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: Read what is given.
We are told the molar conductivity of a $0.4\,\text{M}$ KCl solution is $\Lambda_m = 2.5 \times 10^5\ \Omega^{-1}\,\text{cm}^2\,\text{mol}^{-1}$. We have to find the resistivity.

Step 2: Decide the path.
Resistivity is just one over conductivity, so first we need the conductivity $\kappa$. We will get $\kappa$ from the molar conductivity, then flip it.

Step 3: Recall the link between the two conductivities.
The handy relation is $\Lambda_m = \dfrac{1000\,\kappa}{C}$, so turning it around gives $\kappa = \dfrac{\Lambda_m \times C}{1000}$. Here $C = 0.4\,\text{M}$.

Step 4: Plug in the numbers.
\[ \kappa = \frac{2.5 \times 10^5 \times 0.4}{1000} = \frac{1.0 \times 10^5}{1000} = 100\ \Omega^{-1}\,\text{cm}^{-1} \]
Step 5: Flip it to get resistivity.
Resistivity is the reciprocal of conductivity, so \[ \rho = \frac{1}{\kappa} = \frac{1}{100} = 0.01\ \Omega\,\text{cm} \]
Step 6: Write it in clean form and pick the option.
In powers of ten, $0.01 = 1 \times 10^{-2}$. This matches option (C).
\[ \boxed{\rho = 1 \times 10^{-2}\ \Omega\,\text{cm}} \]