Question

What is the molar conductivity of 0.1 M NaCl if its conductivity is \(1.06 \times 10^{-2} \, \Omega^{-1} \, \text{cm}^{-1}\text{cm}^{-1}\)?

• A. \(1.06 \times 10^{2} \, \Omega^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\)
• B. \(1.06 \times 10^{-2} \, \Omega^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\)
• C. \(9.4 \times 10^{-2} \, \Omega^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\)
• D. \(5.3 \times 10^{3} \, \Omega^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\)

Hint:

Molar conductivity is calculated by dividing the conductivity by the concentration of the solution.

Answer 1

The Correct Option is A

Step 1: Understanding Molar Conductivity.
Molar conductivity (\(\lambda_m\)) is related to the conductivity (\(\kappa\)) by the equation: \[ \lambda_m = \frac{\kappa}{c} \] where \(c\) is the concentration of the solution. In this case, the concentration of NaCl is 0.1 M, and the conductivity is \(1.06 \times 10^{-2} \, \Omega^{-1} \, \text{cm}^{-1}\text{cm}^{-1}\). Thus, the molar conductivity is: \[ \lambda_m = \frac{1.06 \times 10^{-2}}{0.1} = 1.06 \times 10^{2} \, \Omega^{-1} \, \text{cm}^{2} \, \text{mol}^{-1} \]
Step 2: Analyzing the options.
(A) \(1.06 \times 10^{2} \, \Omega^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\): Correct. This is the correct molar conductivity.
(B) \(1.06 \times 10^{-2} \, \Omega^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\): Incorrect. This is the given conductivity, not the molar conductivity.
(C) \(9.4 \times 10^{-2} \, \Omega^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\): Incorrect. This does not match the correct molar conductivity.
(D) \(5.3 \times 10^{3} \, \Omega^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\): Incorrect. This is too large.

Step 3: Conclusion.
The correct molar conductivity is \(1.06 \times 10^{2} \, \Omega^{-1} \, \text{cm}^{2} \, \text{mol}^{-1}\), corresponding to option (A).