JEE Chemistry 2023 Question with Solution

Given: Degree of dissociation of weak monobasic acid is $\alpha = 0.3$.

Find: The percentage by which the observed freezing point effect differs from the expected/theoretical value.

For a weak monobasic acid:

$$i = 1 + \alpha$$

Substituting $\alpha = 0.3$,

$$i = 1 + 0.3 = 1.3$$

The observed depression in freezing point is:

$$(\Delta T_f)_{\text{obs}} = iK_f m = 1.3K_f m$$

The theoretical value without dissociation is:

$$(\Delta T_f)_{\text{cal}} = K_f m$$

Hence,

$$\%\,\text{increase} = \frac{(\Delta T_f)_{\text{obs}} - (\Delta T_f)_{\text{cal}}}{(\Delta T_f)_{\text{cal}}} \times 100$$ $$= \frac{1.3K_f m - K_f m}{K_f m} \times 100$$ $$= \frac{0.3}{1} \times 100 = 30\%$$

Therefore, the required nearest integer is 30.

Given: Weak monobasic acid in aqueous solution with degree of dissociation $\alpha = 0.3$.

Find: Percentage increase in observed freezing point depression over the theoretical value for no dissociation.

A weak monobasic acid dissociates as:

$$\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$$

If the initial number of particles is taken as $1$, then after dissociation the total number of particles becomes:

$$1 - \alpha + \alpha + \alpha = 1 + \alpha$$

So the van't Hoff factor is:

$$i = 1 + \alpha = 1.3$$

Now freezing point depression is proportional to $i$:

$$\Delta T_f = iK_f m$$

Thus,

$$\Delta T_f(\text{obs}) = 1.3K_f m$$

and

$$\Delta T_f(\text{theoretical}) = 1.0K_f m$$

Therefore, the percentage higher value is:

$$\frac{1.3K_f m - 1.0K_f m}{1.0K_f m} \times 100 = 30\%$$

Therefore, the answer is 30.