Question

Resistance of $0.2\, M$ solution of an electrolyte is $50 \,\Omega$. The specific conductance of the solution of $0.5 \,M$ solution of same electrolyte is $1.4\, S \, m^{-1}$ and resistance of same solution of the same electrolyte is $280 \,\Omega$ . The molar conductivity of $0.5\, M$ solution of the electrolyte in $S\,m^2 \,mol^{-1}$ is

• A. $5 \times10^{-4}$
• B. $5 \times10^{-3}$
• C. $5 \times10^{3}$
• D. $5 \times10^{2}$

Answer 1

The Correct Option is A

For $0.2 \,M$ solution

$R =50\, \Omega$
$\sigma=1.4 \,S \,m ^{-1}=1.4 \times 10^{-2} S \,cm ^{-1}$
$\Rightarrow \rho=\frac{1}{\sigma}=\frac{1}{1.4 \times 10^{-2}} \Omega \,cm$

Now, $R =\rho \frac{l}{ a }$
$\Rightarrow \frac{l}{ a }=\frac{ R }{\rho}=50 \times 1.4 \times 10^{-2}$

For $0.5$ M solution

$R =280 \,\Omega$
$\sigma=?$
$\frac{l}{ a }=50 \times 1.4 \times 10^{-2}$
$\Rightarrow R =\rho \frac{l}{ a } $
$\Rightarrow \frac{1}{\rho}=\frac{1}{ R } \times \frac{l}{ a } $
$\Rightarrow \sigma=\frac{1}{280} \times 50 \times 1.4 \times 10^{-2} $
$=\frac{1}{280} \times 70 \times 10^{-2}$
$=2.5 \times 10^{-3} \,S \,cm ^{-1} $

Now , $\lambda_{ m }=\frac{\sigma \times 1000}{ M }$
$=\frac{2.5 \times 10^{-3} \times 1000}{0.5} $
$=5\, S \,cm ^{2} \,mol ^{-1} $
$=5 \times 10^{-4} \,S \,m ^{2} \,mol ^{-1}$