्तर-
ब्यास तथा सतलुज दरियाओं के पहले शब्दों ‘बि’ तथा ‘सत’ को मिला. कर बिस्त शब्द बना है।

प्रश्न 7.
हरीके झील से राजस्थान को पानी ले जाने वाली दो नहरें कौन-कौन सी हैं ?
उत्तर-
राजस्थान फीडर नहर जिसे इंदिरा गांधी कमांड नहर भी कहा जाता है।

प्रश्न 8.
पंजाब की कौन-सी नहर हरियाणा को जल प्रदान करती है ?
उत्तर-
घग्गर नदी।

प्रश्न 9.
अपर बारी दोआब नहर का स्त्रोत क्या है ?
उत्तर-
माधोपुर हैडवर्कस।

प्रश्न 10.
पौंग डैम का निर्माण कौन-सी नदी पर किया गया है ?
उत्तर-
ब्यास दरिया।

(ग) प्रश्नों के संक्षेप उत्तर दें:

प्रश्न 1.
ब्यास व रावी की सहायक नदियों की सारणी बनाएं।
उत्तर-
व्यास-ब्यास की सहायक नदियां हैं-सुकन्तरी, पार्वती, सोहां, उहल तथा काली बेईं।
रावी-ऊज, सक्की किरन नाला रावी की प्रमुख नदियां हैं।

प्रश्न 2.
चोअ क्या होते हैं ? किन्हीं चार के नाम लिखें।
उत्तर-
चोअ छोटी मौसमी नदियां होती हैं जो वर्षा की ऋतु में पानी से पूर्णतया भर जाती हैं। बहुत सी चौएं कटारधार तथा सोलहासिंगी पहाड़ियों से शुरू होती हैं। पंजाब के कण्डी क्षेत्र में बहुत सी मौसमी चौएं हैं। बाणा चोअ, टोसां चोअ, बलाचौर चोअ, गढ़शंकर चोअ, नरियाला चोअ, मैली चोअ इत्यादि कुछ प्रमुख चौ हैं।

प्रश्न 3.
पंजाब के नदी, नहरों के प्रदूषण से अवगत करवायें।
उत्तर-
जब पानी में अनावश्यक वस्तुएं मिला दी जाती हैं जिस से पानी प्रयोग करने लायक नहीं रहता, इसे जल प्रदूषण कहते हैं। इसमें कोई शंका नहीं है कि पंजाब की सभी नदियों तथा नहरों में काफ़ी अधिक जल प्रदूषण है। भारत सरकार के कई विभागों तथा मन्त्रालयों का भी मानना है कि पंजाब की नहरों में काफ़ी अधिक जल प्रदूषण है तथा इनमें खतरनाक ज़हर भर रहा है। यह ज़हर पानी की सहायता से हमारी भोजन प्रणाली में पहुँच रहा है तथा लोग इससे खतरनाक बिमारियों का शिकार हो रहे हैं। उदाहरण के लिए बुड्ढा नाला पूर्णतया तेजाबी हो चुका है। हमें सभी नदियों को बचाने की आवश्यकता है ताकि हम पानी के साथ-साथ अपने जीवन को बचा कर रख सकें।

(घ) निम्न प्रश्नों के विस्तृत उत्तर दें

प्रश्न 1.
सतलुज नदी, उस पर बनाये डैमों तथा उसकी सहायक नदियों की जानकारी दें।
उत्तर-
सतलुज नदी तिब्बत में 4630 मीटर की ऊँचाई पर स्थित मानसरोवर झील से रक्षताल नामक स्थान से शुरू होता है। जब यह हिमालय पर्वत को पार कर रहा होता है तो गहरी खाइयां बनाता है। सतलुज मैदानों में भाखड़ा में दाखिल होता है तथा यहां ही भाखड़ा डैम बनाया गया है। नंगल से सतलुज दरिया दक्षिण दिशा की तरफ बढ़ता है तथा जब यह रोपड़ पहुँचता है तो इसमें सुआं, सरसा नदियां, मौसमी चोअ मिल जाते हैं। फिरोजपुर जिले में यह हरीके पत्तण से 60 किलोमीटर की दूरी पर स्थित सुलेमानकी नामक स्थान से पाकिस्तान में चला जाता है। सतलुज दरिया पर भाखड़ा बाँध के साथ-साथ कोटला बाँध, नाथपा झाखड़ी तथा नंगल बाँध भी बनाए गए हैं। सुआं, ब्यास तथा चिट्टी बेईं सतलुज की सहायक नदियां हैं। मक्खु में गिद्दड़ पिण्डी तथा चिट्टी बेईं सतलुज में मिल जाती है। सतलुज दरिया पर कई बाँधों के साथ-साथ रोपड़ तथा हरीके हैडवर्क्स भी बनाए गए हैं।

प्रश्न 2.
पंजाब के नहर प्रबन्ध के विषय में लिखें। इससे कृषि को क्या लाभ हैं ?
उत्तर-
पंजाब की अधिकतर जनता कृषि या इससे संबंधित कार्यों में लगी हुई है तथा पंजाब में ही 1960 के दशक में हरित क्रान्ति की शुरूआत हुई। हरित क्रान्ति में सिंचाई की बहुत बड़ी भूमिका थी क्योंकि किसान फसलों की सिंचाई के लिए केवल वर्षा पर निर्भर नहीं रह सकता। इस कारण पंजाब ने समय-समय पर अपनी नहरी व्यवस्था को काफी विकसित किया। पंजाब में 14500 किलोमीटर लंबी नहरें तथा 5 हैडवर्क्स हैं। यहां 10 नहरें भी हैं जिनके नाम हैंसरहिन्द नहर, अपर बारी दोआब नहर, बिस्त दोआब नहर, भाखड़ा मेन लाइन नहर, फिरोज़पुर/सरहिन्द फीडर प्रबन्ध, कश्मीर नहर, मक्खु, नहर, शाह नहर, राजस्थान फीडर तथा बीकानेर नहर। इन 10 नहरों में से 8 नहरों का वर्णन इस प्रकार है-

कृषि को लाभ-इस नहरी व्यवस्था से पंजाब की कृषि को काफी लाभ हुआ जिसका वर्णन इस प्रकार है-

1. इन नहरों से पंजाब की कृषि को सारा साल पानी मिलता रहता है।
2. सिंचाई के साधन बढ़ने से किसान साल में दो या अधिक फसलें उगाने में सफल हो गए हैं।
3. अधिक फसलें उगाने से किसानों को काफ़ी फायदा हुआ है तथा उनकी आय भी बढ़ गई है।
4. दरियाओं तथा नहरों पर डैम बना कर पानी को रोका गया ताकि वर्षा न होने की स्थिति में किसानों तक पानी पहुँचाया जा सके।
5. डैमों से बिजली तैयार की गई जिससे उद्योगों तथा घरों को 24 घण्टे बिजली प्राप्त हुई।

प्रश्न 3.
पंजाब के चोअ और रौअ कौन-कौन से हैं ? विस्तत नोट लिखें।
उत्तर-
चौएं छोटी छोटी वर्षा पर आधारित तथा मौसमी नदियां होती हैं जो वर्षा के मौसम में पानी से भर जाती हैं। पंजाब में एक कण्डी क्षेत्र है जहां बहुत-सी चौएं मौजूद हैं। इनमें से कई चौओं का जन्म कटारधार तथा सेलासिंगी की पहाड़ियों में होता है। जब वर्षा आती है तो इन चौओं में पानी भर जाता है। पंजाब सरकार ने इनमें से कई चौओं को बंद करने में सफलता प्राप्त कर ली है तथा इनमें आने वाले वर्षा के पानी को कृषि अथवा अन्य कार्यों के लिए प्रयोग किया जा रहा है।

होशियारपुर जिले के दक्षिण पश्चिम में 93 चौएं मौजूद हैं जिनमें से बहुत से काली बेईं तथा चिट्टी बेईं में जाकर मिल जाते हैं। होशियारपुर में बहुत से चौएं हैं जिनमें कुछ काफी प्रमुख हैं जैसे कि टोसां चौ, बणा चौ, गढ़शंकर चौ, बलाचौर चौ, मैली चौ, नरियाला चौ, नंगल शहीदां चौ, गोंदपुर चौ, दसूहा चौ इत्यादि। इन पर नियन्त्रण रखने के लिए पंजाब सरकार ने कण्डी क्षेत्र विकास (Kandi Area Development) को भी शुरू किया है। पंजाब में कुछ बरसाती नाले भी हैं जैसे कि पटियाला की राव, जैंतिया देवी की रौ, बुड्ढ़ा नाला इत्यादि।

PSEB 9th Class Social Science Guide पंजाब : जलतन्त्र Important Questions and Answers

बहुविकल्पीय प्रश्न (Multiple Choice Questions)

प्रश्न 1.
पंजाब कौन-से दो शब्दों से मिलकर बना है ?
(क) पंज + आब
(ख) पंजा + आहब
(ग) पंज + अहाब
(घ) पं + जाहब।
उत्तर-
(क) पंज + आब

प्रश्न 2.
अब पंजाब में कितने दरिया हैं ?
(क) दो
(ख) तीन
(ग) चार
(घ) पाँच।
उत्तर-
(ख) तीन

प्रश्न 3.
इनमें से कौन-सा मौसमी दरिया है ?
(क) घग्गर
(ख) सकी किरन
(ग) काली बेई
(घ) उपर्युक्त सभी।
उत्तर-
(घ) उपर्युक्त सभी।

प्रश्न 4.
इनमें से कौन-सा बारहमासी दरिया है ?
(क) रावी
(ख) ब्यास
(ग) सतलुज
(घ) उपर्युक्त सभी।
उत्तर-
(घ) उपर्युक्त सभी।

प्रश्न 5.
रणजीत सागर डैम किस दरिया पर बना है ?
(क) रावी
(ख) सतलुज
(ग) ब्यास
(घ) चिनाब।
उत्तर-
(क) रावी

प्रश्न 6.
पौंग डैम किस दरिया पर बना है ?
(क) रावी
(ख) सतलुज
(ग) ब्यास
(घ) जेहलम।
उत्तर-
(ग) ब्यास

प्रश्न 7.
होशियारपुर में कितने चौएं हैं ?
(क) 70
(ख) 93
(ग) 84
(घ) 54.
उत्तर-
(ख) 93

रिक्त स्थान की पूर्ति करें (Fill in the Blanks)

1. सन् …………. में भारत तथा पाकिस्तान के विभाजन का सबसे बड़ा नुकसान …….. को हुआ।
2. रावी, ब्यास तथा सतलुज ……….. दरिया हैं।
3. रणजीत सागर डैम का कार्य ………… में पूर्ण हुआ था।
4. सुकन्तरी ………. की प्रमुख सहायक नदी है।
5. ………… किसी समय सरस्वती नदी का हिस्सा थी।

उत्तर-

1. 1947, पंजाब,
2. बारहमासी,
3. 2001,
4. ब्यास,
5. घग्गर

सही/ग़लत (True/False)

1. जेहलम, चिनाब तथा सिन्धु पाकिस्तान वाले पंजाब में रह गए।
2. रावी ककझ मंझ नाम के स्थान पर पाकिस्तान में प्रवेश करता है।
3. रणजीत सागर डैम से 1600 वाट बिजली उत्पन्न होती है।
4. ब्यास दरिया पर पौंग डैम बनाया गया है।
5. रावी दरिया से राजस्थान फीडर नहर निकाली गई है।

उत्तर-

1. ✓
2. ✓
3. ✗
4. ✓
5. ✗

अति लघु उत्तरों वाले प्रश्न।।

प्रश्न 1.
पंजाब शब्द का क्या अर्थ है ?
उत्तर-
पंजाब शब्द दो शब्दों ‘पंज-आब’ से मिलकर बना है जिसका अर्थ है पाँच दरियाओं की धरती।

प्रश्न 2.
1947 के पश्चात् कौन-से दरिया पंजाब में रह गए।
उत्तर-
सतलुज, रावी तथा ब्यास।

प्रश्न 3.
1947 के पश्चात् कौन-से दरिया पाकिस्तान वाले पंजाब में चले गए ?
उत्तर-
जेहलम, चिनाब तथा सिन्धु।

प्रश्न 4.
बारहमासी दरिया कौन-से होते हैं ?
उत्तर-
वह दरिया जिनमें सम्पूर्ण वर्ष पानी रहता है उन्हें बारहमासी दरिया कहते हैं।

प्रश्न 5.
बारहमासी दरियाओं में सम्पूर्ण वर्ष पानी कहां से आता है ?
उत्तर-
बारहमासी दरियाओं में पहाड़ों से पिघली बर्फ का पानी सम्पूर्ण वर्ष आता रहता है।

प्रश्न 6.
पंजाब के कुछ मौसमी दरियाओं के नाम लिखो।
उत्तर-
घग्गर, काली बेईं, चिट्टी बेईं, ऊज, चक्की खड्ड, स्वात इत्यादि।

प्रश्न 7.
किन्हीं दो अवशेषी दरियाओं के नाम लिखें।
उत्तर-
बुड्डा नाला तथा सक्की किरन नाला।

प्रश्न 8.
रावी दरिया का जन्म कहां पर होता है ?
उत्तर-
रावी दरिया कुल्लू की पहाड़ियों में स्थित रोहतांग दर्रे के उत्तर में 4116 मीटर की ऊंचाई से शुरू होता है।

प्रश्न 9.
रावी दरिया पर कौन-सा डैम बनाया गया है तथा इसमें से कौन-सी नहर निकाली गई है ?
उत्तर-
रावी दरिया पर रणजीत सागर डैम बनाया गया है तथा इससे अपर बारी दोआब नहर निकाली गई है।

प्रश्न 10.
रावी दरिया पर कौन-से हैडवर्क्स बनाए गए हैं ?
उत्तर-
शाहपुर कण्डी के नज़दीक धाना या बसन्तपुर, कटारधार, माधोपुर हैडवर्क्स तथा माधोपुर ब्यास लिंक पर कठुआ फीडर।

प्रश्न 11.
रणजीत सागर डैम के बारे में बताएं।
उत्तर-
यह रावी दरिया पर बनाया गया डैम है जिससे 600 मेगावाट बिजली पैदा होती है। यह 1981 में शुरू हुआ था तथा मार्च 2011 में इसका कार्य पूर्ण हुआ था।

प्रश्न 12.
ब्यास दरिया कहां से निकलता है ?
उत्तर-
ब्यास दरिया ब्यास कुण्ड से निकलता है जो हिमाचल प्रदेश में रोहतांग दर्रे के पास 4060 मीटर की ऊंचाई पर स्थित है।

प्रश्न 13.
व्यास दरिया पर कौन-से डैम बनाए गए हैं ?
उत्तर-
हिमाचल प्रदेश में पंडोह तथा पंजाब में पौंग डैम।

प्रश्न 14.
ब्यास से कौन सी नहर निकाली गई है ?
उत्तर-
ब्यास से राजस्थान फीडर नहर निकाली गई है जिसे इंदिरा गांधी कमांड नहर का नाम भी दिया गया है।

प्रश्न 15.
ब्यास की सहायक नदियों के नाम लिखें।
उत्तर–
पार्वती, सुकन्तरी, सौहां, उम्मन तथा काली बेईं।

प्रश्न 16.
सतलुज दरिया कहां पर शुरू होता है ?
उत्तर-
सतलुज दरिया तिब्बत में मानसरोवर झील के नज़दीक स्थित रक्षताल से शुरू होता है।

प्रश्न 17.
सतलुज दरिया कहां पर पाकिस्तान में प्रवेश करता है ?
उत्तर-
सतलुज दरिया फिरोज़पुर में सुलेमान की नामक स्थान से पाकिस्तान में प्रवेश करता है।

प्रश्न 18.
सतलुज दरिया पर कौन-से डैम बनाए गए हैं ?
उत्तर-
नाथपा झाखड़ी, नंगल डैम, कौटला डैम।

प्रश्न 19.
घग्गर किस प्रकार की नदी है ?
उत्तर-
घग्गर दक्षिणी पंजाब में बहने वाली एक मौसमी नदी है ।

प्रश्न 20.
घग्गर कहां से निकलती है ?
उत्तर-
घग्गर नदी सिरमौर की पहाड़ियों से निकलती है।

प्रश्न 21.
पंजाब के किस क्षेत्र में बहुत से चोअ मिलते हैं ?
उत्तर-
कण्डी क्षेत्र में।

प्रश्न 22.
चोअ क्या होता है ?
उत्तर-
चोअ एक छोटी सी नदी होती है जो वर्षा के मौसम में पानी से भर जाती है।

प्रश्न 23.
पंजाब के किस जिले में बहुत से चोअ हैं ?
उत्तर-
होशियारपुर जिले में।

प्रश्न 24.
पंजाब की नहरों की लंबाई बताएं।
उत्तर-
पंजाब की नहरों की लंबाई 14500 किलोमीटर है।

प्रश्न 25.
पंजाब की सबसे लंबी नहर कौन-सी है ?
उत्तर-
भाखड़ा मेन लाइन जिसकी लंबाई 161.36 किलोमीटर है।

प्रश्न 26.
कौन-सा दरिया किसी समय सरस्वती नदी का सहायक होता था ?
उत्तर-
घग्गर दरिया किसी समय सरस्वती नदी का सहायक होता था।

लघु उत्तरों वाले प्रश्न

प्रश्न 1.
पंजाब के जलतन्त्र के बारे में बताएं।
उत्तर-
पंजाब दो शब्द ‘पंज’ तथा ‘आब’ से मिलकर बना है जिसका अर्थ है पाँच दरियाओं की धरती। पंजाब में 1947 से पहले कई दरिया होते थे परन्तु देशों के विभाजन के कारण जेहलम, चिनाब, सिन्धु तथा बहुत सी नदियां पाकिस्तान में चली गईं। अब पंजाब में केवल तीन दरिया रावी, ब्यास तथा सतलुज ही हैं। यह तीनों दरिया बारहमासी हैं जिनमें पहाड़ों की बर्फ पिघलने के कारण सारा साल पानी रहता है। यहां बहुत से मौसमी दरिया भी हैं जैसे कि घग्गर, ऊज, काली बेईं, चिट्टी बेईं, स्वान, नूरपुर बेदी चोअ इत्यादि। यहां अवशेषी दरिया, जैसे कि बुड्ढा नाला तथा सक्की किरन नाला भी मिलते हैं।

प्रश्न 2.
रावी की सहायक नदियों के बारे में बताएं।
उत्तर-
जब रावी दरिया माधोपुर पहुँचता है तो इसमें कई सहायक नदियां आकर मिल जाती हैं। इनमें सबसे महत्त्वपूर्ण ऊज नदी है। इसके साथ ही सक्की किरन नाला रावी के साथ-साथ चलता है तथा भारत और पाकिस्तान की सरहद पर इसमें मिल जाता है। रावी दरिया पर चार हैडवर्क्स भी बनाए गए हैं जिनके नाम हैं माधोपुर ब्यास लिंक पर कठुआ फीडर, शाहपुर कण्डी के नज़दीक धाना या बसन्तपुर, माधोपुर हैडवर्क्स तथा कटारधार।

प्रश्न 3.
ब्यास दरिया की सहायक नदियों के बारे में बताएं।
उत्तर-
सुकन्तरी-उग्मन, पारबती, काली बेईं तथा सौहां ब्यास की कुछ सहायक नदियां हैं। तलवाड़ा पहुँच कर सौहां ब्यास में मिल जाती हैं। हरीके के नज़दीक काली बेईं होशियारपुर तथा कपूरथला से होते हुए ब्यास में मिल जाती है। ब्यास दरिया पर पौंग डैम तथा पंडोह डैम को भी बनाया गया है।

प्रश्न 4.
घग्गर पर एक नोट लिखें।
उत्तर-
पंजाब में काफी पहले सरस्वती नदी बहती थी तथा घग्गर भी उसका ही हिस्सा थी। परन्तु अब घग्गर एक मौसमी नदी है जो दक्षिण पंजाब में बहती है। यह सिरमौर की पहाड़ियों से निकलती है। मुबारकपुर नाम के स्थान पर यह मैदानी इलाकों में आ जाती है। इसके पश्चात् यह पटियाला, घनौर तथा हरियाणा के इलाकों को पार करती हैं। इसके पश्चात् यह राजस्थान के रेगिस्तान में जाकर खत्म हो जाती है।

दीर्घ उत्तर वाला प्रश्न

प्रश्न-रावी दरिया पर एक नोट लिखें।
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Punjab State Board PSEB 11th Class Chemistry Book Solutions Chapter 7 Equilibrium Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Chemistry Chapter 7 Equilibrium

PSEB 11th Class Chemistry Guide Equilibrium InText Questions and Answers

Question 1.
A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased.
(a) What is the initial effect of the change on vapour pressure?
(b) How do rates of evaporation and condensation change initially?
(c) What happens when equilibrium is restored finally and what will be the final vapour pressure?
Answer:
(a) If the volume of the container is suddenly increa50sed, then the vapour pressure would decrease initially. This is because the amount of vapour remains the same, but the volume increases suddenly. As a result, the same amount of vapour is distributed in a larger volume.
(b) Since the temperature is constant, the rate of evaporation also remains constant. When the volume of the container is increased, the density of the vapour phase decreases. As a result, the rate of collisions of the vapour particles also decreases. Hence, the rate of condensation decreases initially.
(c) When equilibrium is restored finally, the rate of evaporation becomes equal to the rate of condensation. In this case, only the volume changes while the temperature remains constant. The vapour pressure depends on temperature and not on volume. Hence, the final vapour pressure will be equal to the original vapour pressure of the system.

Question 2.
What is K_c for the following equilibrium when the equilibrium concentration of each substance is: [SO₂] = 0.60M,[O₂] = 0.82M and [SO₃] = 1.90 M?
2SO₂(g) + O₂(g) ↔ 2SO₃(g)
Answer:
The given reaction is
2SO₂(g) + O₂(g) ↔ 2SO₃(g)
Equilibrium constant
K_c = \(\frac{\left[\mathrm{SO}_{3}\right]^{2}}{\left[\mathrm{SO}_{2}\right]^{2}\left[\mathrm{O}_{2}\right]}=\frac{(1.90 \mathrm{M})^{2}}{(0.60 \mathrm{M})^{2}(0.82 \mathrm{M})}\)
= 12.238 M^-1

Question 3.
At a certain temperature and total pressure of 10⁵ Pa, iodine vapour contains 40% by volume of I atoms I₂(g) ⇌ 2I(g)
Calculate K_p for the equilibrium.
Answer:
Given, I₂(g) ⇌ 2I(g)
I atoms in iodine vapours = 40% by volume
So, iodine vapours of I₂ molecules = 60% by volume

Question 4.
Write the expression for the equilibrium constant, Kc for each of the following reactions:
(i) 2NOCl(g) ⇌ 2NO (g) + Cl₂(g)
(ii) 2CU(NO₃)₂(S) ⇌ 2CuO(s) + 4NO₂(g) + O₂(g)
(iii) CH₃COOC₂H₅(oq) + H₂O(l) ⇌ CH₃COOH(aq) + C₂H₅OH(ag)
(iv) Fe³⁺(aq) + 3OH^–(aq) ⇌ Fe(OH)₃(s)
(v) I₂(s) + 5F₂ ⇌ 2IF₅
Answer:

Question 5.
Find out the value of K_c for each of the following equilibria from the value of Kp.
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); K_p = 1.8 x 10^-2 at 500 K
(ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); K_p = 167 at 1073 K
Answer:
The relation between K_p and K_c is given as
K_p = K_c(RT)^Δn
(i) 2NOCl(g) ⇌ 2NO(g) + Cl₂(g); K_p = 1.8 x 10^-2 at 500 K.
Δn = 3 – 2 = 1
R = 0.0831 bar L mol^-1K^-1
T = 500 K
K_p =1.8 x 10^-2
K_p = K_c( RT)^Δn
1.8 x 10^-2 = K_c(0.0831 x 500)¹
K_c = \(\frac{1.8 \times 10^{-2}}{0.0831 \times 500}\) = 4.33 x 10^-4

(ii) CaCO₃(s) ⇌ CaO(s) + CO₂(g); K_p = 167 at 1073 K
Δn = 2 -1 = 1
R = 0.0831 bar L mol^-1K^-1
T = 1073 K
K_p =167
Now, K_p = K_c(RT)^Δn
⇒ 167 = K_c(0.0831 x 1073)¹
⇒ K_c = \(\frac{167}{0.0831 \times 1073}\) = 1.87

Question 6.
For the following equilibriuih, K_c = 6.3 x 10¹⁴ at 1000 K
NO(g) + O₃(g) ⇌ NO₂(g) + O₂(g)
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is K_c, for the reverse reaction?
Answer:
It is given that K_c for the forward reaction is 6.3 x 10¹⁴ at 1000 K.
Then, K_c for the reverse reaction will be,
NO₂(g) + O₃(g) ⇌ NO(g) + O₃(g)
K_c = \(\frac{1}{K_{c}}=\frac{1}{6.3 \times 10^{14}}\) = 1.59 x 10^-15

Question 7.
Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?
Answer:
For a pure substance (both solids and liquids),

Now, the molecular mass and density (at a particular temperature) of a pure substance is always fixed and is accounted for in the equilibrium constant. Therefore, the values of pure substances are not mentioned in the equilibrium constant expression.

Question 8.
Reaction between N₂ and O₂ takes place as follows:
2N₂(g) + O₂(g) ⇌ 2N₂O(g)
If a mixture of 0.482 mol of N₂ and 0.933 mol of O₂ is placed in a 10 L reaction vessel and allowed to form N₂O at a temperature for which K_c = 2.0 x 10^-37 , determine the composition of equilibrium mixture.
Answer:
Let the concentration of N₂O at equilibrium be x.
The given reaction is :

The value of equilibrium constant i.e., K_c = 2.0 x 10^-37 is very small which means negligible amounts of N₂ and O₂ react.

Question 9.
Nitric oxide reacts with Br₂ and gives nitrosyl bromide as per reaction given below:
2NO(g) + Br₂(g) ⇌ 2NOBr(g)
When 0.087 mol of NO and 0.0437 mol of Br₂ are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO andBr₂.
Answer:
The balanced chemical equation is

Given, 2x = 0.0518
x = 0.0259 mol
Moles of NO at equilibrium = 0.087 – 2x
= 0.087-0.0518
= 0.0352 mol
Moles of Br2 at equilibrium = 0.0437 – x
= 0.0437 – 0.0259
= 0.0178 mol

Question 10.
At 450 K, K_p = 2.0 x 10¹⁰/bar for the given reaction at equilibrium.
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
What is K_c at this temperature?
Answer:
The given reaction is
2SO₂(g) + O₂(g) ⇌ 2SO₃Cg)
Δn = 2 – 3 = -1
T = 450 K
R = 0.0831 bar L K^-1 mol^-1
K_p = 2.0 x 10¹⁰ bar^-1
We know that,
K_p = K_c(RT)^Δn
=> 2.0 x 10¹⁰ bar^-1 = k_c(0.0831 L bar K^-1 mol^-1 x 450 K)^-1
\(K_{c}=\frac{2.0 \times 10^{10} \mathrm{bar}^{-1}}{\left(0.0831 \mathrm{~L} \mathrm{barK} \mathrm{K}^{-1} \mathrm{~mol}^{-1} \times 450 \mathrm{~K}\right)^{-1}}\)
K_c = (2.0 x 10¹⁰ bar^-1) (0.0831 L bar K^-1mol^-1450 K)
= 74.79 x 10¹⁰ L mol^-1 = 7.48 x 10¹¹ L mol^-1

Question 11.
A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is K_p for the given equilibrium?
2HI(g) ⇌ H₂(g) + I₂(g)
Answer:
The given reaction is

∵ Decrease is pressure of HI = 0.2 – 0.04 = 0.16 atm;
So equilibrium pressure of H₂ is \(\frac{0.16}{2}\) = 0.08 atm and for I₂ is \(\frac{0.16}{2}\) = 0.08 atm
as two moles of HI on dissociation gives 1 mol of H₂ and 1 mol of I₂.
Therefore,
K_p = \(\frac{p_{\mathrm{H}_{2}} \times p_{\mathrm{I}_{2}}}{\left(p_{\mathrm{HI}}\right)^{2}}=\frac{0.08 \times 0.08}{(0.04)^{2}}=\frac{0.0064}{0.0016}\) = 4.0
Hence, the value of K_p is 4.0.

Question 12.
A mixture of 1.57 mol of N₂,1.92 mol of H₂ and 8.13 mol of NH₃ is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is 1. 7 x 10².
Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction?
Answer:
The given reaction is :
N₂(g) + 3H₂(g) 2NH₃(g)
Given, [N₂] = \(\frac{1.57}{20}\) = 0.0785 M
[H₂] = \(\frac{1.92}{20}\) = 0.096 M
[NH₃] = \(\frac{8.13}{20}\) = 0.4065 M
Now, reaction quotient Q_c. is :
Q_c = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}=\frac{(0.4065)^{2}}{(0.0785)(0.096)^{3}}\) = 2.4 x 10³M^-2
Since Q_c ≠ K_c the reaction mixture is not in equilibrium.
Again Q_c > K_c. Hence, the reaction will proceed in the reverse direction.

Question 13.
The equilibrium constant expression for a gas reaction is, K_c = \(\frac{\left[\mathrm{NH}_{3}\right]^{4}\left[\mathrm{O}_{2}\right]^{5}}{\left[\mathrm{NO}^{4}\left[\mathrm{H}_{2} \mathrm{O}\right]^{6}\right.}\)
Write the balanced chemical equation corresponding to this expression.
Answer:
The balanced chemical equation corresponding to the given expression can be written as :
4NO(g) + 6H₂O(l) ⇌ 4NH₃(g) + 5O₂(g)

Question 14.
One mole of H₂O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation,
H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)
Calculate the equilibrium constant for the reaction.
Answer:
The given reaction is

H₂O reacted = 40% of 1 mol of H₂O = 0.4 mol
x = 0.4 mol 1 – x = 1 – 0.4 = 0.6 mol
Therefore, the equilibrium constant for the reaction,
K_c = \(\) = 0.444

Question 15.
At 700 K, equilibrium constant for the reaction
H₂(g) + I₂(g) ⇌ 2HI(g)
is 54.8. If 0.5 mol L^-1 of Hl(g) is present at equilibrium at 700 K, what are the concentration of H₂(g) and I₂(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700 K?
Answer:
The given reaction is
H₂(g) + I₂(g) ⇌ 2HI(g); K_c = 54.8
Or the reaction
2HI(g) ⇌ H₂(g) + I₂(g); K_c‘ = \(\)
Given, [HI] = 0.5 mol L^-1
According to equation
[H₂] = [I₂] = x mol L^-1
Therefore,
\(\frac{\left[\mathrm{H}_{2}\right]\left[\mathrm{I}_{2}\right]}{[\mathrm{HI}]^{2}}=K_{c}^{\prime}\)
⇒ \(\frac{x \times x}{(0.5)^{2}}=\frac{1}{54.8}\)
⇒ x² = \(\frac{0.25}{54.8}\)
⇒ x = 0.06754
x = 0.068 mol L^-1
Hence, at equilibrium, [H₂] = [I₂] = x = 0.068 mol L^-1

Question 16.
What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of IC1 was 0.78 M?
2ICl(g) ⇌ I₂(g) + Cl₂(g); K_c = 0.14
Answer:

Question 17.
K_p = 0.04 atm at9 K for the equilibrium shown below. What is the equilibrium concentration of C₂H₆ when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
Answer:

Question 18.
Ethyl acetate is formed by the reaction between ethanol acid and acetic acid and the equilibrium is represented as :
CH₃COOH(l) + C₂H₅OH (l) ⇌ CH₃COOC₂H₅(Z) + H₂O(l)
(i) Write the concentration ratio (reaction quotient), Q_c, for this reaction (note: water is not in excess and is not a solvent in this reaction)
(ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant.
(iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached?
Answer:

Question 19.
A sample of pure PCl₅ was introduced into an evacuated vessel at 473K. After equilibrium was attained, concentration of PCl₅ was found to be 0.5 x 10^-1 mol L^-1. If value of K is 8.3 x 10^-3, what are the concentrations of PCl₃ and Cl₂ at equilibrium?
PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g)
Answer:
The given reaction is

It is given that the value of equilibrium constant, K = 8.3 x 10^-3.
K_c = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)
[Given, [PCl₅]_equili = 0.5 x 10^-1 mol L^-1]
\(\frac{x \times x}{0.5 \times 10^{-1}}\) = 8.3 x 10^-3
⇒ x² = 4.15 x10^-4
⇒ x = 2.04 x 10^-2 = 0.0204 mol L^-1 = 0.02 mol L^-1
Therefore, at equilibrium,
[Pcl₃] = [Cl₂] = 0.02 mol L^-1

Question 20.
One of the reaction that takes place in producing steel from iron ore is the reduction of iron (H) oxide by carbon monoxide to give iron metal and CO₂.
FeO(s) + CO(g) ⇌ Fe(s) + CO₂ (g); K_p = 0.265 atm at 1050 K. What are the equilibrium partial pressures of CO and ₂ at 1050 K if the initial partial pressures are P_Co = 1.4 atm and pCO₂ = 0.80 atm?
Answer:
(i) The given reaction is

Since Q_p > K_p, the reaction will proceed in the backward direction.
Therefore, we can say that the pressure of CO will increase while the pressure of CO₂ will decrease.
Now,let the increase in pressure of CO = decrease in pressure of CO₂ be p.
Hence pCO₂ = 0.80 – p and P_CO = 1.4 + p
and K_p = \(\frac{p_{\mathrm{CO}_{2}}}{p_{\mathrm{CO}}}\)
0.265 = \(\frac{0.80-p}{1.4+p}\)
0.371 + 0.265p = 0.80 — p= 1.265p= 0.429
p = 0.339atm
Hence, at equilibrium
P_CO2 = 0.80 – 0.339 = 0.461 atm
And, equilibrium partial pressure of
P_CO = 1.4 + 0.339 = 1.739 atm.

Question 21.
Equilibrium constant, K_c for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K is 0.06 1.
At a particular time, the analysis shows that composition of the
reaction mixture is 3.0 mol L^-1 N₂,2.0 mol L^-1 H₂ and 0.5 mol L^-1 NH₃ Is the reaction at equilibrium? if not in which direction
does the reaction tend to proceed to reach equilibrium?
Answer:
The given reaction is
N₂(g) + 3H₂(g) ⇌ 2NH₃(g);K_c = 0.061 at 500K
Given, [N₂] = 3.0mol L^-1, [H₂] = 2.0 mol L^-1, [NH₃] = 0.5 mol L^-1
So, Q = \(\frac{\left[\mathrm{NH}_{3}\right]^{2}}{\left[\mathrm{~N}_{2}\right]\left[\mathrm{H}_{2}\right]^{3}}=\frac{(0.5)^{2}}{(3.0)(2.0)^{3}}\) = 0.0104
It is given that K_c = 0.06 1
Since Q_c ≠ K_c, the reaction is not at equilibrium.
Since Q_c < K_c, the reaction will proceed in the forward direction to reach equilibrium.

Question 22.
Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium:
2BrCl(g) ⇌ Br₂(g) + Cl₂(g)
for which K_c = 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 x 10^-3 mol L^-1, what is its molar concentration in the mixture at equilibrium?
Answer:

x = 11.312(3.30 x 10^-3 – x)
x = 0.03732 – 11.312x
x + 11.312x = 0.03732
x = \(\frac{0.03732}{12.312}\)= 3.0321 x 10^-3 mol L^-1
[BrCl]_equili = (3.30 x 10^-3 – 3.032 x 10^-3) mol L^-1
= 2.68 x 10^-4 mol L^-1

Question 23.
At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO₂ in equilibrium with solid carbon has 90.55% CO by mass
C(s) + CO₂(g) ⇌ 2CO(g)
Calculate K_c for this reaction at the above temperature.
Answer:
Let the total mass of the gaseous mixture be 100g.
Mass of CO = 90.55 g
and, mass of CO₂ = (100 – 90.55) = 9.45 g
Now, number of moles of CO,
n_CO = \(\frac{90.55}{28}\) = 3.234 mol
(Molar mass of CO = 28 g mol^-1 )
Now, number of moles of CO₂,
n_CO = \(\)
(Molar mass of CO₂ = 44 g mol^-1 )

For the given reaction, Δn = 2 -1 = 1
We know that,
K_p = K_c(RT)^Δn
⇒ 14.19 = K_c(0.0831 x 1127)¹
⇒ K_c = 0.154 (approximately)

Question 24.
Calculate (a) \(\Delta \boldsymbol{G}^{\ominus}\) and (b) the equilibrium constant for the formation of N02 from NO and Oa at 298K
NO(g) + \(\frac{1}{2}\)O₂(g) ⇌ NO₂(g)
where \(\Delta_{f} \boldsymbol{G}^{\ominus}\) (N02)= 52.0 kJ/mol; \(\Delta_{f} \boldsymbol{G}^{\ominus}\) (NO) = 87.0kJ/mol;
\(\Delta_{f} \boldsymbol{G}^{\ominus}\) (O2) = 0 kJ/mol
Answer:
(a) The given reaction is
N0(g) + \(\frac{1}{2}\)O₂(g) ⇌ NO₂(g)

= (52.0 – 87.0 + \(\frac{1}{2}\) x 0 )kJ mol^-1 = -35.0 kJ mol^-1

(b) \(\Delta_{r} G^{\ominus}\) = – 2.303 RT logK_c
-35.0 = – 2.303 x 0.0831 x 298 log K_c
∴ log K_c= \(\frac{35}{5.7058}\)= 6.134
∴ K_c = antilog 6.134 = 1.361 x 10⁶.

Question 25.
Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume?
(a) PCl₅(g) ⇌ PCl₃(g) +Cl₂(g)
(b) CaO (s) + CO₂ (g) ⇌ CaCO₃ (s)
(c) 3Fe(s) + 4H₂O (g) ⇌ Fe₃O₄ (s) + 4H₅(g)
Answer:
(a)The number of moles of reaction products will increase. According to Le-Chatelier’s principle, if pressure is decreased, then the equilibrium shifts in the direction in which the number of moles of gases is more. In the given reaction, the number of moles of gaseous products is more than that of gaseous reactants. Thus, the reaction will proceed in the forward direction. As a result, the number of moles of reaction products will increase.
(b) The number of moles of reaction products will decrease.
(c) The number of moles of reaction products remains the same

Question 26.
Which of the following reactions will get affected hy increasing the pressure?
Also, mention whether change will cause the reaction to go into forward or backward direction.
(i) COCl₂(g) ⇌ CO(g) +Cl₂(g)
(ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g)
(iii) CO₂(g) + C(S) ⇌ 2CO(g)
(iv) 2H₂(g) +CO(g) ⇌ CH₃OH(g)
(v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
(vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)
Answer:
In all the above reactions, the reaction no. (ii) proceeds with the same no. of moles on both sides
i.e., n_p = n_r = 3 .
∴ This reaction will not be affected by the increase in pressure i. e., the direction of equilibrium will not be affected by the increase in pressure. All other reactions will be affected by the increase in pressure.
(i) COCl₂(g) ⇌ CO(g) +Cl₂(g)
n_p > n_r , n_p = 2; nr = 1
∴ Equilibrium will shift to the left increasing pressure.
(iii) CO₂(g) + C(S) ⇌ 2CO(g)
Here, n_r – 1; n_p = 2, therefore n_p > n_r
∴ Equilibrium will go to left on increase of pressure.
(iv) 2H₂(g) +CO(g) ⇌ CH₃OH(g)
Here, n_r = 3; n_p = 1 therefore n_p < n_r
∴ Equilibrium will shift to the right on increasing pressure.
(v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
Here n_r = 0; n_p = 1, therefore n_p > n_r
∴ Equilibrium will shift backwards (left) on increasing the pressure.
(vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)
Here n_r = 9; n_p = 10, therefore n_p > n_r
∴ Equilibrium will shift backwards on increasing the pressure.

Question 27.
The equilibrium constant for the following reaction is 1.6 x 10⁵ at 1024 K.
H₂(g) + Br₂(g) ⇌ 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K.
Answer:
Given reaction is H₂(g) + Br₂(g)⇌ 2HBr(g); K_p = 1.6 x 10⁵ at 1024 K
Therefore, for the reaction 2HBr(g) ⇌ H₂(g)+Br₂(g), the equilibrium constant will be,

p = 2.5 x 1^-2(5.0 x 10^-3)p
p+(5.0 x 10^-3)p = 2.5 x 10^-2
(1005 x 10^-3)p = 2.5 x 10^-2
p = 2.49 x 10^-2 bar = 2.5 x 10^-2 bar
rherefore, at equilibrium,
[H₂] = [Br₂] = 2.49 x 10^-2 bar
[HBr] =10 — 2 x (2.49 x 10^-2) bar
= 9.95 bar = 10 bar

Question 28.
Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction:
CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)
(a) Write an expression for K_p for the above reaction.
(b) How will the values of k_p and composition of equilibrium mixture be affected by
(i) increasing the pressure
(ii) increasing the temperature
(iii) Using a catalyst?
Answer:
(a) The given reaction is
CH₄(g) + H₄O(g) ⇌ CO(g) + 3H₂(g)
\(K_{p}=\frac{p_{\mathrm{CO}} \times p_{\mathrm{H}_{2}}^{3}}{p_{\mathrm{CH}_{4}} \times p_{\mathrm{H}_{2} \mathrm{O}}}\)
(b) (1) According to LeChatelier’s principle, the equilibrium will shift in the backward direction.
(ii) According to Le-Chatelier’s principle, as the reaction is endothermic, the equilibrium will shift in the forward direction.
(iii) The equilibrium of the reaction is not affected by the presence of a catalyst. A catalyst only increases the rate of a reaction. Thus, equilibrium will be attained quickly.

Question 29.
Describe the effect of:
(a) addition of H₂
(b) addition of CH₃OH
(e) removal of CO (d) removal of CH₃OH
on the equilibrium of the reaction:
2H₂(g) + CO (g) ⇌ CH₃OH(g)
Answer:
2H₂(g) + CO(g) ⇌ CH₃OH(g)
According to Le Chatelier’s principle,
(a) Addition of H₂ (increase in concentration of reactants) shifts the equilibrium in forward direction (more product is formed).
(b) Addition of CH₃OH (increase in concentration of product) shifts the equilibrium in backward direction.
(c) Removal of CO also shifts the equilibrium in backward direction.
(d) Removal of CH₃OH shifts the equilibrium in forward direction.

Question 30.
At 473K, equilibrium constant K_c for decomposition of phosphorus pentachloride, PCl₅ is 8.3 x 10^-3. If decomposition is depicted as,
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); \(\Delta_{\boldsymbol{r}} \boldsymbol{H}^{\ominus}\) = 1240 kJ mol^-1
(a) Write an expression for K_c for the reaction.
(b) What is the value of K_c for the reverse reaction at the same temperature?
(c) What would be the effect on K_c if
(i) more PCl₅ is added
(ii) pressure is increased?
(iii) the temperature is increased?
Answer:
PCl₅(g) ⇌ PCl₃(g) + Cl₂(g); Kc = 8.3 x 10^-3
(a) K_c = \(\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}\)
(b) Value of K_c for the reverse reaction at the same temperature is
K’_c = \(\frac{1}{K_{c}}=\frac{1}{8.3 \times 10^{-3}}\) = 1.2048 x 10² = 120.48
(c) (i) Addition pf PCl₅ have no effect on K_c because K_c is constant at constant temperature.
(ii) K_c does not change with pressure.
(iii) The given reaction is endothermic, hence on increasing the temperature, K_c will increase.

Question 31.
Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H₂. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,
CO(g) +H₂O (g) ⇌ CO₂(g) + H₂(g)
If a reaction vessel at 400° C is charged with an equimolar mixture of CO and steam such that p_CO = P_H2O = 4.0 bar, what will be the partial pressure of H₂ at equilibrium? K_P = 10.1 at 400°C
Answer:
The given reaction is
CO(g) + H20(g) ⇌ C02(g) + H2(g)

p = 12.71 – 3.17p
4.17 p = 12.71
p = \(\frac{12.71}{4.17}\) = 3.04 bar
Hence PH₂ = 3.04 bar

Question 32.
Predict which of the following reaction will have appreciable concentration of reactants and products:
(a) Cl₂(g) ⇌ 2Cl(g);K_c = 5 x 10^-39
(b) Cl₂(g) + 2NO(g) ⇌ 2NOCl(g); Kc = 3.7 x 10⁸
(c) Cl₂(g) + 2NO₂(g) ⇌ 2NO₂Cl(g); K_c = 1.8
Answer:
Following conclusions can be drawn from the values of K_c:
(a) Since the value of K_c is very small, this means that the molar concentration of the products is very small as compared to that of the reactants.
(b) Since the value of K_c is quite large, this means that the molar concentration of the products is very large as compared to that of the reactants.
(c) Since the value of K_c is 1.8, this means that both the products and reactants have appreciable concentration.

Question 33.
The value of K_c for the reaction
3O₂(g) ⇌ 2O₃(g)
is 2.0 x 10^-50 at 25°C. If the equilibrium concentration of O₂ in air at 25°C is 1.6 x 10^-2, what is the concentration of O₃?
Answer:
The given reaction is
3O₂(g) ⇌ 2O₃(g)
Then K
It is given that Kc = 2.0 x 10^-50 and [02(g)] = 1.6 x 10^-2
Then, we have,
\(2.0 \times 10^{-50}=\frac{\left[\mathrm{O}_{3}\right]^{2}}{\left[1.6 \times 10^{-2}\right]^{3}}\)
⇒ [O₃]² = 2.0 x 10^-50 x (1.6 x 10^-2)³
⇒ [O₃]² = 8.192 x 10^-56
⇒ [O₃] = 2.86 x 10^-28 M
Hence, the concentration of O₃ is 2.86 x 10^-28 M.

Question 34.
The reaction, CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H20 and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90.
Answer:
The given equation is
CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g)
Therefore,
\(\frac{\left[\mathrm{CH}_{4}\right]\left[\mathrm{H}_{2} \mathrm{O}\right]}{[\mathrm{CO}]\left[\mathrm{H}_{2}\right]^{3}}=K_{c}\)
Given, K_c = 3.90, [CO] = 0.30 mol, [H₂] = 0.10 mol and [H₂O] \(\frac{\left[\mathrm{CH}_{4}\right] \times 0.02}{0.3 \times(0.1)^{3}}\) = 3.90
[CH₄] = \(\frac{3.90 \times 0.3 \times(0.1)^{3}}{0.02}=\frac{0.00117}{0.02}\)
= 0.0585 M= 5.85 x 10^-2M
Hence, the concentration of CH4 at equilibrium is 5.85 x 10^-2 M.

Question 35.
What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species :
\(\mathrm{HNO}_{2}, \mathrm{CN}^{-}, \mathrm{HClO}_{4}, \mathrm{~F}^{-}, \mathrm{OH}^{-}, \mathrm{CO}_{3}^{2-} \text { and } \mathrm{S}^{2-}\)
Answe:
A conjugate acid-base pair is a pair that differs only by one proton.
The conjugate acid-base for the given species is mentioned in the table below:

Question 36.
Which of the followings are Lewis acids
\(\mathbf{H}_{2} \mathbf{O}, \mathbf{B F}_{3}, \mathrm{H}^{+} \text {and } \mathrm{NH}_{4}^{+}\)
Answer:
Lewis acids are those acids which can accept a pair of electrons. For example, BF₃, H⁺ and \(\mathrm{NH}_{4}^{+}\) are Lewis acids.

Question 37.
What will be the conjugate bases for the Bronsted acids : HF, H₂SO₄ and \(\mathrm{HCO}_{3}^{-}\)?
Answer:
The table below lists the conjugate bases for the given Bronsted acids :

Question 38.
Write the conjugate acids for the following Bronsted bases: \(\mathbf{N H}_{2}^{-}\), NH₃ andHCOC^–.
Answer:
The table below lists the conjugate acids for the given Bronsted bases : Bronsted base Conjugate acid

Question 39.
The species : H₂O, HCO^–₃, HSO^–₄ and NH₃ can act both as Bronsted acids and bases. For each case give the corresponding conjugate acid and base.
Answer:
The table below lists the conjugate acids and conjugate bases for the given species :

Question 40.
Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH^– (b)F^– (c)H⁺ (d) BCl₃
OH^– and F^– are electron rich species and can donate electron pair. Hence, these act as Lewis base.

H⁺ and BCl₃ are electron deficient species and can accept electron pair. Hence, these act as Lewis acid.

Question 41.
The concentration of hydrogen ion in a sample of soft drink is 3.8 x 10^-3 M. What is its pH?
Answer:
Given,
[H⁺] = 3.8 x 10^-3 M
∴ pH value of soft drink = – log[H⁺] = – log(3.8 x 10^-3)
= – log3.8 – log10^-3 = – log3.8 + 3 log10
= – log3.8 + 3
= -0.58 + 3
= 2.42

Question 42.
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it.
Answer:
Given, pH = 3.76
We know that,
pH = – log[H⁺]
⇒ log[H⁺] = -pH
⇒ [H⁺] = antilog (-pH)
= antilog (-3.76) -1 +1 = antilog \(\overline{4} .24\) = 1.74 x 10-4 M Hence, the concentration of hydrogen ion in the given sample of vinegar is 1.74 x 10^-4 M.

Question 43.
The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 x 10^-4, 1.8 x 10^-4 and 4 8 x 10^-9 respectively. Calculate the ionization constants of the corresponding conjugate base.
Answer:
If K_a is the ionization constant of a weak acid and Kb is the ionization constant of its conjugate base then K_a.K_b = K_w
or K_b = \(\frac{K_{w}}{K_{a}}\)
Given, K_a of HF = 6.8 x 10^-4
Hence, K_b of its conjugate base F^–
= \(\frac{K_{w}}{K_{a}}=\frac{1 \times 10^{-14}}{6.8 \times 10^{-4}}\)= 1.5 x 10^-11
(K_w = ionic product of water =1 x 10^-14 at 298 K)
Given, K_a of HCOOH = 1.8 x 10^-4
= \(\frac{K_{w}}{K_{a}}=\frac{1 \times 10^{-14}}{1.8 \times 10^{-4}}\) = 5.6 x 10^-11
Hence, K_b of its coagulate base CN^–
Given, K_a of HCN = 4.8 x 10^-9
Hence, K_b of its coagulate base HCOO^–
= \(\frac{K_{w}}{K_{a}}=\frac{1 \times 10^{-14}}{4.8 \times 10^{-9}}\) = 2.08 x 10^-6

Question 44.
The ionization constant of phenol is 1.0 x 10^-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01 M in sodium phenolate?
Answer:
Ionization of phenol :
C₆H₅OH + H₂O ⇌ C₆H₅O^– + H₃O⁺

Question 45.
The first ionization constant of H2S is 9.1 x 10^-8. (i) Calculate the concentration of HS^– ion in its 0.1 M solution. (ii) How will this concentration be affected if the solution is 0.1 M in HCI also? (ifi) If the second dissociation constant of H₂S is 1.2 x 10^-13, calculate the concentration of S^2- under both conditions.
Answer:

Hence, concentration of [HS^–] is decreased in the presence of 0.1 M
HCI due to common-ion effect.
(iii) For second dissociation constant,
HS + H₂O ⇌ H₃O⁺ + S^2- (In absence of HCl)
[HS^–] = 9.54 x 10^-5 M
\(K_{a_{2}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{S}^{2-}\right]}{\left[\mathrm{HS}^{-}\right]}\)

Question 46.
The ionization constant of acetic acid is 1.74 x 10^-5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.
Answer:
CH₃COOH CH₃COO^– + H⁺
K_a for CH₃COOH = 1.74 x 10^-5
[CH₃COOH] = c = 0.05 M
CH₃COOH CH₃COO^– + H⁺ [where α = degree of dissociation and c = molar concentration]

[CH₃COO^–] = 0.933 x 10^-3 = 9.33 x 10^-4 M
pH = – log[H⁺] = – log (9.33 x 10^-4)
= – (-4) – log9.3 = 4 – 0.9 = 3.03

Question 47.
It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_a.
Answer:
Let the organic acid be HA.
⇒ HA ⇌ H⁺ + A^–
Concentration of HA = 0.01 M
pH = 4.15
-log[H⁺] = pH= 4.15
log[H⁺] = – 4.15
log[H⁺] = 5.85
[H⁺] = antilog \(\overline{5} .85\)
= 7.080 x 10^-5
K_a = \(\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}\)
Now, [H⁺] = [A^–] = 7.08 x 10^-5 M
Then K_a = \(
K_a = 5.01 x 10^-7
PK_a = – logK_a = – log(5.01 x 10^-7)
pK_a = 7 – 0.699 = 6.301

Question 48.
Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH
Answer:
(a) HCl (aq) ⇌ H⁺ (aq) + Cl^–(aq)
[HCl]= 0.003 M
As HC1 is completely dissociated into H⁺ ions
∴ [H⁺] = [HCl] = 0.003 M
pH = – log[H⁺] = – log [3 x 10^-3]
= 3 + (-0.4771) = 2.523
(b) NaOH(aq) ⇌ Na⁺(aq) + OH^– (aq)
[NaOH] = 0.005 = 5 x 10^-3 M
[OH^–] = [NaOH] = 5 x 10^-3 M
∴ [latex]\left[\mathrm{H}^{+}\right]=\frac{K_{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{5.0 \times 10^{-3}}\)
[H⁺]= 2.0 x 10^-12
∴ pH = – log(2 x 10^-12) = – (-12) – log2
= 12 – 0.30 = 11.70
[log2 = 0.30]

(c)

[HBr] = 0.002 M
[H⁺]= [HBr] = 0.002 M= 2.0 x 10^-3 M
pH = – log[H⁺] = – log[2 x 10^-3]
=- (-3) – log2
= 3 – log2
= 3 – 0.3 = 2.70

(d)

[OH^–] = 0.002 M
[H⁺] = \(\frac{K_{w}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.0 \times 10^{-14}}{0.002}\) = 2 x 10^-12
pH = – log[H⁺] = -(-12) – log 5 = 12 – 0.70 = 11.30

Question 49.
Calculate the pH of the following solutions:
(a) 2 g of TIOH dissolved in water to give 2 litre of solution.
(b) 0.3 g of Ca(OH)₂ dissolved in water to give 500 niL of solution.
(c) 0.3 g of NaOH dissolved in water to give 200 mL of solution.
(d) 1 mL of 13.6 M HC1 is diluted with water to give 1 litre of solution.
Answer:

(d) 1 mL of 13.6 M HC1 is diluted with water to give 1 litre of solution HC1 is completely dissociated to give H+ ions
[HCl] = ?
M₁V₁ = M₂V₂
1 mL of 13.6 M HCl = 1000 mL of M₂
M₂ = \(\frac{1 \times 13.6}{1000}\) = 0.0136 M
[HC1] = [H+] = 0.0136 M pH = – log[H+] = – log(1.36 x 10-2)
= – (-2) – log 1.36 = 2 – 0.13 = 1.87

Question 50.
The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pK_a of bromoacetic acid.
Answer:
α (Degree of ionization) = 0.132
c (molar cone.) = 0.1 M

∴ H⁺ = c x α = 0.1 x 0.132 = 0.0132
pH = – logH⁺ = – log(1.32 x 10^-2)
= – (-2) – log 1.32 = 2 – 0.12 = 1.88
pK_a = -logK_a
Now, K_a = cα²
K_a = 0.1 x (0.132)² = 1.74 x 10^-3
∴ pK_a = – log (1.74 x 10^-3) = – (-3) – log1.74 = 3 – 0.24 = 2.76

Question 51.
The pH of 0.005 M codeine (C₁₈H₂₁NO₃) solution is 9.95.
Calculate its ionization constant and pK_b.
Answer:
Molar cone, of codeine, c = 0.005 = 5 x 10^-3
pH = 9.95
pOH = 14 – 9.95 = 4.05 (∵ pH + pOH = 14)
pOH = – log [OH^–]
log[OH^–] = -4.05= \(\overline{5} .95\)
[OH^–] = antilog \(\overline{5} .95\)
= 8.91 x 10-5
k_b = \(\left(\frac{8.91 \times 10^{-5}}{5 \times 10^{-3}}\right)^{2}\) = 1.588 x 10^-6
pK_b = – logK_b = – log(1.588 x 10^-6)
= 6 + (-0.2009) = 5.7991 = 5.80

Question 52.
What is the pH of 0.001 M aniline solution? The ionization constant of aniline can be taken from table 7.7 (427 x 10^-10). Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline.
Answer:
Given, K_b = 4.27 x 10^-10, c = 0.001 M
\(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-}\)

[0H] = 6.534 x 10^-7
pOH = — log(6.534 x 10^-7)
= 7+ (-0.8152)= 6.18
pH + pOH =14
pH 14—6.18=7.82

Thus, the ionization constant of the conjugate acid of aniline is 2.34 x 10^-5.

Question 53.
Calculate the degree of ionization of 0.05 M acetic acid if its pK_a value is 4.74.
How is the degree of dissociation affected when its solution also contains (a) 0.01 M (b) 0.1 M in HCl?
Answer:
PK_a = – log K_a,
4.74 =-logK_a
log K_a = -4.74 = \(\overline{5} .26\)
Ka = antilog \(\overline{5} .26\) = 1.82 x 10^-5

[CH₃COOH is a weak acid and HC1 is a strong acid, so we can assume
that (cα + 0.01) 0.01]

In the presence of strong acid, dissociation of weak acid i.e., CH₃COOH decreases due to common ion effect.

Q.54. The Ionization constant of dimethylanilne is 54 x 10.
Calculate Its degree of ionization in its 0.02 M solution. What
percentage of dimethylamine is ionized if the solution is also
0.1MInNaOH?
Ans. Given, Kb = 5.4 x 10
c=0.02M

It means that in the presence of 0.1 M NaOH, 0.54% of dimethylamine will get dissociated.

Question 55.
Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below :
(a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2
(c) Human blood, 7.38 (d) Human saliva, 6.4.
Answer:
(a) pH of Human muscle fluid = 6.83
pH = – log[H+] log[H⁺] = -6.83 = \(\overline{7} .17\)
[H⁺] = antilog \(\overline{7} .17\)
[H⁺] = 1.48 x 10^-7 M

(b) pH of Human stomach fluid =1.2
log[H⁺] = -1.2 = \(\overline{2} .80\)
[H⁺] = antilog \(\overline{2} .80\)
.-. [H⁺] = 6.3 x 10^-2 M

(c) pH of Human blood = 7.38
log[H⁺] = – 7.38 = \(\overline{8} .62\)
.-. [H⁺] = antilog \(\overline{8} .62\) = 4.17 x 10^-8 M

(d) pH of Human saliva = 6.4
log[H⁺] =-6.4 = \(\overline{7} .60\)
[H⁺] = antilog \(\overline{7} .60\) = 3.98 x 10^-7 M

Question 56.
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each.
Answer:
The hydrogen ion concentration in the given substances can be calculated by using the given relation: pH = – log[H⁺]
(i) pH of milk = 6.8
Since, pH = -log[H⁺]
6.8 = -log[H+] log[H⁺] = -6.8 = \(\overline{7} .20\)
[H+] = antilog (\(\overline{7} .20\)) = 1.5 x 10~7 M

(ii) pH of black coffee = 5.0
Since, pH = – log[H⁺]
5.0 = – log[H+] log[H⁺] = – 5.0
[H⁺] = antilog (-5.00) = 10-5 M

(iii) pH of tomato juice = 4.2
Since, pH = – log[H⁺]
4.2 = – log[H⁺]
log[H⁺] = – 4.2 = \(\overline{5} .80\)
[H⁺] = antilog (\(\overline{5} .80\)) = 6.31 x 10^-5M

(iv) pH of lemon juice = 2.2
Since, pH = – log[H⁺]
2.2 = – log[H+]
log[H⁺] = -2.2 = \(\overline{3} .8\)
[H⁺] – antilog (\(\overline{3} .8\)) = 6.31 x 10^-3 M

(v) pH of egg white = 7.8
Since, pH = -log[H⁺]
7.8 = – log[H⁺]
log[H⁺]= -7.8 = \(\overline{8} .20\)
[H⁺] = antilog (\(\overline{8} .20\)) = 1.58 x 10^-8 M

Question 57.
0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH?
Answer:
Molar cone, of KOH = \(\frac{0.561 \times 1000}{56.1 \times 200}\) = 0.05M
56.1 x 200
KOH being a strong electrolyte, is completely ionized in aqueous solution.
KOH(aq) ⇌ K⁺(aq) + OH^–(aq)
[OH^–] = 0.05 M = [K⁺]
[H⁺][OH^–] = k_w
[H⁺] = \(\) = 2 x 10^-13
pH = – log[H⁺] = – log[2 x 10^-13]
= – (-13) – log2 = 13 – 0.03
∴ pH = 12.70

Question 58.
The solubility of Sr(OH)₂ at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.
Answer:
Solubility of Sr(OH)₂ = 19.23 g/L
Then, concentration of Sr(OH)₂ = \(\frac{19.23}{121.63 \times 1}\) M = 0.1581 M
Sr(OH)₂(aq) Sr²⁺(aq) + 2(OH^–)(aq)
∴ [Sr²⁺] = 0.1581 M
[OH^–] – 2 x 0.1581 M = 0.3162 M
Now
K_w = [OH^–] [H⁺]
\(\frac{1 \times 10^{-14}}{0.3162}\) = [H⁺]
[H⁺] = 3.16 x 10^-14
pH = – log[H⁺]
pH = 14 – 0.4997 = 13.5003 ≈ 13.5

Question 60.
The pH of 0.1 M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution.
Answer:
Given, pH = 2.34
Molar cone, (c) = 0.1 M
HCNO H⁺ + CNO^–
pH = – log[H⁺]
2.34 = -log[H⁺]
log[H⁺] =-2.34 = \(\overline{3} .66\)
.-. [H⁺] = antilog \(\overline{3} .66\) = 4.57 x 10^-3 M
[H⁺] = \(\sqrt{K_{a}^{c}}\)
4.57 x 10^-3 = \(\sqrt{K_{a}^{c}}\)
Ionization constant,
K_a = 2.088 x 10^-4
Degree of ionization α = \(\sqrt{\frac{K_{a}}{c}}=\sqrt{\frac{2.088 \times 10^{-14}}{0.1}}\)
α = 0.0457

Question 61.
The ionization constant of nitrous acid is 45 x 104. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis.
Answer:
Hydrolysis constant K_h = \(\frac{K_{w}}{K_{a}}\)
where K_w = Ionic product of water, K_a = Ionisation constant of the acid

pOH = -log(9.42 x 10^-7)= 7-0.97= 6.03
∴ pH = 14 – pOH = 14 – 6.03 = 7.97

Question 62.
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
Answer:
Given, pH = 3.44
We know that,
PH = – log[H⁺]
.-. [H⁺]= 3.63 x 10^-4
Then K_h = \(\frac{\left(3.63 \times 10^{-4}\right)^{2}}{0.02}\) (Concentration = 0.02M)
=> K_h = 6.6 x 10^-6
Now, K_h = \(\frac{K_{w}}{K_{a}}\)
K_a = \(\frac{K_{w}}{K_{h}}=\frac{1 \times 10^{-14}}{6.6 \times 10^{6}}\)
= 1.51 x 10^-9

Question 63.
Predict if the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF
Answer:

Question 64.
The ionization constant of chloroacetic acid is 1.35 x 10^-3 . What will be the pH of 0.1 M acid and its 0.1 M sodium salt solution?
Answer:
Given that Ka = 1.35 x 10^-3.
=> Ka = cα²
α = \(\sqrt{\frac{K_{a}}{c}}=\sqrt{\frac{1.35 \times 10^{-3}}{0.1}}\)
(∵ Concentration of acid = 0.1 M)
= \(\sqrt{1.35 \times 10^{-2}}\) =0.116
.-. [H⁺]= cα = 0.1 x 0.116 = 0.0116
=> pH = – log[H⁺] = – log[0.0116] = 1.94
To find pH of 0.1 M sodium salt, we use the formula
pH = – \(\frac{1}{2}\)[log C_w + log K_a – log c]
= –\(\frac{1}{2}\)[log1 x 10^-14 + log(1.35 x 10^-3) – log(0.1)]
= –\(\frac{1}{2}\)[-14 + (-3 + 0.1303) – (-1)]
= – \(\frac{1}{2}\) [-15.8697] = 7.93485 ≈ 7.94

Question 65.
Ionic product of water at 310 K is 2.7 x 10^-14. What is the pH of neutral water at this temperature?
Answer:
Ionic product,
K_w = [H₃O⁺] [OH^–]
= 2.7 x 10^-14 at 310 K
H₂O + H₂O *=* [H₃0⁺][OH^–]
[H₃0⁺]= [OH^–]
Therefore, [H₃0⁺] = \(\sqrt{2.7 \times 10^{-14}}\)
⇒ = 1.64 x 10^-7 M
⇒ [H₃0⁺] = 1.64 x 10^-7
⇒ pH = – log[H₃0⁺] = – log[1.64 x 10^-7 = 6.78
Hence, the pH of neutral water is 6.78.

Question 66.
Calculate the pH of the resultant mixtures :
(a) 10 mL of 0.2 M Ca(OH)₂ + 25 mL of 0.1 M HC1
(b) 10 mL of 0.01 M H₂SO₄ + 10 mL of 0.01 M Ca(OH)₂
(c) 10 mL of 0.1 M H₂SO₄ + 10 mL of 0.1 M KOH
Answer:

Question 67.
Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298 K from their solubility product constants.
[K_sp(Ag₂CrO₄) = 1.1 x 10^-12, K_sp(BaCrO₄) = 1.2 x 10^-10,
K_sp[Fe(OH)₃] = 1.0 x 10^-38, K_sp(PbCl₂) = 1.6 x 10^-5
K_sp(Hg₂I₂) = 4.5 x 10^-29]
Determine also the molarities of individual ions.
Answer:
(i) Silver chromate : Ag₂CrO₄⇌ 2Ag⁺ + CrO₄^2-, K_sp = 1.1 x 10^-12
Then, K_sp = [Ag⁺]²[CrO₄^2-]
Let the solubility of Ag₂CrO₄ be s.
⇒ [Ag⁺] = 2s and [CrO₄^2-] = s
Then, K_sp = (2s)² s – 4s³
⇒ 1.1 x 10^-12 = 4s³
0.275 x 10^-12 = s³
s = 0.65 x 10^-4 M
Molarity of Ag⁺ = 2s = 2 x 0.65 x 10^-4
= 1.30 x 10^-4 M
Molarity of CrO₄^2- = s = 0.65 x 10^-4 M

(ii) Barium chromate : BaCrO₄ ⇌ Ba²⁺ + \(\mathrm{CrO}_{4}^{2-}\); K_sp = 1.2 x 10^-10
Then, K_sp = [Ba²⁺] [latex]\mathrm{CrO}_{4}^{2-}[/latex]
Let s be the solubility of BaCrO₄.
⇒ [Ba²⁺] = s and [latex]\mathrm{CrO}_{4}^{2-}[/latex] = s
⇒ K_sp = s²
⇒ 1.2 x 10^-10 = s²
⇒ s = 1.09 x 10^-5 M
Molarity of [Ba²⁺] = Molarity of [latex]\mathrm{CrO}_{4}^{2-}[/latex] = s = 1.09 x 10^-5 M

(iii) Ferric hydroxide: Fe(OH)₃ ⇌ Fe²⁺ + 3OH^–; K_sp = 1.0.x 10^-38

K_sp = [Fe²⁺][OH^–]³
Let s be the solubility of Fe(OH)₃
⇒ [Fe³⁺] = s and [OH^–] = 3s
⇒ K_sp = s. (3s)³ = s x 27x³
K_sp = 27x⁴
1.0 x 10^-38 = 27x⁴
0.037 x 10^-38 = s⁴
0.00037 x 10^-36 = s⁴
s = 1.39 x 10^-10 M
Molarity of [Fe³⁺] = s = 1.39 x 10^-10 M
Molarity of [OH^–] = 3s = 4.17 x 10^-10 M

(iv) Lead chloride : PbCl₂ ⇌ Pb²⁺ + 2Cl^–; K_sp = 1.6 x 10^-5
K_sp = [Pb²⁺][Cl^–]²
Let s be the solubility of PbCl₂.
⇒ [Pb²⁺] = s and [Cl^–] = 2s
Thus, K_sp = s. (2s)² = 4s³
⇒ 1.6 x 10^-5 = 4s³
⇒ 0.4 x 10^-5 = s³
4 x 10^-6 = s³
⇒ s = 1.59 x 10^-2 M
Molarity of [Pb²⁺] = s = 1.59 x 10^-2 M
Molarity of [Cl^–] = 2s = 3.18 x 10^-2 M

(v) Mercurous iodide : Hg₂I₂ ⇌ \(\mathrm{Hg}_{2}^{2+}\) + = 4.5 x 10^-29
K_sp = [\(\mathrm{Hg}_{2}^{2+}\) ] []I^–]
Let s be the solubility of [Hg²I²]
⇒ [Hg₂] = s and [I^–] = 2s
K_sp = (s).(2s)² = 4s³
⇒ 4s³ = 4.5 x 10^-29
⇒ s³ = 1.125 x 10^-29
s = 2.24 x 10^-10 M
Molarity of [ \(\mathrm{Hg}_{2}^{2+}\) ] = s = 2.24 x 10^-10 M
Molarity of [I^–] = 2s = 4.48 x 10^-10 M.

Question 68.
The solubility product constant of Ag₂CrO₄ and AgBr are
1.1 x 10^-12 and 5.0 x 10^-13 respectively. Calculate the ratio of the molarities of their saturated solutions.
Answer:
Let s be the solubility of Ag₂CrO₄
Thus, Ag₂CrO₄ ⇌ 2Ag²⁺ + \(\mathrm{CrO}_{4}^{-}\); K_sp = 1.1 x 10^-12
K_sp = [Ag²⁺]². [latex]\mathrm{CrO}_{4}^{-}[/latex]
=> [Ag²⁺] = (2s)² and [latex]\mathrm{CrO}_{4}^{-}[/latex] = s
K_sp = (2s)². s= 4s³
1.1 x 10^-12 = 4s³
s = 6.5 x 10^-5 M

Let s be the solubility of AgBr.
AgBr(s) ⇌ Ag⁺ + Br^–; K_sp = 5.0 x 10^-13
K_sp = s² = 5.0 x 10^-13
s = \(\sqrt{5.0 \times 10^{-13}}\)
∴ s = 7.07 x 10^-7 M

Therefore, the ratio of the molarities of their saturated solution is
\(\frac{s\left(\mathrm{Ag}_{2} \mathrm{CrO}_{4}\right)}{s(\mathrm{AgBr})}=\frac{6.5 \times 10^{-5} \mathrm{M}}{7.07 \times 10^{-7} \mathrm{M}}\) = 9.19

Question 69.
Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate, K_sp = 7.4 x 10^-8).
Ans. 2NaIO₃ + Cu(ClO₃)3 → 2NaClO₃ + Cu(IO₃)₂
Molar cone, of both solutions before mixing = 0.002 M
Molar cone, of both solution after mixing
\(\left[\mathrm{IO}_{3}^{-}\right]=\left[\mathrm{Cu}^{2+}\right]=\frac{0.002}{2}\)= 0.001 M
Cu(IO₃)₂ ⇌ Cu²⁺ + \(2 \mathrm{IO}_{3}^{-}\)
[Cu²⁺] = 0.001 M
\(\) = 0.001 M
Ionic product = [Cu²⁺]\(\left[\mathrm{IO}_{3}^{-}\right]^{2}\)
– 1 x 10^-3 x [1 x 10^-3]² = 1 x 10^-9
K_sp = 7.4 x 10^-8
Cu(IO₃)₂ is precipitated if [Cu²⁺] .\(\left[\mathrm{IO}_{3}^{-}\right]^{2}\) > K_sp
Since, the ionic product is less than the solubility product. Hence there will be no precipitation.

Question 70.
The ionization constant of benzoic acid is 6.46 x 10^-5 and Ksp for silver benzoate is 2.5 x 10 . How many times is silver benzoate
more soluble in a buffer of pH 3.19 compared to its solubility in pure water?
Answer:

where s is the solubility of C₆H₅COOAg
pH = 3.19
pH = -log[H⁺]
log[H⁺] = – pH = -3.19 = \(\overline{4} .81\)
[H⁺] = antilog \(\overline{4} .81\) = 6.46 x 10^-4

C₆H₅COOAg is 3.2 times more soluble in buffer than in pure water.

Punjab State Board PSEB 12th Class Maths Book Solutions Chapter 3 Matrices Ex 3.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4

Direction (1 – 17) Using elementary transformations, find the inverse of each of the matrices.

Question 1.
\(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{cc}
1 & -1 \\
2 & 3
\end{array}\right]\)
We know that A = IA

Question 2.
\(\left[\begin{array}{ll}
2 & 1 \\
1 & 1
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ll}
2 & 1 \\
1 & 1
\end{array}\right]\)

We know that A = IA

Question 3.
\(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ll}
1 & 3 \\
2 & 7
\end{array}\right]\)
We know that A = IA

Question 4.
\(\left[\begin{array}{ll}
2 & 3 \\
5 & 7
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ll}
2 & 3 \\
5 & 7
\end{array}\right]\)
We know that A = IA

Question 5.
\(\left[\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right]\)
Solution.
\(\left[\begin{array}{ll}
2 & 1 \\
7 & 4
\end{array}\right]\)
We know that A = IA

Question 6.
\(\left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]\)
Solution.
Let A = \left[\begin{array}{ll}
2 & 5 \\
1 & 3
\end{array}\right]
We know that A = IA

Question 7.
\(\left[\begin{array}{ll}
\mathbf{3} & 1 \\
5 & 2
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ll}
\mathbf{3} & 1 \\
5 & 2
\end{array}\right]\)

We know that A = IA

Question 8.
\(\left[\begin{array}{ll}
4 & 5 \\
3 & 4
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ll}
4 & 5 \\
3 & 4
\end{array}\right]\)
We know that A = IA

Question 9.
\(\left[\begin{array}{cc}
3 & 10 \\
2 & 7
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{cc}
3 & 10 \\
2 & 7
\end{array}\right]\)
We know that A = IA

Question 10.
\(\left[\begin{array}{cc}
3 & -1 \\
-4 & 2
\end{array}\right]\)
Solution.
\(\left[\begin{array}{cc}
3 & -1 \\
-4 & 2
\end{array}\right]\)
We know that A = IA.

Question 11.
\(\left[\begin{array}{rr}
2 & -6 \\
1 & -2
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{rr}
2 & -6 \\
1 & -2
\end{array}\right]\)
We know that A = IA.

Question 12.
\(\left[\begin{array}{cc}
6 & -3 \\
-2 & 1
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{cc}
6 & -3 \\
-2 & 1
\end{array}\right]\)
We know that A = IA.

Now, in the above equation, we can see all the zeroes in the second law of the matrix on the L.H.S.
Threrefore, A^-1 does not exist.

Question 13.
\(\left[\begin{array}{cc}
\mathbf{2} & -\mathbf{3} \\
-1 & 2
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{cc}
\mathbf{2} & -\mathbf{3} \\
-1 & 2
\end{array}\right]\)

We know that A = IA.

⇒ \(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right] A\) (applying R₁ → R₁ + R₂)

∴ A^-1 = \(\left[\begin{array}{ll}
2 & 3 \\
1 & 2
\end{array}\right]\)

Question 14.
\(\left[\begin{array}{ll}
2 & 1 \\
4 & 2
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ll}
2 & 1 \\
4 & 2
\end{array}\right]\)
We know that A = IA
∴ \(\left[\begin{array}{ll}
2 & 1 \\
4 & 2
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] A\)

Applying R₁ → R₁ – \(\frac{1}{2}\) R₂, we have
\(\left[\begin{array}{ll}
0 & 0 \\
4 & 2
\end{array}\right]=\left[\begin{array}{rr}
1 & -\frac{1}{2} \\
0 & 1
\end{array}\right] A\)

Now, in the above equation, we can see all the zeroes in the first row of the matrix on the L.H.S.
Therefore, A^-1 does not exist.

Question 15.
\(\left[\begin{array}{rrr}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right]\)
Solution.
let A = \(\left[\begin{array}{rrr}
2 & -3 & 3 \\
2 & 2 & 3 \\
3 & -2 & 2
\end{array}\right]\)
We know that A = IA

Question 16.
\(\left[\begin{array}{ccc}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ccc}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]\)
We know that A = IA

Question 17.
\(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
Solution.
Let A = \(\left[\begin{array}{ccc}
2 & 0 & -1 \\
5 & 1 & 0 \\
0 & 1 & 3
\end{array}\right]\)
We know that A = IA

Question 18.
Matrices A and B will be inverse of each other only if
(A) AB = BA
(B) AB = BA = 0
(C) AB = 0,BA = I
(D) AB = BA = I
Solution.
We know that if A is a square matrix of order m, and if there exists another square matrix B of the same order m, such that AB = BA = I, then B is said to be the inverse of A.
In this case, it is clear that A is the inverse of B.
Thus, matrices A and B will be inverse of each other only if AB = BA = I. Hence, the correct answer is (D).