JEE Chemistry 2025 Question with Solution

Given: resistance for the higher concentration solution is $100 \, \Omega$. From the graph, the molar conductivity values are $100$ and $150$ for the concentrated and dilute solutions respectively.

Find: the resistance $x$ of the same cell for the dilute solution.

For the same conductivity cell, the cell constant remains unchanged. Using

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

and

$$\kappa = \frac{1}{R} \times \text{cell constant}$$

we use the values shown in the extracted working.

For higher concentration:

$$100 = \frac{K \times 1000}{0.0225}$$

So,

$$K = \frac{0.0225}{10}$$

Also,

$$K = \frac{1}{R} \times \frac{\ell}{A}$$

With $R = 100 \, \Omega$,

$$\frac{\ell}{A} = \frac{0.0225}{10} \times 100 = 0.225$$

For lower concentration:

$$150 = \frac{K \times 1000}{0.01}$$

So,

$$K = \frac{0.15}{100}$$

Again,

$$K = \frac{1}{R} \times \frac{\ell}{A}$$

Hence,

$$\frac{0.15}{100} = \frac{1}{R} \times 0.225$$

Therefore,

$$R = \frac{22.5}{0.15} = \frac{2250}{15} = 150 \, \Omega$$

So, the value of $x$ is $150$.

Given: the same conductivity cell is used, and the resistance at higher concentration is $100 \, \Omega$.

Find: the resistance for the dilute solution.

From the graph:

• at higher concentration, molar conductivity is $100$
• at lower concentration, molar conductivity is $150$

For the same cell, conductivity is inversely proportional to resistance, and the graphical comparison used in the provided solution gives a ratio of $1.5$ between dilute and concentrated cases.

Thus,

$$\frac{R_{\text{dilute}}}{R_{\text{concentrated}}} = \frac{150}{100} = 1.5$$

So,

$$R_{\text{dilute}} = 1.5 \times 100 = 150 \, \Omega$$

Therefore, $x = 150$.