Question

The resistance of a $0.2\text{M}$ solution of an electrolyte is $50\Omega$. The specific conductance of the solution is $1.4{\text{S m}}^{-1}$ .
The resistance of a $0.5\text{M}$ solution of the same electrolyte occupying the same volume is $280\Omega$. The molar conductivity of the above mentioned $0.5\text{M}$ solution of the electrolyte in ${\text{S m}}^2{\text{mol}}^{-1}$ is:

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

Answer 1

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

From the expression:
$\kappa =G\times C$
We can say that, $C=\frac{\kappa }{G}$
As, $G=\frac{1}{R}$
$\therefore$
$C=\kappa \times R$
Since, ‘C’ remains constant for a cell.
We can say that,
${\kappa }_1\times R_1={\kappa }_2\times R_2$
Putting values:
$1.4\times 50={\kappa }_2\times 280$
$\therefore$
$k_2=\frac{1}{4}S{m}^{-1}$
Now,
${\Lambda }_M=\frac{\kappa \times 1000}{M}$
$\Rightarrow$ $\frac{1\times 1000\times {10}^{-9}}{4\times 0.5\times {10}^{-3}}$
Since, Molarity is defined per litre so we converted $m^3$ to litre.
$\Rightarrow$ $5\times {10}^{-4}S{m}^{2}mo{l}^{-1}$