Question

Resistance of 0.2 M solution of an electrolyte is 50 $\Omega$. The specific conductance of the solution is $1.4S{m}^{-1}.$ The resistance of 0.5 M solution of the same electrolyte is 280 $\Omega$. The molar conductivity of 0.5 M solution of the electrolyte in $S{m}^{2}mo{l}^{-1}$ is:
(IIT-JEE-2014)

• A. $5\times {10}^3$
• B. $5\times {10}^2$
• C. $5\times {10}^{-4}$
• D. $5\times {10}^{-3}$

Answer 1

The correct option is C $5\times {10}^{-4}$

R = $\frac{1}{K}\times \frac{1}{a}(\frac{1}{a}=cellconstant)$

For 0.2 M solution
$50=\frac{1}{k}\times \frac{1}{a}\Rightarrow 50=\frac{1}{1.4}\times \frac{1}{a}$
$\frac{1}{a}=70m^{-1}$

For 0.5 M solution
$280=\frac{1}{k}\times 70$
$K=\frac{1}{4}S{m}^{-1}$

${\wedge }_m=\left(\frac{K\times 1000}{M}\right)({10}^{-2}m{)}^3$
${\wedge }_m=\frac{1}{4}\times \left(\frac{1000}{M}\right)({10}^{-2}m{)}^3$
$=\frac{1}{4}\times \frac{1000}{0.5}\times {10}^{-6}$
$=500\times {10}^{-6}=5\times {10}^{-4}S{m}^{2}mo{l}^{-1}$