Question

Resistance of $0.2\, M$ solution of an electrolyte is $50 \,\Omega$. The specific conductance of the solution of $0.5 \,M$ solution of same electrolyte is $1.4\, S \, m^{-1}$ and resistance of same 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. $5 \times10^{-4}$
• B. $5 \times10^{-3}$
• C. $5 \times10^{3}$
• D. $5 \times10^{2}$

Answer 1

Answer: (A)

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,
$k=1.4\, Sm ^{-1}, R=50 \, \Omega, M=0.2$
Specific conductance $(\kappa)=\frac{1}{R} \times \frac{l}{A}$
$1.4 Sm ^{-1} =\frac{1}{50} \times \frac{l}{A}$
$\Rightarrow \frac{l}{A} =50 \times 1.4\, m ^{-1}$
For second solution,
$R=280, \frac{l}{A}=50 \times 1.4 \, m ^{-1} $
$\kappa=\frac{1}{280} \times 1.4 \times 50=\frac{1}{4}$
Now, molar conductivity
$\lambda_{m} =\frac{\kappa}{1000 \times m}=\frac{1 / 4}{1000 \times 0.5}=\frac{1}{2000} $
$=5 \times 10^{-4} \, S \, m ^{2}\, mol ^{-1}$