Q25 / 150 · MHT CET 2026 (11 April Shift 1) · Chemistry
At $298$ K, the specific conductance ($\kappa$) of a $0.0020$ M NaCl solution is $2.50 \times 10^{-4} \text{ S cm}^{-1}$. Calculate the molar conductivity ($\Lambda_m$) of the solution in $\text{S cm}^2 \text{ mol}^{-1}$.
- Correct answerA$125$
- B$62.5$
- C$12.5$
- D$250$
Solution
Given:
$\kappa = 2.50 \times 10^{-4} \text{ S cm}^{-1}$
$C = 0.0020 \text{ M}$
The molar conductivity $\Lambda_m$ is given by the formula:
$\Lambda_m = \dfrac{\kappa \times 1000}{C}$
Substituting the given values:
$\Lambda_m = \dfrac{2.50 \times 10^{-4} \times 1000}{0.0020}$
$\Lambda_m = \dfrac{0.250}{0.0020}$
$\Lambda_m = 125 \text{ S cm}^2 \text{ mol}^{-1}$
Answer: $125$
$\kappa = 2.50 \times 10^{-4} \text{ S cm}^{-1}$
$C = 0.0020 \text{ M}$
The molar conductivity $\Lambda_m$ is given by the formula:
$\Lambda_m = \dfrac{\kappa \times 1000}{C}$
Substituting the given values:
$\Lambda_m = \dfrac{2.50 \times 10^{-4} \times 1000}{0.0020}$
$\Lambda_m = \dfrac{0.250}{0.0020}$
$\Lambda_m = 125 \text{ S cm}^2 \text{ mol}^{-1}$
Answer: $125$
Hence, the correct answer is option: 'A'.