2g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene lowers its freezing point by 1.62 K. Molal depression constant is 4.9 K kg/mol. The value of ‘i’ when benzoic acid dimerises would be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.504
When a solute associates, the number of particles in solution falls, so the observed colligative property and the van’t Hoff factor i are smaller than expected:
\( \displaystyle i = \frac{\text{normal molar mass}}{\text{abnormal (observed) molar mass}} \)
First the observed molar mass from the freezing point depression (w2 = 2 g, w1 = 25 g, Kf = 4.9 K kg mol−1, ΔTf = 1.62 K):
\( \displaystyle M_2 = \frac{1000\times w_2\times K_f}{\Delta T_f\times w_1} = \frac{1000\times2\times4.9}{1.62\times25} = 241.98\ \mathrm{g\,mol^{-1}} \)
The normal molar mass of benzoic acid, \( \ce{C6H5COOH} \), is 7(12) + 6(1) + 2(16) = 122 g mol−1:
\( \displaystyle i = \frac{122}{241.98} = 0.504 \)
Benzoic acid dimerises in benzene through hydrogen bonding, \( \ce{2C6H5COOH} \) ⇌ \( \ce{(C6H5COOH)2} \), so i is close to 0.5.
Shortcut: For complete dimerisation i = 0.5; only 0.504 is just above it (i = 1 − x/2 with x = 0.992).
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27 edition):
Abnormal Molar Masses – “Example 1.12 2 g of benzoic acid (C6H5COOH) dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molal depression constant for benzene is 4.9 K kg mol−1. What is the percentage association of acid if it forms dimer in solution?
Solution The given quantities are: w2 = 2 g; Kf = 4.9 K kg mol−1; w1 = 25 g, ΔTf = 1.62 K
Substituting these values in equation (1.36) we get:
\( \displaystyle M_2 = \frac{4.9\ \mathrm{K\,kg\,mol^{-1}}\times2\ \mathrm{g}\times1000\ \mathrm{g\,kg^{-1}}}{25\ \mathrm{g}\times1.62\ \mathrm{K}} = 241.98\ \mathrm{g\,mol^{-1}} \)
Thus, experimental molar mass of benzoic acid in benzene is = 241.98 g mol−1
Now consider the following equilibrium for the acid:
2 C6H5COOH ⇌ (C6H5COOH)2
If x represents the degree of association of the solute then we would have (1 – x) mol of benzoic acid left in unassociated form and correspondingly x/2 as associated moles of benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium is: 1 − x + x/2 = 1 − x/2
Thus, total number of moles of particles at equilibrium equals van’t Hoff factor i.
\( \displaystyle i = \frac{\text{Normal molar mass}}{\text{Abnormal molar mass}} = \frac{122\ \mathrm{g\,mol^{-1}}}{241.98\ \mathrm{g\,mol^{-1}}} \)
or \( \displaystyle \frac{x}{2} = 1 - \frac{122}{241.98} = 1 - 0.504 = 0.496 \)
or x = 2 × 0.496 = 0.992
Therefore, degree of association of benzoic acid in benzene is 99.2 %.”