Question

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

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