Kohlrausch's Law | Electrochemistry | 10 MCQs for CSIR NET GATE | Download PDF

Kohlrausch Law
MCQs for CSIR NET & GATE

Electrochemistry
Topic wise Assignment-3

QUIZ              PDF

Q1. The molar conductivity at infinite dilution for
CaCl₂= x, CH₃COOH= y, HCl=z
What will be the value for [(CH₃COO)₂Ca]
(a) 2y + x - (1/2) z
(b) y - 2x + 2z
(c) 2y + x - 2z
(d) None

Q2. If the molar conductivity at infinite dilution for
Fe₂(SO₄)₃ = x₁, H₂SO₄ = x₂, HCl = x₃; Calculate value for [FeCl₃] at infinite dilution
(a) 3x₃ + 2x₁ - (1/2)x₂
(b) 3x₃ + (1/2)x₁ - (3/2)x₂
(c) (3/2)x₃ - (1/2)x₁ + 2x₂
(d) (1/2)x₁ - (3/2)x₂ - 3x₃

Q3. In each of the following pairs, which will show greater conduction of electricity?
(i) A= Silver wire at 20°C B= Same wire at 50°C
(ii) C=NH₄OH solution at 20°C D= Same solution T 50°C
(iii) E= 0.1M CH₃COOH solution F= 1M CH₃COOH solution
(a) A, C, E
(b) A, D, F
(c) B, D, E
(d) A, D, E

Q4. Considering Al₂(SO₄)₃, the relation between molar and equivalent conductivity will be
(a) Λ_eq = 6 Λ_m
(b) Λ_m = 6 Λ_eq
(c) Λ_eq = 3 Λ_m
(d) Λ_m = Λ_eq

Q5. In 0.1M HCl and 0.1M NaCl solution, which will have greater transport number for Cl^- ion?
(a) NaCl
(b) Same for both
(c) HCl
(d) Insufficient data

Q6. Equivalent conductivities at infinite dilution are given for following electrolytes. Calculate for FeSO₄;

FeCl₂ = x, HCl = y, H₂SO₄ = z

(a) x+z-2y
(b) x+z-y
(c) (1/2)x+z-y
(d) (1/2)x+(1/2)z-2y

Q7. Λ°_eq for BaCl₂, H₂SO₄ and HCl are x₁, x₂ and x₃ S cm² eq^-1 respectively. If conductivity of saturated BaSO4 solution is y S cm^-1, then Ksp for BaSO₄ is

Q8. If transport number of Na⁺ is 0.463 in infinitely dilute solution of NaCl. Calculate transport no. of H+ in dilute solution of HCl. Given that
Λ°m (NaCl)= 96.9 S cm² mol^-1 and Λ°m (HCl)= 96.9 S cm² mol^-1
(a) 0.728
(b) 0.76
(c) 0.272
(d) 0.24

Q9. Calculate the fraction of current carried by sodium ions in a solution which is 0.1M in Na₂SO₄ and 0.01M in H₂SO₄. Given that
λ_m(Na⁺)=50.1, λ_m(H⁺)=349.8 and λ_m(SO₄^2-)=160 S cm² mol^-1
(a) 0.39
(b) 0.51
(c) 0.29
(d) 0.7

Q10. A dilute solution of KCl is placed between two electrodes 10cm apart across which a potential difference of 10V was applied. Calculate
(i) ionic mobility (K⁺)
(ii) Potential Gradient
(iii) Speed of K⁺ ions
(iv) distance travelled by K⁺ ions in 5000 seconds.
Given that, λ°_m(K⁺) = 96.5 S cm² mol^-1

Answer Key

Q1.c Q2.b Q3.d Q4.b Q5.a Q6.b Q7.d Q8.a Q9.c Q10. Ans

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