JEE Chemistry 2023 Question with Solution

Given: Resistivity $\rho = 5 \times 10^{-3} \, \Omega \, \text{cm}$ and molarity $M = 0.8 \, \text{M}$.

Find: Molar conductivity in the form _____ \times 10^4 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}.

Use the relation between conductivity and resistivity:

$$\kappa = \frac{1}{\rho}$$

and then

$$\Lambda_m = \frac{\kappa \times 1000}{M}$$

Substituting $\rho = 5 \times 10^{-3}$ and $M = 0.8$,

$$\Lambda_m = \frac{1}{5 \times 10^{-3}} \times \frac{1000}{0.8}$$ $$= 200 \times 1250$$ $$= 250000 = 25 \times 10^4 \; \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}$$

Therefore, the required nearest integer is 25.

Given: $\rho = 5 \times 10^{-3} \, \Omega \, \text{cm}$ and $M = 0.8 \, \text{M}$.

Find: The coefficient multiplying $10^4$ in molar conductivity.

From the extracted solution,

$$\Lambda_m = \frac{\kappa \times 1000}{M}$$

with

$$\kappa = \frac{1}{\rho}$$

Hence,

$$\Lambda_m = \frac{1}{\rho} \times \frac{1000}{M}$$

Now substitute the values:

$$\Lambda_m = \frac{1}{5 \times 10^{-3}} \times \frac{1000}{0.8}$$ $$= \frac{1000}{5 \times 10^{-3}} \times 0.8$$

This evaluates to

$$\Lambda_m = 25 \times 10^4 \; \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}$$

So, the correct numerical answer is 25.