Question

The conductivity of $0.1MNaCl$ is $3\times {10}^{-3}Sc{m}^{-1}$ and its resistance in an electrolytic cell is $500ohm$. Resistance of $\frac{1}{20}MCuS{O}_4$ solution measured in the same cell at the same temperature is found to be $70ohm$. The molar conductivity of the given $CuS{O}_4$ solution is ?

• A. $428Sc{m}^{2}mo{l}^{-1}$
  
• B. $120.2Sc{m}^{2}mo{l}^{-1}$
  
• C. $20.5Sc{m}^{2}mo{l}^{-1}$
  
• D. $12.50Sc{m}^{2}mo{l}^{-1}$
  

Answer 1

The correct option is A $428Sc{m}^{2}mo{l}^{-1}$


Cell constant is constant for a particular cell.
For $0.1MNaCl$ solution,
$\text{Resistance,}\left(R\right)=500\Omega$
$\text{Cell constant,}\frac{l}{A}=\text{Conductivity of KCl}\times \text{Resistance of KCl}$
$=3\times {10}^{-3}\times 500c{m}^{-1}$

For $0.05MCuS{O}_4$ solution
$\text{Conductivity,}\kappa =\frac{\text{Cell constant}}{\text{Resistance}}$

$\kappa =3\times {10}^{-3}\times 500\times \frac{1}{70}$

$\text{Molar conductance,}{\Lambda }_m=\text{Conductivity}\times \frac{1000}{C}$
${\Lambda }_m=\frac{3\times {10}^{-3}\times 500}{70}\times \frac{1000}{0.05}{\Lambda }_m=428oh{m}^{-1}c{m}^{2}mo{l}^{-1}$