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Question #44524

Difficulty: 8/10 | 🟢 Free

The Question

An aqueous solution is prepared by dissolving $0.08\text{ mol}$ of $\ce{KMnO4}$ and $0.03\text{ mol}$ of $\ce{Na2S2O3}$ in $1\text{ kg}$ of pure water. The solution is allowed to react completely at $298\text{ K}$ in a neutral medium. After the reaction is complete, the solid precipitate formed is filtered off. Assuming complete dissociation of all soluble electrolytes and neglecting any change in the mass of water, what is the depression in freezing point ( $\Delta T_f$ ) of the resulting filtrate?

(Given: $K_f$ of water = $1.86\text{ K kg mol}^{-1}$ )

Options

A.
$0.409\text{ K}$
B.
$0.465\text{ K}$
C.
$0.493\text{ K}$
D.
$0.521\text{ K}$

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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

Colligative Property Definition|KMnO₄ Redox pH Dependency|Degree of Dissociation (Alpha)

Why This Question Is Tricky

Common Mistakes

The problem tests the integration of redox stoichiometry in specific pH mediums with colligative properties.

The 'hidden variable' is the pH-dependent oxidation state of the products.

Students often mistakenly assume that thiosulfate is oxidized to tetrathionate ( $\ce{S4O6^{2-}}$ ) as it does with weak oxidants like iodine, or they apply the acidic medium n-factor (5) for permanganate.

Furthermore, balancing the reaction in neutral medium reveals the generation of $\ce{KOH}$ , which contributes to the total solute particle count.

Failing to account for the exact stoichiometry and the resulting $\ce{OH-}$ ions leads to incorrect Van't Hoff factor applications.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Solutions - Class 12
Concept	Colligative Property Definition
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free