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Question #44533

Difficulty: 8/10 | 🟢 Free

The Question

When $2.44 \text{ g}$ of benzoic acid ( $\ce{C6H5COOH}$ ) is dissolved in $100 \text{ g}$ of benzene ( $\ce{C6H6}$ ), the freezing point of the solution is depressed by $0.6144 \text{ K}$ . Assuming benzoic acid undergoes dimerization in benzene, calculate the equilibrium constant $K_m$ (in $\text{kg mol}^{-1}$ ) for the dimerization process.

(Given: $K_f$ for benzene $= 5.12 \text{ K kg mol}^{-1}$ , Molar mass of benzoic acid $= 122 \text{ g mol}^{-1}$ )

Options

A.
50
B.
10
C.
100
D.
25

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Freezing Point Depression (Pure Solvent)|Degree of Dissociation/Association|Inductive Effect Trends

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Colligative Properties and Chemical Equilibrium.

The 'Hidden Variable' is the solvent-dependent aggregation: benzoic acid dimerizes in non-polar benzene due to intermolecular hydrogen bonding, making $i < 1$ .

The primary trap lies in the calculation of the equilibrium constant $K_m$ .

Students frequently memorize dimensionless formulas like $K = \frac{\alpha}{2(1-\alpha)^2}$ (yielding 10), forgetting that the equilibrium constant for $2A \rightleftharpoons A_2$ is dimensionally dependent on the initial concentration ( $1/m$ factor).

This forces the student to derive the actual equilibrium concentrations rather than relying on rote formulas.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Solutions - Class 12
Concept	Freezing Point Depression (Pure Solvent)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free