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Question #44547

Difficulty: 8/10 | 🟢 Free

The Question

A $2.665 \text{ g}$ sample of a hydrate isomer of $\ce{CrCl3 \cdot 6H2O}$ (Molar mass = $266.5 \text{ g mol}^{-1}$ ) is dissolved in water and passed through a cation-exchange resin in the $\ce{H+}$ form. The effluent requires $20.0 \text{ mL}$ of $1.0 \text{ M NaOH}$ for complete neutralization. A separate $0.10 \text{ m}$ aqueous solution of this specific isomer exhibits a degree of dissociation $\alpha = 0.85$ . Assuming no hydrolysis of the complex cation, what is the freezing point of this $0.10 \text{ m}$ solution? (Given: $K_f$ for water = $1.86 \text{ K kg mol}^{-1}$ )

Options

A.
$-0.502 ^\circ\text{C}$
B.
$-0.660 ^\circ\text{C}$
C.
$-0.344 ^\circ\text{C}$
D.
$-0.558 ^\circ\text{C}$

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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/Association|Hydrate and Ionization Isomerism|Van’t Hoff Factor (i) Application

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Coordination Chemistry (Werner's Theory/Hydrate Isomerism) and Colligative Properties.

The 'Hidden Variable' is the charge of the complex cation, which dictates the stoichiometry of dissociation ( $n$ ).

Students often blindly assume $\ce{CrCl3 \cdot 6H2O}$ dissociates into 4 ions ( $\ce{Cr^3+}$ and $3\ce{Cl-}$ ), ignoring the coordination sphere, or they misinterpret the cation exchange stoichiometry.

The trap also includes the partial dissociation factor ( $\alpha$ ), preventing the simple use of integer Van't Hoff factors.

Option B assumes $n=4$ , Option C assumes $n=2$ , and Option D assumes complete dissociation ( $\alpha=1$ ).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Solutions - Class 12
Concept	Degree of Dissociation/Association
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free