Question

A weak electrolyte having the limiting equivalent conductance of $400$ $S$ ${cm}^2$ $g$ ${equiv}^{-1}$ at $298K$ is $2$% ionized in its $0.1N$ solution. The resistance of this solution (in ohm) in an electrolytic cell of cell constant $0.4$ ${cm}^{-1}$ at this temperature is:

• A. $200$
• B. $300$
• C. $400$
• D. $500$
• E. $600$

Answer 1

Answer: (D)

$500$

Since, the weak electrolyte is $2$% ionised,
$C=\alpha \cdot N$
$=\frac{2}{100}\times 0.1$

$=2\times {10}^{-3}$ $g$ $equiv$ $L^{-1}$

Limiting equivalnet conductance,
$A_{eq}=\frac{K\times 1000}{C}$

$400$ $S$ ${cm}^2$ $g$ ${equiv}^{-1}$ $=\frac{K\times 1000}{2\times {10}^{-3}}$

$K=8\times {10}^{-4}S{cm}^{-1}$

Further, $K=\frac{1}{R}\times \left(\frac{1}{a}\right)$ (where, $\frac{1}{a}=$ cell constant)

$8\times {10}^{-4}S{cm}^{-1}=\frac{1}{R}\times 0.4{cm}^{-1}$

$R=\frac{0.4}{8\times {10}^{-4}}{S}^{-1}=\frac{1000}{2}\Omega =500\Omega$

Hence, option D is correct.