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Question #44535

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The Question

A $33.3 \text{ g}$ sample of a non-reducing tetrasaccharide ( $\ce{C24H42O21}$ , molar mass = $666 \text{ g mol}^{-1}$ ) is dissolved in $100.0 \text{ g}$ of water. The solution is subjected to complete hydrolysis using a catalytic amount of an insoluble solid acid resin, which is subsequently filtered off. The solution is then allowed to stand at $298 \text{ K}$ until mutarotation equilibrium is reached.

The specific rotation of pure $\alpha$ -D-glucose is $+112^\circ$ and pure $\beta$ -D-glucose is $+19^\circ$ . The final equilibrium mixture exhibits a specific rotation of $+52.6^\circ$ .

Assuming ideal behavior, what is the freezing point depression ( $\Delta T_f$ ) of the final equilibrium solution?

(Given: $K_f$ of water = $1.86 \text{ K kg mol}^{-1}$ , Molar mass of water = $18 \text{ g mol}^{-1}$ )

Options

A.
$3.82 \text{ K}$
B.
$3.72 \text{ K}$
C.
$7.65 \text{ K}$
D.
$2.44 \text{ K}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Freezing Point Depression (Pure Solvent)|Reducing Sugar Criteria|Epimerization vs Anomerization

Why This Question Is Tricky

Common Mistakes

This question tests two major hidden variables.

First, the 'Stoichiometric Solvent Consumption' trap: During the hydrolysis of higher oligosaccharides, a significant mass of the solvent (water) is consumed to break the glycosidic bonds.

Failing to subtract this mass from the initial solvent mass leads to an incorrect, lower molality (Option B).

Second, the 'Biomolecular Equilibrium' trap: Students often assume that because a sugar undergoes mutarotation to form an equilibrium mixture of alpha and beta anomers, the number of solute particles increases (analogous to dissociation, $i > 1$ ), or they erroneously use the specific rotation data to calculate the concentration of only the major anomer (Option D).

In reality, isomerization does not change the total number of moles of solute; the equilibrium mixture acts with an effective van't Hoff factor of $i=1$ .

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Solutions - Class 12
Concept	Freezing Point Depression (Pure Solvent)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free