Question

Calculate the molar conductivity of 0.2 M KCl solution at 298 K (\(k = 0.0248 \Omega ^{-1}\)cm\(^{-1}\))

• A. \(143 \Omega ^{-1}\) cm\(^2\) mol\(^{-1}\)
• B. \(98 \Omega ^{-1}\) cm\(^2\) mol\(^{-1}\)
• C. \(124 \Omega ^{-1}\) cm\(^2\) mol\(^{-1}\)
• D. \(87 \Omega ^{-1}\) cm\(^2\) mol\(^{-1}\)

Hint:

Molar conductivity equals conductivity times 1000 divided by molarity.

Answer 1

The Correct Option is C

Step 1: Convert concentration.
0.2 M means 0.2 mol per 1000 cm$^3$, so 0.0002 mol per cm$^3$.

Step 2: Divide.
Molar conductivity = conductivity / (moles per cm$^3$).
\[ \Lambda_m = \frac{0.0248}{0.0002} = 124\ \Omega^{-1}\text{cm}^2\text{mol}^{-1} \]

Final Answer:
Option (C) matches. \[ \boxed{124\ \Omega^{-1}\text{cm}^2\text{mol}^{-1}} \]