Question

The specific conductivity of $0.02MKCl$ solution at ${25}^{o}C$ is $2.768\times {10}^{-3}oh{m}^{-1}c{m}^{-1}.$
The resistance of this at ${25}^{o}C$ when measured with a particular cell was $250.2ohm$. The resistance of $0.01MCuS{O}_4$ solution at ${25}^{o}C$ measured with the same cell was $8331ohm$. Calculate the molar conductivity of the copper sulphate solution.

• A. $2.42oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• B. $14.24oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• C. $6.02oh{m}^{-1}c{m}^{2}mo{l}^{-1}$
• D. $8.31oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

Answer 1

The correct option is D $8.31oh{m}^{-1}c{m}^{2}mo{l}^{-1}$

Cell constant is same for a particular cell. It is a constant.
For $0.02MKCl$ solution,

$\text{Resistance,}\left(R\right)=250.2\Omega$
$\text{Cell constant,}\frac{l}{A}=\text{Conductivity of KCl}\times \text{Resistance of KCl}$
$=2.768\times {10}^{-3}\times 250.2$

For $0.01MCuS{O}_4$ solution
$\text{Conductivity,}\kappa =\frac{\text{Cell constant}}{\text{Resistance}}$

$\kappa =2.768\times {10}^{-3}\times 250.2\times \frac{1}{8331}$

$\text{Molar conductance,}{\Lambda }_m=\text{Conductivity}\times \frac{1000}{C}$
${\Lambda }_m=\frac{2.768\times {10}^{-3}\times 250.2}{8331}\times \frac{1000}{1/100}=8.312oh{m}^{-1}c{m}^{2}mo{l}^{-1}$