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

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

Answer 1

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

In order to solve the problem, calculate the value of cell constant of the first solution and then use this value of cell constant to calculate the value of k of second solution. Afterwards, finally calculate molar conductivity using value of k and m.
For first solution,
$k=1.4S{m}^{-1},R=50\Omega ,M=0.2$
Specific conductance $\left(K\right)=\frac{1}{R}\times \frac{l}{A}$
$1.4S{m}^{-1}=\frac{1}{50}\times \frac{l}{A}\Rightarrow \frac{l}{A}=50\times 1.4m^{-1}$
For second solution, $R=280,\frac{l}{A}=50\times 1.4m^{-1}$
$K=\frac{1}{280}\times 1.4\times 50=\frac{1}{4}$
Now, molar conductivity
${\lambda }_m=\frac{K}{1000\times m}=\frac{\frac{1}{4}}{1000\times 0.5}=\frac{1}{2000}=5\times {10}^{-4}S{m}^{2}mo{l}^{-1}$