Resistance of 0.2 M solution of an electrolyte is 50Ω. The specific conductance of the solution is $1.4 {~S}{~m}^{-1}$. The resistance of $0.5 {M}$ solution of the same electrolyte is $280 \Omega$.

Question

Resistance of $0.2 {M}$ solution of an electrolyte is $50 \Omega$. The specific conductance of the solution is $1.4 {~S}{~m}^{-1}$. 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 ${S} {m}^2 {~mol}^{-1}$ is

(a) $5 \times 10^2$

(b) $5 \times 10^{-4}$

(c) $5 \times 10^{-3}$

(d) $5 \times 10^3$

Answer 1

Correct Option : (b)

Explanation—

Case I : $C=0.2 \mathrm{M}, R=50 \Omega, \kappa=1.4 \mathrm{~S} \mathrm{~m}^{-1}$

$\kappa=\frac{l}{A} \cdot \frac{1}{R} \text { or } 1.4=\frac{l}{A} \cdot \frac{1}{50}$

$\Rightarrow \frac{l}{A}=1.4 \times 50=70 \mathrm{~m}^{-1}$

Case II : $\frac{l}{A}=70 \mathrm{~m}^{-1}, C=0.5 \mathrm{M}, R=280 \Omega$, $R=\rho \frac{l}{A}$ or $\frac{1}{\rho}=\frac{1}{R} \times \frac{l}{A} \Rightarrow \frac{1}{\rho}=\frac{1}{280} \times 70$

$\kappa=\frac{1}{\rho}=0.25 \mathrm{~S} \mathrm{~m}^{-1}$

Now, $\Lambda_m=\kappa \times \frac{1000}{C}$

If molarity is in $\mathrm{mol} \mathrm{L}^{-1}$, then

$\Lambda_m\left(\mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}\right)=\frac{\kappa\left(\mathrm{S} \mathrm{m}^{-1}\right)}{1000 \mathrm{~L} \mathrm{~m}^{-3} \times \text { Molarity }\left(\mathrm{mol} \mathrm{L}^{-1}\right)}$

$=\frac{0.25 \mathrm{~S} \mathrm{~m}^{-1}}{1000 \mathrm{~L} \mathrm{~m}^{-3} \times 0.5 \mathrm{~mol} \mathrm{~L}^{-1}}=5 \times 10^{-4} \mathrm{~S} \mathrm{~m}^2 \mathrm{~mol}^{-1}$