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Vidaara.orgClass 12 · Chemistry
CodeVID-C12-03-T2-01
Conductance of Electrolytes — Assignment
Chapter: Electrochemistry
Topic: Conductance of Electrolytes
Maximum Marks: 30
Time: 60 minutes
Name: ____________________ Roll No.: __________ Date: ____________

General Instructions

  • All questions are compulsory.
  • Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
  • Show all steps in numericals and state units clearly. For full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions 5 × 1 = 5 marks
1.
The reciprocal of resistivity is called:
  • A.conductance
  • B.conductivity
  • C.cell constant
  • D.molar conductivity
2.
The cell constant has units of:
  • A.$\text{S cm}^{-1}$
  • B.$\text{cm}^{-1}$
  • C.$\text{S cm}^2\,\text{mol}^{-1}$
  • D.$\Omega$
3.
Molar conductivity increases on dilution because:
  • A.ions slow down
  • B.more ions become available / move freely
  • C.temperature drops
  • D.conductivity rises
4.
Kohlrausch's law is strictly valid at:
  • A.high concentration
  • B.infinite dilution
  • C.high temperature
  • D.low temperature
5.
Which is a weak electrolyte?
  • A.KCl
  • B.NaCl
  • C.CH3COOH
  • D.HCl
Section B — Short Answer (2 marks) 3 × 2 = 6 marks
6.
Define conductivity and give its SI unit.
7.
Write the formula relating molar conductivity to conductivity.
8.
A $0.05\,\text{M}$ solution has $\kappa = 0.0124\,\text{S cm}^{-1}$. Find $\Lambda_m$.
Section C — Short Answer (3 marks) 2 × 3 = 6 marks
9.
State Kohlrausch's law and give two of its applications.
10.
Distinguish between the variation of $\Lambda_m$ with concentration for strong and weak electrolytes.
Section D — Long Answer (5 marks) 1 × 5 = 5 marks
11.
Define molar conductivity. Explain how it varies with concentration for strong and weak electrolytes and how Kohlrausch's law is used to obtain $\Lambda_m^0$ of a weak acid such as acetic acid.

Answer Key

Section A — Multiple Choice Questions
  1. (B) conductivity
  2. (B) $\text{cm}^{-1}$
  3. (B) more ions become available / move freely
  4. (B) infinite dilution
  5. (C) CH3COOH
Section B — Short Answer (2 marks)
  1. Conductance of a solution between electrodes of unit area unit distance apart; unit $\text{S m}^{-1}$ (or $\text{S cm}^{-1}$).
  2. $\Lambda_m = \frac{\kappa \times 1000}{C}$, with $\kappa$ in $\text{S cm}^{-1}$ and $C$ in $\text{mol L}^{-1}$.
  3. $\Lambda_m = \frac{0.0124 \times 1000}{0.05} = 248\,\text{S cm}^2\,\text{mol}^{-1}$.
Section C — Short Answer (3 marks)
  1. $\Lambda_m^0 = \nu_+\lambda_+^0 + \nu_-\lambda_-^0$; used to find $\Lambda_m^0$ of weak electrolytes and degree of dissociation $\alpha = \Lambda_m/\Lambda_m^0$.
  2. Strong: small linear rise with $\sqrt{C}$, extrapolates to $\Lambda_m^0$. Weak: steep rise near zero C, no direct extrapolation.
Section D — Long Answer (5 marks)
  1. $\Lambda_m = \frac{\kappa \times 1000}{C}$; strong electrolytes give $\Lambda_m = \Lambda_m^0 - A\sqrt{C}$, weak ones rise steeply; $\Lambda_m^0(\text{CH}_3\text{COOH}) = \Lambda_m^0(\text{CH}_3\text{COONa}) + \Lambda_m^0(\text{HCl}) - \Lambda_m^0(\text{NaCl})$.
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