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MCQEasyJEE 2024Colligative Properties

JEE Chemistry 2024 Question with Solution

If the boiling point of a solution containing 11 mole of glucose in 1000 g1000 \, \text{g} of water is 100.52∘C100.52^\circ \text{C}, the ebullioscopic constant (KbK_b) of water is:

  • A

    0.52 K kg/mol0.52 \, \text{K kg/mol}

  • B

    1.52 K kg/mol1.52 \, \text{K kg/mol}

  • C

    2.52 K kg/mol2.52 \, \text{K kg/mol}

  • D

    3.52 K kg/mol3.52 \, \text{K kg/mol}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The solution contains 11 mole of glucose in 1000 g1000 \, \text{g} of water, and the boiling point is 100.52∘C100.52^\circ \text{C}.

Find: The ebullioscopic constant KbK_b of water.

Using the boiling point elevation relation:

ΔTb=iKbm\Delta T_b = iK_b m

For glucose, i=1i = 1 because it is a non-electrolyte. The mass of solvent is 1000 g=1 kg1000 \, \text{g} = 1 \, \text{kg}, so the molality is:

m=11=1 mol kg−1m = \frac{1}{1} = 1 \, \text{mol kg}^{-1}

The elevation in boiling point is:

ΔTb=100.52−100=0.52 K\Delta T_b = 100.52 - 100 = 0.52 \, \text{K}

Substituting into the formula:

0.52=1⋅Kb⋅10.52 = 1 \cdot K_b \cdot 1

Therefore,

Kb=0.52 K kg/molK_b = 0.52 \, \text{K kg/mol}

The correct option is A. Final answer: 0.52 K kg/mol0.52 \, \text{K kg/mol}.

Common mistakes

  • Using the total mass of solution instead of the mass of solvent for molality is incorrect. Molality is defined using kilograms of solvent only. Here, use 1000 g1000 \, \text{g} of water as 1 kg1 \, \text{kg} solvent.

  • Forgetting that glucose is a non-electrolyte leads to taking i≠1i \neq 1, which is wrong. Since glucose does not dissociate in water, use i=1i = 1.

  • Taking the boiling point elevation as 100.52100.52 instead of 0.520.52 is incorrect. The elevation is the increase over the normal boiling point of water, so compute 100.52−100=0.52100.52 - 100 = 0.52.

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