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Question #44844

Difficulty: 8/10 | 🟢 Free

The Question

A $0.6 \text{ mol kg}^{-1}$ aqueous solution of a radioactive weak monoprotic acid \ce{HA} ( $K_a = 0.1$ ) has an initial freezing point depression of $\Delta T_{f,0}$ . \ce{HA} undergoes first-order radioactive decay with a half-life $t_{1/2}$ to form a stable, non-volatile, non-electrolyte product \ce{B}, which remains completely dissolved in the solution. Assuming that the decay product \ce{B} does not interfere with the acidic dissociation of \ce{HA}, what is the ratio of the freezing point depression of the solution after a time $t = 3 t_{1/2}$ to its initial freezing point depression ( $\frac{\Delta T_{f,t}}{\Delta T_{f,0}}$ )?

Options

A.
$\frac{13}{16}$
B.
$\frac{25}{32}$
C.
$\frac{9}{16}$
D.
$\frac{5}{32}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Radioactive Decay as Kinetics|Pseudo-First Order Kinetics|Van’t Hoff Factor (i) Application|Freezing Point Depression (Pure Solvent)

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of nuclear kinetics, chemical equilibrium, and colligative properties.

The 'Hidden Variable' is the dynamic nature of the degree of dissociation ( $\alpha$ ).

As the radioactive weak acid decays, its concentration decreases, which according to Ostwald's dilution law, increases its degree of dissociation.

Students commonly fall into the trap of assuming $\alpha$ remains constant (leading to 25/32) or assuming the acid is strong (leading to 9/16).

Furthermore, the approximation $\alpha = \sqrt{K_a/c}$ fails here since $\alpha$ is large, forcing the use of the exact quadratic equation.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Kinetics - Class 12
Concept	Radioactive Decay as Kinetics
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free