JEE Chemistry 2026 Question with Solution

Given: For a strong electrolyte,

$$\Lambda_m = \Lambda_m^0 - A\sqrt{c}$$

The given data are molar conductivity values at different concentrations.

Find: The value of constant $A$.

Concept: For strong electrolytes, a plot of $\Lambda_m$ versus $\sqrt{c}$ is linear. From

$$\Lambda_m = \Lambda_m^0 - A\sqrt{c}$$

the magnitude of the slope is $A$.

Step 1: Tabulate values of $\sqrt{c}$.

$$\sqrt{0.04} = 0.2, \quad \sqrt{0.09} = 0.3, \quad \sqrt{0.16} = 0.4, \quad \sqrt{0.25} = 0.5$$

Step 2: Use two data points to calculate the slope. Take $c = 0.04$ and $c = 0.25$.

$$\Lambda_m(0.04) = 96.1, \quad \Lambda_m(0.25) = 94.9$$

So,

$$\Delta \Lambda_m = 96.1 - 94.9 = 1.2$$ $$\Delta \sqrt{c} = 0.5 - 0.2 = 0.3$$

Hence,

$$A = \frac{\Delta \Lambda_m}{\Delta \sqrt{c}} = \frac{1.2}{0.3} = 4.0$$

Step 3: Verify with another pair of points. Between $c = 0.09$ and $c = 0.16$,

$$\Delta \Lambda_m = 95.7 - 95.3 = 0.4, \quad \Delta \sqrt{c} = 0.4 - 0.3 = 0.1$$ $$A = \frac{0.4}{0.1} = 4.0$$

Conclude: Therefore, the constant is $A = 4.0$, so the numerical answer is 4.

Given:

$$\Lambda_m = \Lambda_m^0 - A\sqrt{c}$$

Find: The value of $A$.

Because the relation is linear in $\sqrt{c}$, any two points can be used directly. Convert only the endpoint concentrations:

$$\sqrt{0.04} = 0.2, \quad \sqrt{0.25} = 0.5$$

Now compare the corresponding molar conductivities:

$$96.1 - 94.9 = 1.2$$

So,

$$A = \frac{1.2}{0.5 - 0.2} = \frac{1.2}{0.3} = 4$$

This works because $\Lambda_m$ changes linearly with $\sqrt{c}$, so the slope remains constant.

Conclude: The numerical answer is 4.