MCQMediumJEE 2026Colligative Properties

JEE Chemistry 2026 Question with Solution

Consider the following aqueous solutions.

I. 2.2g2.2 \, \text{g} Glucose in 125mL125 \, \text{mL} of solution. II. 1.9g1.9 \, \text{g} Calcium chloride in 250mL250 \, \text{mL} of solution. III. 9.0g9.0 \, \text{g} Urea in 500mL500 \, \text{mL} of solution. IV. 20.5g20.5 \, \text{g} Aluminium sulphate in 750mL750 \, \text{mL} of solution.

The correct increasing order of boiling point of these solutions will be: [Given: Molar mass in g mol1\text{g mol}^{-1}: H = 11, C = 1212, N = 1414, O = 1616, Cl = 35.535.5, Ca = 4040, Al = 2727 and S = 3232]

  • A

    I < III < IV < II

  • B

    III < I < II < IV

  • C

    I < II < III < IV

  • D

    III < II < I < IV

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Four aqueous solutions of glucose, calcium chloride, urea and aluminium sulphate are given.

Find: The increasing order of boiling point.

Concept: Elevation in boiling point is a colligative property and depends on effective molality:

ΔTb=iKbm\Delta T_b = i K_b m

where ii is the van't Hoff factor and mm is the molality.

So, compare imi m for all four solutions.

I. Glucose is a non-electrolyte, so i=1i = 1.

Molar mass=180g mol1\text{Molar mass} = 180 \, \text{g mol}^{-1} Moles=2.21800.0122\text{Moles} = \frac{2.2}{180} \approx 0.0122 Molality=0.01220.1250.0976\text{Molality} = \frac{0.0122}{0.125} \approx 0.0976

Hence, effective molality =0.0976= 0.0976.

II. Calcium chloride, CaCl2\mathrm{CaCl_2} dissociates into Ca2+\mathrm{Ca^{2+}} and 2Cl2\mathrm{Cl^-}, so i=3i = 3.

Molar mass=111g mol1\text{Molar mass} = 111 \, \text{g mol}^{-1} Moles=1.91110.0171\text{Moles} = \frac{1.9}{111} \approx 0.0171 Molality=0.01710.250.0684\text{Molality} = \frac{0.0171}{0.25} \approx 0.0684 Effective molality=3×0.06840.205\text{Effective molality} = 3 \times 0.0684 \approx 0.205

III. Urea is a non-electrolyte, so i=1i = 1.

Molar mass=60g mol1\text{Molar mass} = 60 \, \text{g mol}^{-1} Moles=960=0.15\text{Moles} = \frac{9}{60} = 0.15 Molality=0.150.5=0.30\text{Molality} = \frac{0.15}{0.5} = 0.30

Hence, effective molality =0.30= 0.30.

IV. Aluminium sulphate, Al2(SO4)3\mathrm{Al_2(SO_4)_3} dissociates into 2Al3+2\mathrm{Al^{3+}} and 3SO423\mathrm{SO_4^{2-}}, so i=5i = 5.

Molar mass=342g mol1\text{Molar mass} = 342 \, \text{g mol}^{-1} Moles=20.53420.0599\text{Moles} = \frac{20.5}{342} \approx 0.0599 Molality=0.05990.750.0799\text{Molality} = \frac{0.0599}{0.75} \approx 0.0799 Effective molality=5×0.07990.399\text{Effective molality} = 5 \times 0.0799 \approx 0.399

Now compare the effective molalities:

I=0.0976,II=0.205,III=0.30,IV=0.399\text{I} = 0.0976, \quad \text{II} = 0.205, \quad \text{III} = 0.30, \quad \text{IV} = 0.399

Since boiling point increases with increasing ΔTb\Delta T_b, the increasing order of boiling point is:

I<II<III<IV\text{I} < \text{II} < \text{III} < \text{IV}

Therefore, the correct option is C.

The solution marks option B, but its own working and final order clearly give I < II < III < IV, which matches option C.

Common mistakes

  • Using only molality and ignoring the van't Hoff factor. This is wrong because boiling point elevation depends on the total number of dissolved particles. Always compare i×mi \times m, not only mm.

  • Assuming all solutes behave as non-electrolytes. This is wrong because calcium chloride and aluminium sulphate dissociate into multiple ions in water. Count the ions correctly to get i=3i = 3 for CaCl2\mathrm{CaCl_2} and i=5i = 5 for Al2(SO4)3\mathrm{Al_2(SO_4)_3}.

  • Calculating molarity instead of molality. This is wrong because colligative property relations here use molality. Divide moles of solute by mass of solvent in kilograms as done in the solution.

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