Q125 / 200 · MHT CET - Medical 2026 (26 April Shift 1) · Chemistry
Calculate the conductivity of $0.025$ M $\text{BaCl}_2$ solution if the molar conductivity of salt at $25^\circ\text{C}$ is $325\ \Omega^{-1}\ \text{cm}^2\ \text{mol}^{-1}$.
- A$8.125 \times 10^{-2}\ \Omega^{-1}\text{cm}^{-1}$
- Correct answerB$8.125 \times 10^{-3}\ \Omega^{-1}\text{cm}^{-1}$
- C$7.6923 \times 10^{-2}\ \Omega^{-1}\text{cm}^{-1}$
- D$8.372 \times 10^{-2}\ \Omega^{-1}\text{cm}^{-1}$
Solution
Given:
Concentration, $C = 0.025\ \text{M}$
Molar conductivity, $\Lambda_m = 325\ \Omega^{-1}\ \text{cm}^2\ \text{mol}^{-1}$
The relationship between molar conductivity and conductivity ($\kappa$) is given by:
$\Lambda_m = \dfrac{\kappa \times 1000}{C}$
Rearranging the formula to solve for $\kappa$:
$\kappa = \dfrac{\Lambda_m \times C}{1000}$
Substituting the given values:
$\kappa = \dfrac{325 \times 0.025}{1000}$
$\kappa = \dfrac{8.125}{1000}$
$\kappa = 8.125 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$
Answer: $8.125 \times 10^{-3}\ \Omega^{-1}\text{cm}^{-1}$
Concentration, $C = 0.025\ \text{M}$
Molar conductivity, $\Lambda_m = 325\ \Omega^{-1}\ \text{cm}^2\ \text{mol}^{-1}$
The relationship between molar conductivity and conductivity ($\kappa$) is given by:
$\Lambda_m = \dfrac{\kappa \times 1000}{C}$
Rearranging the formula to solve for $\kappa$:
$\kappa = \dfrac{\Lambda_m \times C}{1000}$
Substituting the given values:
$\kappa = \dfrac{325 \times 0.025}{1000}$
$\kappa = \dfrac{8.125}{1000}$
$\kappa = 8.125 \times 10^{-3}\ \Omega^{-1}\ \text{cm}^{-1}$
Answer: $8.125 \times 10^{-3}\ \Omega^{-1}\text{cm}^{-1}$
Hence, the correct answer is option: 'B'.