Question

Resistance of $0.2M$ solution of an electrolyte is $50\Omega$. The specific conductance of the solution is $1.3S{m}^{-1}$. If resistance of the $0.4M$ solution of the same electrolyte is $260\Omega$ , its molar conductivity is

• A. $6250S{m}^{2}mo{l}^{-1}$
• B. $6.25\times 10^{-4}S{m}^{2}mo{l}^{-1}$
• C. $625\times 10^{-4}S{m}^{2}mo{l}^{-1}$
• D. $62.5S{m}^{2}mo{l}^{-1}$

Answer 1

Answer: (B)

For $0.2M$ solution,
Specific conductance (k ) = Conductance $\left(\frac{1}{R}\right)\times$ Cell constant$\left(la\right)$
$\Rightarrow K\frac{1}{50}\times \frac{l}{a}=1.3S{m}^{-1}$
$\Rightarrow \frac{1}{a}=1.3\times 50m^{-1}$
$\frac{1}{a}=1.3\times 50\times 10^{-2}c{m}^{-1}$
for $0.4$ solution $R=260\Omega$
$R=\rho \frac{1}{a}$
$\Rightarrow \frac{1}{\rho }=k=\frac{1}{R}\times \frac{1}{a}$
$k=\frac{1}{260}\times 1.3\times 50\times 10^{-2}SC{m}^{-1}$
Molar conductivity of the solution is given by
${\wedge }_m=\frac{k\times 1000}{M}$
$=\frac{1}{260}\times \frac{1.3\times 50\times 10^{-2}\times 1000}{4}$
$=6.25Sc{m}^{2}mo{l}^{-1}$
$=6.25\times 10^{-4}S{m}^{2}mo{l}^{-1}$