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: Get conductivity first.
Molar conductivity links to conductivity by $\Lambda_m = \dfrac{1000\,\kappa}{C}$, so $\kappa = \dfrac{\Lambda_m\, C}{1000}$.

Step 2: Plug in values.
$$\kappa = \frac{2.5 \times 10^5 \times 0.4}{1000} = 100\ \Omega^{-1}\text{cm}^{-1}$$

Step 3: Flip for resistivity.
$$\rho = \frac{1}{\kappa} = \frac{1}{100} = 1 \times 10^{-2}\ \Omega\,\text{cm}$$
\[ \boxed{1 \times 10^{-2}} \]