Question

$0.1M$ $NaCl$ solution has resistance of $500$ ohm. This solution known to have conductivity of $3\times {10}^{-3}S$ ${cm}^{-1}$. Resistance of $\frac{1}{20}M$ $Cu{SO}_4$ measured in the same cell at the same temperature is found $70$ ohm. The molar conductivity of the $Cu{SO}_4$ solutions is?

• A. $1250S{cm}^2$ ${mol}^{-1}$
• B. $0.428S{cm}^2$ ${mol}^{-1}$
• C. $428S{cm}^2$ ${eq}^{-1}$
• D. $12.50S{cm}^2$ ${eq}^{-1}$

Answer 1

Answer: (A)

$1250S{cm}^2$ ${mol}^{-1}$

We know, $K=K\times \frac{1}{R}$

$\therefore K_1{R}_1=K_2{R}_2$

$3\times {10}^{-3}\times 500=K_2\times 70$

$\left(Cu{SO}_4\right)K_2=2.14\times {10}^{-3}S{cm}^{-1}$

Molar conductivity $=\frac{K}{C}$

$=\frac{2.14\times {10}^{-3}}{1/20}$

$=428{Scm}^2{eq}^{-1}$