Solutions: Edge Cases and Subtle Traps (1)

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Question

A solution is prepared by dissolving 18 g18 \text{ g} of glucose (M=180 g/molM = 180 \text{ g/mol}) in 200 g200 \text{ g} of water. Calculate the molality and the depression in freezing point. KfK_f for water =1.86 K kg/mol= 1.86 \text{ K kg/mol}. Then explain a subtle JEE trap: why molality (not molarity) is used in colligative-property calculations.

Solution — Step by Step

Moles of glucose:

n=18180=0.1 moln = \frac{18}{180} = 0.1 \text{ mol}

Mass of solvent in kg:

msolvent=0.2 kgm_{\text{solvent}} = 0.2 \text{ kg}

Molality:

m=nsolutemsolvent=0.10.2=0.5 mol/kgm = \frac{n_{\text{solute}}}{m_{\text{solvent}}} = \frac{0.1}{0.2} = 0.5 \text{ mol/kg}

ΔTf=Kf⋅m⋅i\Delta T_f = K_f \cdot m \cdot i

For glucose (a non-electrolyte), van’t Hoff factor i=1i = 1:

ΔTf=1.86×0.5×1=0.93 K\Delta T_f = 1.86 \times 0.5 \times 1 = 0.93 \text{ K}

So the solution freezes at 0−0.93=−0.93 °C0 - 0.93 = -0.93 \text{ °C}.

Colligative properties (freezing point depression, boiling point elevation, osmotic pressure, vapor pressure lowering) depend on the number ratio of solute particles to solvent molecules. Molality measures moles of solute per kg of solvent — a quantity that doesn’t change with temperature.

Molarity uses volume of solution, which expands or contracts with temperature. So molarity drifts as the solution heats up, but molality stays constant. For freezing/boiling experiments where temperature changes significantly, only molality gives reliable results.

Why This Works

Each colligative property formula has the same structure: Δ(property)=K⋅m⋅i\Delta(\text{property}) = K \cdot m \cdot i, where KK is a solvent-specific constant, mm is molality, and ii is the van’t Hoff factor accounting for dissociation.

Glucose doesn’t dissociate, so i=1i = 1. NaCl dissociates into two ions, so i=2i = 2 (theoretically; slightly less in reality). CaCl2\text{CaCl}_2 gives i=3i = 3. NEET often hides this distinction in the problem statement — “NaCl” means i=2i = 2, period.

Alternative Method

You can compute the depression directly via the molal depression formula written out: ΔTf=Kf⋅(nsolute/msolvent in kg)=(1.86 K kg/mol)(0.1 mol)/(0.2 kg)=0.93 K\Delta T_f = K_f \cdot (n_{\text{solute}}/m_{\text{solvent in kg}}) = (1.86 \text{ K kg/mol})(0.1 \text{ mol})/(0.2 \text{ kg}) = 0.93 \text{ K}. Skips the intermediate molality calculation. Useful for quick MCQs.

Memorise: KfK_f (water) =1.86= 1.86, KbK_b (water) =0.512= 0.512. These appear in 80%80\% of board exam colligative-property problems. The other solvent values (benzene, camphor) are usually given in the problem.

Common Mistake

Three traps in this template:

  1. Using molarity instead of molality. They differ for dilute aqueous solutions by only a few percent (because 1 kg1 \text{ kg} of water ≈1 L\approx 1 \text{ L}), but for concentrated or non-aqueous solutions they differ by much more. Colligative properties always use molality.

  2. Forgetting the van’t Hoff factor for ionic solutes. For NaCl, i≈2i \approx 2. If you treat NaCl as a non-electrolyte, you’ll get half the right depression.

  3. Mass of solvent vs mass of solution. Molality uses mass of solvent only — so 200 g200 \text{ g} in this problem (water alone), not 218 g218 \text{ g} (water + glucose). For dilute solutions the difference is small; for concentrated solutions it’s large.

Final answer: Molality =0.5 mol/kg= 0.5 \text{ mol/kg}, ΔTf=0.93 K\Delta T_f = 0.93 \text{ K}, freezing point ≈−0.93 °C\approx -0.93 \text{ °C}.

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