Electrochemistry Numerical Problems

Class 12 Physical Chemistry

Comprehensive numerical problems with step-by-step solutions for Willer Academy students

Willer Academy - Excellence in Chemistry Education

1

Cell Constant Calculation

The resistance of a conductivity cell with 0.01 M KCl solution is 210 Ω at 298 K. If the conductivity of 0.01 M KCl is 0.141 S m⁻¹, find the cell constant.

298 K पर 0.01 M KCl विलयन वाले चालकता सेल का प्रतिरोध 210 Ω है। यदि 0.01 M KCl की चालकता 0.141 S m⁻¹ है, तो सेल स्थिरांक ज्ञात कीजिए।

Formula Hint

Cell constant (G*) = Conductivity (κ) × Resistance (R)

1

Use the formula: Cell constant = Conductivity × Resistance

2

Substitute values: G* = 0.141 S m⁻¹ × 210 Ω

3

Calculate: G* = 29.61 m⁻¹

Final Answer: 29.61 m⁻¹

उत्तर: 29.61 m⁻¹

2

Molar Conductivity

The conductivity of 0.02 M KCl solution is 2.48 × 10⁻² S cm⁻¹. Calculate its molar conductivity.

0.02 M KCl विलयन की चालकता 2.48 × 10⁻² S cm⁻¹ है। इसकी मोलर चालकता की गणना कीजिए।

Formula Hint

Λ_m = (κ × 1000) / C

Where κ is in S cm⁻¹, C is in mol L⁻¹

1

Apply the formula: Molar conductivity = (Conductivity × 1000) / Concentration

2

Substitute values: Λ_m = (2.48 × 10⁻² × 1000) / 0.02

3

Calculate: Λ_m = 1240 S cm² mol⁻¹

Final Answer: 1240 S cm² mol⁻¹

उत्तर: 1240 S cm² mol⁻¹

3

Degree of Dissociation

The molar conductivity of 0.1 M acetic acid is 5.2 S cm² mol⁻¹. Given Λ_m° = 390.5 S cm² mol⁻¹, find its degree of dissociation.

0.1 M एसिटिक अम्ल की मोलर चालकता 5.2 S cm² mol⁻¹ है। यदि Λ_m° = 390.5 S cm² mol⁻¹ है, तो वियोजन की मात्रा ज्ञात कीजिए।

Formula Hint

α = Λ_m / Λ_m°

1

Use the formula: Degree of dissociation (α) = Molar conductivity / Limiting molar conductivity

2

Substitute values: α = 5.2 / 390.5

3

Calculate: α = 0.0133

Final Answer: 0.0133

उत्तर: 0.0133

4

Nernst Equation

Calculate EMF for: Zn | Zn²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu. Given E°_Zn²⁺/Zn = -0.76 V, E°_Cu²⁺/Cu = +0.34 V.

निम्न सेल का EMF ज्ञात कीजिए: Zn | Zn²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu. दिया है: E°_Zn²⁺/Zn = -0.76 V, E°_Cu²⁺/Cu = +0.34 V

Formula Hint

E = E° - (0.059/n) log([Products]/[Reactants])

1

Calculate E°_cell = E°_cathode - E°_anode = 0.34 - (-0.76) = 1.10 V

2

Apply Nernst equation (n = 2): E = 1.10 - (0.059/2) log([Zn²⁺]/[Cu²⁺])

3

Substitute values: E = 1.10 - 0.0295 log(0.1/0.01) = 1.10 - 0.0295 × 1 = 1.0705 V

Final Answer: 1.0705 V

उत्तर: 1.0705 V

5

Faraday's First Law

Calculate the mass of Ag deposited at the cathode when 0.5 A current flows for 10 minutes through AgNO₃ solution. (Molar mass of Ag = 108 g mol⁻¹)

जब AgNO₃ विलयन में 0.5 A धारा 10 मिनट तक प्रवाहित होती है, तो कैथोड पर निक्षेपित Ag का द्रव्यमान ज्ञात कीजिए। (Ag का मोलर द्रव्यमान = 108 g mol⁻¹)

Formula Hint

Mass = (I × t × M) / (n × F)

Where F = 96500 C mol⁻¹

1

Convert time to seconds: t = 10 × 60 = 600 s

2

Apply formula (n = 1 for Ag⁺ → Ag): Mass = (0.5 × 600 × 108) / (1 × 96500)

3

Calculate: Mass = 32400 / 96500 = 0.3358 g

Final Answer: 0.336 g

उत्तर: 0.336 g

6

Conductivity from Resistance

A conductivity cell has a cell constant of 0.5 cm⁻¹. Its resistance with 0.01 M NaOH is 200 Ω. Calculate κ.

एक चालकता सेल का सेल स्थिरांक 0.5 cm⁻¹ है। 0.01 M NaOH के साथ इसका प्रतिरोध 200 Ω है। κ ज्ञात कीजिए।

Formula Hint

κ = Cell constant / Resistance

1

Use the formula: Conductivity (κ) = Cell constant / Resistance

2

Substitute values: κ = 0.5 / 200

3

Calculate: κ = 0.0025 S cm⁻¹

Final Answer: 0.0025 S cm⁻¹

उत्तर: 0.0025 S cm⁻¹

7

Equilibrium Constant

Calculate K for: Cu + 2Ag⁺ ⇌ Cu²⁺ + 2Ag. Given E°_Ag⁺/Ag = 0.80 V, E°_Cu²⁺/Cu = 0.34 V.

निम्न अभिक्रिया के लिए K ज्ञात कीजिए: Cu + 2Ag⁺ ⇌ Cu²⁺ + 2Ag. दिया है: E°_Ag⁺/Ag = 0.80 V, E°_Cu²⁺/Cu = 0.34 V

Formula Hint

log K = (n × E°) / 0.059

1

Calculate E°_cell = E°_cathode - E°_anode = 0.80 - 0.34 = 0.46 V

2

Apply formula (n = 2): log K = (2 × 0.46) / 0.059 = 15.593

3

Calculate K = 10^15.593 = 3.91 × 10¹⁵

Final Answer: 3.91 × 10¹⁵

उत्तर: 3.91 × 10¹⁵

8

Kohlrausch's Law

The limiting molar conductivity of NaCl, HCl, and NaAc are 126.4, 425.9, and 91.0 S cm² mol⁻¹. Find Λ_m° for HAc.

NaCl, HCl, और NaAc की सीमांत मोलर चालकताएँ क्रमशः 126.4, 425.9, और 91.0 S cm² mol⁻¹ हैं। HAc के लिए Λ_m° ज्ञात कीजिए।

Formula Hint

Λ_m°(HAc) = Λ_m°(HCl) + Λ_m°(NaAc) - Λ_m°(NaCl)

1

Apply Kohlrausch's law for acetic acid:

2

Λ_m°(HAc) = Λ_m°(H⁺) + Λ_m°(Ac⁻) = [Λ_m°(HCl) + Λ_m°(NaAc) - Λ_m°(NaCl)]

3

Substitute values: Λ_m°(HAc) = 425.9 + 91.0 - 126.4 = 390.5 S cm² mol⁻¹

Final Answer: 390.5 S cm² mol⁻¹

उत्तर: 390.5 S cm² mol⁻¹

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© 2023 Physical Chemistry Numerical Problems | Class 12

Designed for Excellence in Science Education

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1

Rate Constant from Half-life

The half-life of a first-order reaction is 30 minutes. Calculate the rate constant.

एक प्रथम कोटि अभिक्रिया का अर्ध-जीवन काल 30 मिनट है। दर स्थिरांक की गणना कीजिए।

Formula Hint

t_1/2 = 0.693 / k

1

For a first-order reaction, the half-life (t_1/2) is given by:

t_1/2 = 0.693 / k

2

Rearrange the formula to solve for k:

k = 0.693 / t_1/2

3

Substitute the given value: t_1/2 = 30 minutes

k = 0.693 / 30

Final Answer: 0.0231 min⁻¹

उत्तर: 0.0231 min⁻¹

2

First-order Reaction Time

For a first-order reaction, calculate the time required for 99% completion. (Given: k = 0.001 s⁻¹)

एक प्रथम कोटि अभिक्रिया के लिए, 99% पूर्णता के लिए आवश्यक समय की गणना कीजिए। (दिया है: k = 0.001 s⁻¹)

Formula Hint

t = (2.303/k) log(a/(a-x))

1

For 99% completion, if initial concentration a = 100%, then remaining concentration (a-x) = 1%

2

Use the first-order kinetics equation:

t = (2.303/k) log(a/(a-x))

3

Substitute values: a = 100, a-x = 1, k = 0.001 s⁻¹

t = (2.303/0.001) log(100/1) = 2303 × 2 = 4606 seconds

Final Answer: 4606 seconds

उत्तर: 4606 सेकंड

3

Second-order Reaction

For a second-order reaction, the rate constant is 0.002 M⁻¹s⁻¹. Calculate the half-life when the initial concentration is 0.5 M.

एक द्वितीय कोटि अभिक्रिया के लिए, दर स्थिरांक 0.002 M⁻¹s⁻¹ है। जब प्रारंभिक सांद्रता 0.5 M है तो अर्ध-जीवन काल की गणना कीजिए।

Formula Hint

t_1/2 = 1 / (k × [A]₀)

1

For a second-order reaction, the half-life is given by:

t_1/2 = 1 / (k × [A]₀)

2

Where k is the rate constant and [A]₀ is the initial concentration

3

Substitute the given values: k = 0.002 M⁻¹s⁻¹, [A]₀ = 0.5 M

t_1/2 = 1 / (0.002 × 0.5) = 1 / 0.001 = 1000 seconds

Final Answer: 1000 seconds

उत्तर: 1000 सेकंड

4

Activation Energy

The rate constant of a reaction is 0.005 s⁻¹ at 300 K and 0.04 s⁻¹ at 320 K. Calculate activation energy.

एक अभिक्रिया का दर स्थिरांक 300 K पर 0.005 s⁻¹ और 320 K पर 0.04 s⁻¹ है। सक्रियण ऊर्जा की गणना कीजिए।

Formula Hint

log(k₂/k₁) = (Eₐ/2.303R) × (1/T₁ - 1/T₂)

1

Use the Arrhenius equation:

log(k₂/k₁) = (Eₐ/2.303R) × (1/T₁ - 1/T₂)

2

Given: k₁ = 0.005 s⁻¹ at T₁ = 300 K, k₂ = 0.04 s⁻¹ at T₂ = 320 K, R = 8.314 J/mol·K

3

Calculate log(k₂/k₁) = log(0.04/0.005) = log(8) = 0.9030

Calculate (1/T₁ - 1/T₂) = (1/300 - 1/320) = (320-300)/(300×320) = 20/96000 = 2.083×10⁻⁴

4

Substitute values: 0.9030 = (Eₐ/(2.303×8.314)) × 2.083×10⁻⁴

Solve for Eₐ: Eₐ = (0.9030 × 2.303 × 8.314) / 2.083×10⁻⁴ = 82900 J/mol = 82.9 kJ/mol

Final Answer: 82.9 kJ/mol

उत्तर: 82.9 kJ/mol

5

Average Rate Calculation

The concentration of a reactant decreases from 0.2 M to 0.1 M in 10 minutes. Calculate the average rate.

एक अभिकारक की सांद्रता 10 मिनट में 0.2 M से 0.1 M हो जाती है। औसत दर की गणना कीजिए।

Formula Hint

Average rate = -Δ[Reactant]/Δt

1

Average rate is defined as the change in concentration per unit time

2

Change in concentration Δ[Reactant] = Final concentration - Initial concentration

Δ[Reactant] = 0.1 M - 0.2 M = -0.1 M

3

Time interval Δt = 10 minutes

4

Average rate = -(-0.1 M) / 10 min = 0.1 M / 10 min = 0.01 M min⁻¹

Final Answer: 0.01 M min⁻¹

उत्तर: 0.01 M min⁻¹

6

Rate Law Determination

For the reaction A + B → Products, the following data was obtained:

[A] (M)	[B] (M)	Rate (M/s)
0.1	0.1	0.005
0.2	0.1	0.01
0.1	0.2	0.005

Find the order with respect to A and B.

अभिक्रिया A + B → उत्पाद के लिए, निम्नलिखित आँकड़े प्राप्त हुए:

[A] (M)	[B] (M)	दर (M/s)
0.1	0.1	0.005
0.2	0.1	0.01
0.1	0.2	0.005

A और B के सापेक्ष कोटि ज्ञात कीजिए।

Formula Hint

Rate = k [A]^m [B]ⁿ

1

Compare experiments 1 and 2: When [A] doubles (0.1→0.2) and [B] is constant, rate doubles (0.005→0.01)

Therefore, rate ∝ [A] → order with respect to A is 1

2

Compare experiments 1 and 3: When [B] doubles (0.1→0.2) and [A] is constant, rate remains the same (0.005)

Therefore, rate is independent of [B] → order with respect to B is 0

3

The rate law is: Rate = k [A]¹ [B]⁰ = k [A]

Final Answer: Order w.r.t. A = 1, w.r.t. B = 0

उत्तर: A के सापेक्ष कोटि = 1, B के सापेक्ष कोटि = 0

7

First-order Rate Constant

For a first-order reaction, the concentration reduces to 25% in 50 minutes. Calculate the rate constant.

एक प्रथम कोटि अभिक्रिया में, सांद्रता 50 मिनट में 25% रह जाती है। दर स्थिरांक की गणना कीजिए।

Formula Hint

k = (2.303/t) log(a/(a-x))

1

For a first-order reaction, the rate constant is given by:

k = (2.303/t) log(a/(a-x))

2

Here, initial concentration a = 100%, remaining concentration (a-x) = 25%

Time t = 50 minutes

3

Substitute values: k = (2.303/50) log(100/25) = (0.04606) log(4)

log(4) = 0.6021, so k = 0.04606 × 0.6021 = 0.0277 min⁻¹

Final Answer: 0.0277 min⁻¹

उत्तर: 0.0277 min⁻¹

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© 2023 Physical Chemistry Numerical Problems | Class 12

Designed for Excellence in Science Education

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1

Molarity Calculation

Calculate the molarity of a solution containing 5.85 g of NaCl in 500 mL of solution.

(Molar mass of NaCl = 58.5 g/mol)

500 mL विलयन में 5.85 g NaCl युक्त विलयन की मोलरता की गणना कीजिए।

(NaCl का मोलर द्रव्यमान = 58.5 g/mol)

Formula Hint

Molarity (M) = moles of solute / volume of solution (in L)

1

Calculate moles of NaCl:

Moles = mass / molar mass = 5.85 g / 58.5 g/mol = 0.1 mol

2

Convert volume to liters:

Volume = 500 mL = 0.5 L

3

Calculate molarity:

M = moles / volume = 0.1 mol / 0.5 L = 0.2 M

Final Answer: 0.2 M

उत्तर: 0.2 M

2

Molality Calculation

Calculate the molality of a solution containing 18 g of glucose (C₆H₁₂O₆) in 500 g of water.

(Molar mass of glucose = 180 g/mol)

500 g जल में 18 g ग्लूकोज (C₆H₁₂O₆) युक्त विलयन की मोललता की गणना कीजिए।

(ग्लूकोज का मोलर द्रव्यमान = 180 g/mol)

Formula Hint

Molality (m) = moles of solute / mass of solvent (in kg)

1

Calculate moles of glucose:

Moles = mass / molar mass = 18 g / 180 g/mol = 0.1 mol

2

Convert solvent mass to kg:

Mass of water = 500 g = 0.5 kg

3

Calculate molality:

m = moles / mass of solvent = 0.1 mol / 0.5 kg = 0.2 m

Final Answer: 0.2 m

उत्तर: 0.2 m

3

Mole Fraction

Calculate the mole fraction of glucose in a solution containing 18 g of glucose and 90 g of water.

(Molar mass: glucose = 180 g/mol, water = 18 g/mol)

18 g ग्लूकोज और 90 g जल युक्त विलयन में ग्लूकोज का मोल अंश की गणना कीजिए।

(मोलर द्रव्यमान: ग्लूकोज = 180 g/mol, जल = 18 g/mol)

Formula Hint

Mole fraction (χ) = n_solute / (n_solute + n_solvent)

1

Calculate moles of glucose:

n_glucose = 18 / 180 = 0.1 mol

2

Calculate moles of water:

n_water = 90 / 18 = 5 mol

3

Calculate mole fraction of glucose:

χ_glucose = 0.1 / (0.1 + 5) = 0.1 / 5.1 ≈ 0.0196

Final Answer: 0.0196

उत्तर: 0.0196

4

Mass Percentage

Calculate the mass percentage of a solution containing 25 g of sugar in 225 g of water.

225 g जल में 25 g शक्कर युक्त विलयन का द्रव्यमान प्रतिशत की गणना कीजिए।

Formula Hint

Mass % = (mass of solute / total mass of solution) × 100

1

Calculate total mass of solution:

Total mass = mass of solute + mass of solvent = 25 g + 225 g = 250 g

2

Calculate mass percentage:

Mass % = (25 / 250) × 100 = 10%

Final Answer: 10%

उत्तर: 10%

5

Henry's Law

The solubility of CO₂ in water at 25°C is 0.145 g/100 mL when partial pressure is 0.5 atm. Calculate the solubility at 1.2 atm pressure.

25°C पर जल में CO₂ की विलेयता 0.145 g/100 mL है जब आंशिक दाब 0.5 atm है। 1.2 atm दाब पर विलेयता की गणना कीजिए।

Formula Hint

S = K_H × P

Henry's Law: Solubility proportional to partial pressure

1

According to Henry's Law:

S ∝ P ⇒ S₁/P₁ = S₂/P₂

2

Given: S₁ = 0.145 g/100 mL, P₁ = 0.5 atm, P₂ = 1.2 atm

3

Calculate S₂:

S₂ = (S₁ × P₂) / P₁ = (0.145 × 1.2) / 0.5 = 0.348 g/100 mL

Final Answer: 0.348 g/100 mL

उत्तर: 0.348 g/100 mL

6

Raoult's Law

The vapor pressure of pure water at 25°C is 23.8 mmHg. Calculate the vapor pressure of a solution containing 18 g glucose (C₆H₁₂O₆) in 90 g water.

(Molar mass: glucose = 180 g/mol, water = 18 g/mol)

25°C पर शुद्ध जल का वाष्प दाब 23.8 mmHg है। 90 g जल में 18 g ग्लूकोज (C₆H₁₂O₆) युक्त विलयन का वाष्प दाब ज्ञात कीजिए।

(मोलर द्रव्यमान: ग्लूकोज = 180 g/mol, जल = 18 g/mol)

Formula Hint

P = P° × χ_solvent

Raoult's Law for non-volatile solute

1

Calculate moles of water:

n_water = 90 / 18 = 5 mol

2

Calculate moles of glucose:

n_glucose = 18 / 180 = 0.1 mol

3

Calculate mole fraction of water:

χ_water = n_water / (n_water + n_glucose) = 5 / (5 + 0.1) = 5/5.1 ≈ 0.9804

4

Calculate vapor pressure:

P = P° × χ_water = 23.8 × 0.9804 ≈ 23.33 mmHg

Final Answer: 23.33 mmHg

उत्तर: 23.33 mmHg

7

Boiling Point Elevation

Calculate the boiling point of a solution containing 18 g glucose (C₆H₁₂O₆) in 500 g water. (K_b for water = 0.512 K kg mol⁻¹, boiling point of water = 373 K)

500 g जल में 18 g ग्लूकोज (C₆H₁₂O₆) युक्त विलयन के क्वथनांक की गणना कीजिए। (जल के लिए K_b = 0.512 K kg mol⁻¹, जल का क्वथनांक = 373 K)

Formula Hint

ΔT_b = i × K_b × m

For glucose, i = 1 (non-electrolyte)

1

Calculate molality:

m = (moles of solute) / (kg solvent)

Moles glucose = 18 / 180 = 0.1 mol

Mass solvent = 500 g = 0.5 kg

m = 0.1 / 0.5 = 0.2 m

2

Calculate boiling point elevation:

ΔT_b = i × K_b × m = 1 × 0.512 × 0.2 = 0.1024 K

3

Calculate boiling point:

T_b = T°_b + ΔT_b = 373 + 0.1024 = 373.1024 K

Final Answer: 373.1024 K

उत्तर: 373.1024 K

8

Freezing Point Depression

Calculate the freezing point of a solution containing 5.85 g NaCl in 500 g water. (K_f for water = 1.86 K kg mol⁻¹, freezing point of water = 273 K, Molar mass NaCl = 58.5 g/mol)

500 g जल में 5.85 g NaCl युक्त विलयन के हिमांक की गणना कीजिए। (जल के लिए K_f = 1.86 K kg mol⁻¹, जल का हिमांक = 273 K, NaCl का मोलर द्रव्यमान = 58.5 g/mol)

Formula Hint

ΔT_f = i × K_f × m

For NaCl, i = 2 (van't Hoff factor)

1

Calculate moles of NaCl:

Moles = 5.85 / 58.5 = 0.1 mol

2

Calculate molality:

m = moles / kg solvent = 0.1 / 0.5 = 0.2 m

3

Calculate freezing point depression:

ΔT_f = i × K_f × m = 2 × 1.86 × 0.2 = 0.744 K

4

Calculate freezing point:

T_f = T°_f - ΔT_f = 273 - 0.744 = 272.256 K

Final Answer: 272.256 K

उत्तर: 272.256 K

Willer Academy - Chemistry Department

© 2023 Physical Chemistry Numerical Problems | Class 12

Designed for Excellence in Science Education

Surface Chemistry Numerical Problems

Class 12 Physical Chemistry

Comprehensive numerical problems with step-by-step solutions for Willer Academy students

Willer Academy - Excellence in Chemistry Education

Electrochemistry

15 Problems

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Solutions

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1

Adsorption Isotherm

In an adsorption experiment, 1 g of activated charcoal adsorbs 100 mL of 0.5 M acetic acid solution to form a monolayer. Calculate the surface area of the charcoal if each acetic acid molecule occupies 16.2 Ų.

एक अधिशोषण प्रयोग में, 1 g सक्रिय चारकोल 0.5 M एसिटिक अम्ल विलयन के 100 mL को एक परत के रूप में अधिशोषित करता है। चारकोल का पृष्ठीय क्षेत्रफल ज्ञात कीजिए यदि प्रत्येक एसिटिक अम्ल अणु 16.2 Ų स्थान घेरता है।

Formula Hint

Surface area = (Number of molecules adsorbed) × (Area per molecule)

1

Calculate moles of acetic acid adsorbed:

Moles = Molarity × Volume(L) = 0.5 mol/L × 0.1 L = 0.05 mol

2

Calculate number of molecules:

Number = Moles × Avogadro's number = 0.05 × 6.022 × 10²³ = 3.011 × 10²² molecules

3

Calculate surface area:

Surface area = 3.011 × 10²² × 16.2 × 10⁻²⁰ m² = 4.878 × 10³ m²

(1 Ų = 10⁻²⁰ m²)

Final Answer: 487.8 m²/g

उत्तर: 487.8 m²/g

2

Freundlich Isotherm

For the adsorption of a gas on charcoal, the Freundlich constant K = 4.0 and 1/n = 0.5. Calculate the amount of gas adsorbed per gram of charcoal at a pressure of 4 atm.

चारकोल पर गैस के अधिशोषण के लिए, फ्रेंडलिच स्थिरांक K = 4.0 और 1/n = 0.5 है। 4 atm दाब पर प्रति ग्राम चारकोल द्वारा अधिशोषित गैस की मात्रा की गणना कीजिए।

Formula Hint

x/m = K P^1/n

1

Use Freundlich adsorption isotherm:

x/m = K P^1/n

2

Substitute values: K = 4.0, P = 4 atm, 1/n = 0.5

x/m = 4.0 × (4)^0.5 = 4.0 × 2 = 8.0

Final Answer: 8.0 g/g

उत्तर: 8.0 g/g

3

Langmuir Isotherm

For the adsorption of nitrogen on iron powder at 0°C, the Langmuir constant a = 1.0 L/atm and b = 0.5 L/mol. Calculate the volume of nitrogen adsorbed per gram of iron at 0.5 atm pressure.

0°C पर लोहे के पाउडर पर नाइट्रोजन के अधिशोषण के लिए, लैंगमुइर स्थिरांक a = 1.0 L/atm और b = 0.5 L/mol है। 0.5 atm दाब पर प्रति ग्राम लोहे द्वारा अधिशोषित नाइट्रोजन का आयतन ज्ञात कीजिए।

Formula Hint

V = (V_m a P) / (1 + a P)

Where V_m = b × Molar volume at STP

1

Molar volume at STP = 22.4 L/mol

V_m = b × 22.4 = 0.5 × 22.4 = 11.2 L

2

Apply Langmuir equation:

V = (V_m a P) / (1 + a P)

3

Substitute values: V = (11.2 × 1.0 × 0.5) / (1 + 1.0 × 0.5) = 5.6 / 1.5 = 3.733 L

Final Answer: 3.73 L/g

उत्तर: 3.73 L/g

4

Catalyst Efficiency

A reaction takes 50 minutes for completion at 25°C. When a catalyst is added, the same reaction completes in 10 minutes. Calculate the lowering of activation energy if the reaction temperature is 35°C.

एक अभिक्रिया 25°C पर पूर्ण होने में 50 मिनट लेती है। जब उत्प्रेरक मिलाया जाता है, तो वही अभिक्रिया 10 मिनट में पूर्ण हो जाती है। यदि अभिक्रिया का ताप 35°C है तो सक्रियण ऊर्जा में कमी की गणना कीजिए।

Formula Hint

log(k₂/k₁) = (E_a/2.303R) × (1/T₁ - 1/T₂)

1

Reaction rate ∝ 1/time, so k_uncat/k_cat = t_cat/t_uncat = 10/50 = 0.2

2

At temperature T, using Arrhenius equation:

log(k_cat/k_uncat) = (ΔE_a/2.303R) × (1/T_uncat - 1/T_cat)

But temperatures are same, so:

log(k_cat/k_uncat) = -ΔE_a/(2.303RT)

3

k_cat/k_uncat = 1/0.2 = 5

log(5) = -ΔE_a/(2.303 × 8.314 × 298)

0.6990 = -ΔE_a/5705.8

ΔE_a = -0.6990 × 5705.8 = -3990 J/mol = -3.99 kJ/mol

Final Answer: Lowering of activation energy = 3.99 kJ/mol

उत्तर: सक्रियण ऊर्जा में कमी = 3.99 kJ/mol

5

Gold Number Calculation

To prevent the coagulation of 10 mL of gold sol, 0.5 mL of 10% NaCl solution is required. Calculate the gold number if 1 mL of 1% protective colloid is used.

10 mL गोल्ड सोल के स्कंदन को रोकने के लिए 0.5 mL 10% NaCl विलयन की आवश्यकता होती है। यदि 1 mL 1% सुरक्षात्मक कोलाइड का उपयोग किया जाता है तो गोल्ड संख्या की गणना कीजिए।

Formula Hint

Gold number = (Weight of protective colloid in mg) / (Volume of sol protected in mL) × 10

1

Amount of NaCl required without protective colloid = 0.5 mL of 10% solution

With protective colloid, this amount increases

2

Gold number is defined as the milligrams of protective colloid required to prevent coagulation of 10 mL gold sol when 1 mL of 10% NaCl is added

3

Here, 1 mL of 1% colloid is used to protect 10 mL sol against 0.5 mL 10% NaCl

Weight of colloid = 1 mL × 1% = 0.01 g = 10 mg

4

But standard test uses 1 mL of 10% NaCl, while here only 0.5 mL is used

So colloid is twice as effective as needed for standard test

Gold number = 10 mg × (0.5 mL / 1 mL) = 5

Final Answer: Gold number = 5

उत्तर: गोल्ड संख्या = 5

6

Micelle Formation

The critical micelle concentration (CMC) of sodium dodecyl sulfate is 8.2 mM at 25°C. Calculate the number of micelles formed in 500 mL of 0.1 M solution.

सोडियम डोडेसिल सल्फेट की क्रांतिक माइसिल सांद्रता (CMC) 25°C पर 8.2 mM है। 500 mL 0.1 M विलयन में निर्मित माइसिलों की संख्या की गणना कीजिए।

Formula Hint

Number of micelles = (Total surfactant - CMC) × Volume × N_A / Aggregation number

1

Aggregation number for SDS is approximately 62

Concentration = 0.1 M = 100 mM

CMC = 8.2 mM

2

Concentration above CMC = 100 - 8.2 = 91.8 mM

3

Moles of surfactant in micelles = (91.8 × 10⁻³ mol/L) × 0.5 L = 0.0459 mol

4

Number of surfactant molecules = 0.0459 × 6.022 × 10²³ = 2.764 × 10²²

5

Number of micelles = Total surfactant molecules / Aggregation number

= 2.764 × 10²² / 62 ≈ 4.46 × 10²⁰

Final Answer: 4.46 × 10²⁰ micelles

उत्तर: 4.46 × 10²⁰ माइसिल

7

Coagulation Value

100 mL of arsenic sulphide sol requires 5 mL of 1 M NaCl solution for complete coagulation. Calculate the coagulation value of NaCl.

100 mL आर्सेनिक सल्फाइड सोल के पूर्ण स्कंदन के लिए 5 mL 1 M NaCl विलयन की आवश्यकता होती है। NaCl का स्कंदन मान ज्ञात कीजिए।

Formula Hint

Coagulation value = (Millimoles of electrolyte) / (Volume of sol in liters)

1

Millimoles of NaCl added = Molarity × Volume(mL) = 1 M × 5 mL = 5 millimoles

2

Volume of sol = 100 mL = 0.1 L

3

Coagulation value = Millimoles of electrolyte per liter of sol

= (5 mmol) / 0.1 L = 50 mmol/L

Final Answer: 50 mmol/L

उत्तर: 50 mmol/L

8

Surface Area Calculation

1 g of colloidal particle is divided into cubes of 10 nm edge length. Calculate the surface area of the colloidal particles. (Density = 2 g/cm³)

1 g कोलॉइडी कण को 10 nm भुजा वाले घनों में विभाजित किया जाता है। कोलॉइडी कणों के पृष्ठीय क्षेत्रफल की गणना कीजिए। (घनत्व = 2 g/cm³)

Formula Hint

Surface area = (Surface area of one cube) × (Number of cubes)

1

Volume of one cube = (10 × 10⁻⁹ m)³ = 10⁻²⁷ m³ = 10⁻²¹ cm³

2

Mass of one cube = Density × Volume = 2 g/cm³ × 10⁻²¹ cm³ = 2 × 10⁻²¹ g

3

Number of cubes in 1 g = 1 / (2 × 10⁻²¹) = 5 × 10²⁰

4

Surface area of one cube = 6 × (side)² = 6 × (10⁻⁸ m)² = 6 × 10⁻¹⁶ m²

5

Total surface area = 5 × 10²⁰ × 6 × 10⁻¹⁶ = 3 × 10⁵ m²

Final Answer: 300,000 m²/g

उत्तर: 300,000 m²/g

Willer Academy - Chemistry Department

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