Question

What will be the concentration of solution of electrolyte if its molar conductivity and conductivity are respectively 230 \(\Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}\) and 0.0115 \(\Omega^{-1} \, \text{cm}^{-1}\) at 298 K?

• A. 0.04 M
• B. 0.03 M
• C. 0.01 M
• D. 0.05 M

Hint:

Always ensure units are consistent: \(\kappa\) in S/cm, \(\Lambda_m\) in S cm²/mol gives \(C\) in mol/L. The factor 1000 converts cm³ to L.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
Molar conductivity (\(\Lambda_m\)) and conductivity (\(\kappa\)) are given. We need the concentration (\(C\)) of the electrolyte solution.

Step 2: Key Formula or Approach:
The relationship is \(\Lambda_m = \frac{\kappa \times 1000}{C}\), where \(\Lambda_m\) is in \(\Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}\), \(\kappa\) in \(\Omega^{-1} \, \text{cm}^{-1}\), and \(C\) in mol/L.

Step 3: Detailed Explanation:
Rearrange: \(C = \frac{\kappa \times 1000}{\Lambda_m}\). Substitute values: \(C = \frac{0.0115 \times 1000}{230} = \frac{11.5}{230} = 0.05 \, \text{mol/L}\).

Step 4: Final Answer:
Concentration is 0.05 M, option (D).