Question

The molar conductivity of $0.4\text{ M KCl}$ solution is $2.5 \times 10^5\ \Omega^{-1}\text{ cm}^2\text{ mol}^{-1}$. What is the resistivity of solution?

• A. $2.1 \times 10^2$
• B. $2.5 \times 10^2$
• C. $1 \times 10^{-2}$
• D. $2.8 \times 10^{-2}$

Hint:

Always double-check your units in electrochemistry problems! The factor of $1000$ in the formula $\Lambda_m = \frac{1000 \kappa}{C}$ is specifically used to convert Molarity (mol/L) into $\text{mol/cm}^3$ so it matches the $\text{cm}^{-1}$ in the conductivity term.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given the molar conductivity ($\Lambda_m$) and the concentration ($C$) of a KCl solution, and we are asked to calculate its resistivity ($\rho$).

Step 2: Key Formula or Approach:
There are two key formulas needed here:
1. Molar conductivity relates to standard conductivity ($\kappa$) via the formula: $\Lambda_m = \frac{1000 \times \kappa}{C}$
2. Resistivity ($\rho$) is the mathematical inverse of conductivity ($\kappa$): $\rho = \frac{1}{\kappa}$

Step 3: Detailed Explanation:
First, we rearrange the molar conductivity formula to solve for $\kappa$:
$$\kappa = \frac{\Lambda_m \times C}{1000}$$
Substitute the given values ($\Lambda_m = 2.5 \times 10^5$ and $C = 0.4\text{ M}$):
$$\kappa = \frac{2.5 \times 10^5 \times 0.4}{1000}$$
$$\kappa = \frac{100,000}{1000}$$
$$\kappa = 100\ \Omega^{-1}\text{ cm}^{-1}$$
Now, calculate the resistivity ($\rho$):
$$\rho = \frac{1}{\kappa}$$
$$\rho = \frac{1}{100}$$
$$\rho = 0.01\ \Omega\text{ cm} = 1 \times 10^{-2}\ \Omega\text{ cm}$$

Step 4: Final Answer:
The resistivity of the solution is $1 \times 10^{-2}$, matching option (C).