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AIIMS 201827 May
Exam
AIIMS
Year
2018
Date
27 May
Subject
Chemistry
Chapter
Electrochemistry

AIIMS 2018 · 27 May · Chemistry · Electrochemistry

0.1 M solution is present in a conductivity cell with electrode of...

0.1M0.1 \mathrm{M} solution is present in a conductivity cell with electrode of 100 cm2100 \mathrm{~cm}^2 area placed 1 cm1 \mathrm{~cm} apart and resistance observed is 5×103ohm5 \times 10^3 \mathrm{ohm}. What is the molar conductivity of this solution?

Options

  1. A5×102 S cm2 mol−15 \times 10^2 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
  2. B104 S cm mol−110^4 \mathrm{~S} \mathrm{~cm} \mathrm{~mol}^{-1}
  3. C200 S cm2 mol−1200 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}
  4. D0.02 S cm2 mol−10.02 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

Correct answer

D. 0.02 S cm2 mol−10.02 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}

Step-by-step solution

Molar conductivity, Λm=κ×1000MM=\Lambda_m=\frac{\kappa \times 1000}{M} M= molarity and κ=\kappa= specific conductivity κ=1R× cell constant =1R×ℓA=15×103Ω×1cm100cm2=2×10−6Ω−1cm−1∴Λm=2×10−6×10000.1=0.02Scm2mol−1\begin{aligned} \kappa & =\frac{1}{R} \times \text{ cell constant } \\ & =\frac{1}{R} \times \frac{\ell}{A}=\frac{1}{5 \times 10^3 \Omega} \times \frac{1 \mathrm{ cm}}{100 \mathrm{ cm}^2} \\ & =2 \times 10^{-6} \Omega^{-1} \mathrm{ cm}^{-1} \\ \therefore \quad \Lambda_m & =\frac{2 \times 10^{-6} \times 1000}{0.1}=0.02 \mathrm{ S} \mathrm{ cm}^2 \mathrm{ mol}^{-1}\end{aligned}
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