Q23 / 150 · MHT CET 2026 (15 April Shift 1) · Chemistry
The molar conductivity of $0.04$ M $\text{AB}_2$ type salt solution at $300$ K is $200\,\Omega^{-1}\text{cm}^2\text{mol}^{-1}$. Find the conductivity.
- A$0.006\,\Omega^{-1}\text{cm}^{-1}$
- Correct answerB$0.008\,\Omega^{-1}\text{cm}^{-1}$
- C$0.01\,\Omega^{-1}\text{cm}^{-1}$
- D$0.015\,\Omega^{-1}\text{cm}^{-1}$
Solution
The relation between molar conductivity $\Lambda_m$, conductivity $\kappa$, and molarity $M$ is given by:
$\Lambda_m = \dfrac{\kappa \times 1000}{M}$
Given:
$\Lambda_m = 200\,\Omega^{-1}\text{cm}^2\text{mol}^{-1}$
$M = 0.04$ M
Substituting the given values into the formula:
$200 = \dfrac{\kappa \times 1000}{0.04}$
$\kappa = \dfrac{200 \times 0.04}{1000}$
$\kappa = \dfrac{8}{1000} = 0.008\,\Omega^{-1}\text{cm}^{-1}$
Answer: $0.008\,\Omega^{-1}\text{cm}^{-1}$
$\Lambda_m = \dfrac{\kappa \times 1000}{M}$
Given:
$\Lambda_m = 200\,\Omega^{-1}\text{cm}^2\text{mol}^{-1}$
$M = 0.04$ M
Substituting the given values into the formula:
$200 = \dfrac{\kappa \times 1000}{0.04}$
$\kappa = \dfrac{200 \times 0.04}{1000}$
$\kappa = \dfrac{8}{1000} = 0.008\,\Omega^{-1}\text{cm}^{-1}$
Answer: $0.008\,\Omega^{-1}\text{cm}^{-1}$
Hence, the correct answer is option: 'B'.