Question

When a conductance cell was filled with 0.0025 M solution of $K_{2}S{O}_4$, its resistance was 326 ohm. Length of cell =100cm and cross-sectional area $2.14\times {10}^{-4}c{m}^2$. The specific resistance of $K_{2}S{O}_4$ is:

• A. $4.994\times {10}^{-4}$
• B. $5.997\times {10}^{-4}$
• C. $6.997\times {10}^{-4}$
• D. $7.997\times {10}^{-4}$

Answer 1

Answer: (C)

$6.997\times {10}^{-4}$

$R=\frac{\text{specific resistance}\times l}{A}$
$326=\frac{\text{specific resistance}\times 100}{2.14\times {10}^{-4}}$

Specific Resistance = 0.0006997

Hence, option C is correct.