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Question #44075

Difficulty: 8/10 | 🟢 Free

The Question

An aqueous solution of a weak monoprotic acid \ce{HA} (molar mass = $350 \text{ g mol}^{-1}$ ) has a molality of $2.0 \text{ m}$ and a density of $1.02 \text{ g mL}^{-1}$ at $298 \text{ K}$ . The acid dissociation constant $K_a$ of \ce{HA (aq)} at this temperature is $0.1$ . Given that the cryoscopic constant of water is $K_f = 2.0 \text{ K kg mol}^{-1}$ , what is the freezing point depression ( $\Delta T_f$ ) of the solution? (Assume the standard definition of $K_a$ based on molarity, and that activity coefficients are unity.)

Options

A.
$5.0 \text{ K}$
B.
$4.8 \text{ K}$
C.
$3.0 \text{ K}$
D.
$4.0 \text{ K}$

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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

Degree of Dissociation (α) approximation|Van’t Hoff Factor (i) Application|Concentration Units Dependency

Why This Question Is Tricky

Common Mistakes

This problem tests the rigorous application of concentration units across different thermodynamic contexts.

The 'Hidden Variable' is the discrepancy between molarity and molality in concentrated solutions. $K_a$ is an equilibrium constant defined relative to a standard state of $1 \text{ M}$ , thus requiring Molarity to find the true degree of dissociation ( $\alpha$ ).

Conversely, freezing point depression is a colligative property derived from the solvent's mole fraction, requiring Molality.

Students commonly fall into the trap of using a single concentration unit for the entire problem—either using molality to calculate $\alpha$ (yielding $4.8 \text{ K}$ ) or using molarity to calculate $\Delta T_f$ (yielding $3.0 \text{ K}$ ).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Equilibrium - Class 11 Part 1
Concept	Degree of Dissociation (α) approximation
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free