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Question #44686

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The Question

A saturated solution of \ce{K2SO4} is prepared by dissolving exactly $17.426 \text{ g}$ of the salt in $1.000 \text{ kg}$ of pure water at $298 \text{ K}$ . To this solution, $7.455 \text{ g}$ of solid \ce{KCl} is added. Assuming that the addition of \ce{KCl} does not change the mass of the solvent and that the solution behaves ideally, calculate the freezing point depression ( $\Delta T_f$ ) of the final solution at equilibrium.

Given: $K_f$ of water = $1.86 \text{ K kg mol}^{-1}$ Exact atomic masses (in amu): $\text{K} = 39.10$ , $\text{S} = 32.06$ , $\text{O} = 16.00$ , $\text{Cl} = 35.45$

Options

A.
$0.761 \text{ K}$
B.
$0.762 \text{ K}$
C.
$0.930 \text{ K}$
D.
$0.602 \text{ K}$

Check Your Answer

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

Precision in Atomic Mass Units|Freezing Point Depression (Pure Solvent)|Common Ion Effect on Solubility

Why This Question Is Tricky

Common Mistakes

The problem tests the precise calculation of colligative properties coupled with the common ion effect.

The primary trap is the reliance on exact atomic masses.

Students who approximate atomic masses (e.g., K=39, S=32, Cl=35.5) will calculate an incorrect initial molality ( $0.10015 \text{ m}$ ) and $K_{sp}$ ( $4.018 \times 10^{-3}$ ), leading to a final $\Delta T_f$ of $0.762 \text{ K}$ .

Additional traps include incorrectly formulating the $K_{sp}$ expression by forgetting the stoichiometric coefficient inside the squared term for potassium, which leads to $0.930 \text{ K}$ , or ignoring the stoichiometry entirely (treating it as a 1:1 salt), leading to $0.602 \text{ K}$ .

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Some Basic Concepts of Chemistry - Class 11 Part 1
Concept	Precision in Atomic Mass Units
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free