Question

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

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

Answer 1

Answer: (D)

Since, the weak electrolyte is $2\%$ ionised,

$C=\alpha \cdot N$

$=\frac{2}{100}\times 0.1$

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

Limiting equivalent conductance,

${\Lambda }_{eq}=\frac{\kappa \times 1000}{C}$

$400Sc{m}^{2}g{\text{ equiv }}^{-1}=\frac{\kappa \times 1000}{2\times 10^{-3}}$

$\kappa =8\times 10^{-4}Sc{m}^{-1}$

Further, $\kappa =\frac{1}{R}\times \left(\frac{l}{a}\right)$

(where, $\frac{l}{a}=$ cell constant )

$8\times 10^{-4}Sc{m}^{-1}=\frac{1}{R}\times 0.4c{m}^{-1}$

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