If the degree of dissociation of aqueous solution of weak monobasic acid is determined to be 0.3, then the observed freezing point will be \underline\hspace2cm % higher than the expected/theoretical freezing point. (Nearest integer)

Solutions PYQ — JEE Main 2023 · 10 Apr · Shift 1 · Q23 with Solution

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Correct answer: 30

For a solute in solution,

\Delta T_f = iK_f m

where:

A weak monobasic acid dissociates as:

HA \rightleftharpoons H^+ + A^-

If the degree of dissociation is $\alpha = 0.3$ , then starting from 1 mole of $HA$ :

So total particles in solution:

(1-\alpha) + 2\alpha = 1+\alpha

Hence,

i = 1+\alpha = 1+0.3 = 1.3

If dissociation were ignored, the theoretical freezing point depression would be:

\Delta T_f^{\text{theoretical}} = K_f m

Observed freezing point depression is:

\Delta T_f^{\text{observed}} = 1.3K_f m

Thus observed depression is greater by:

\frac{1.3K_f m - K_f m}{K_f m}\times 100 = 0.3 \times 100 = 30\%

Since the freezing point itself decreases by this extra amount, the observed freezing point depression is $30\%$ higher than the theoretical value.

Therefore, the required percentage is:

\boxed{30}