Q22 / 150 · MHT CET 2026 (19 April Shift 2) · Chemistry
What must be the molarity of $\text{BaCl}_2$ solution to have molar conductivity $240\,\Omega^{-1}\text{cm}^2\text{mol}^{-1}$ and conductivity $0.012\,\Omega^{-1}\text{cm}^{-1}$ at $25^\circ\text{C}$ ?
- A$0.01$ M
- B$0.02$ M
- Correct answerC$0.05$ M
- D$0.1$ M
Solution
The relationship between molar conductivity $\Lambda_m$, conductivity $\kappa$, and molarity $M$ is given by the formula:
$\Lambda_m = \dfrac{\kappa \times 1000}{M}$
Given values are:
$\Lambda_m = 240 \, \Omega^{-1}\text{cm}^2\text{mol}^{-1}$
$\kappa = 0.012 \, \Omega^{-1}\text{cm}^{-1}$
Substituting the given values into the formula:
$240 = \dfrac{0.012 \times 1000}{M}$
$240 = \dfrac{12}{M}$
$M = \dfrac{12}{240}$
$M = 0.05 \, \text{M}$
Answer: $0.05$ M
$\Lambda_m = \dfrac{\kappa \times 1000}{M}$
Given values are:
$\Lambda_m = 240 \, \Omega^{-1}\text{cm}^2\text{mol}^{-1}$
$\kappa = 0.012 \, \Omega^{-1}\text{cm}^{-1}$
Substituting the given values into the formula:
$240 = \dfrac{0.012 \times 1000}{M}$
$240 = \dfrac{12}{M}$
$M = \dfrac{12}{240}$
$M = 0.05 \, \text{M}$
Answer: $0.05$ M
Hence, the correct answer is option: 'C'.