Solutions: Conceptual Doubts Cleared (6)

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Question

A solution of 1.51.5 g of a non-volatile solute in 5050 g of water boils at 100.39°100.39°C. Calculate the molar mass of the solute. Given KbK_b of water =0.52= 0.52 K kg/mol.

Solution — Step by Step

Boiling-point elevation ΔTb=Tb(solution)−Tb(pure water)=100.39−100=0.39\Delta T_b = T_b(\text{solution}) - T_b(\text{pure water}) = 100.39 - 100 = 0.39 K.

 ΔTb=Kb⋅m\Delta T_b = K_b \cdot m

where mm is molality (mol solute per kg solvent).

m=ΔTbKb=0.390.52=0.75 mol/kgm = \frac{\Delta T_b}{K_b} = \frac{0.39}{0.52} = 0.75 \text{ mol/kg}

Mass of solvent =50= 50 g =0.050= 0.050 kg.

nsolute=m×kg solvent=0.75×0.050=0.0375 moln_{\text{solute}} = m \times \text{kg solvent} = 0.75 \times 0.050 = 0.0375 \text{ mol} M=mass of solutemoles of solute=1.50.0375=40 g/molM = \frac{\text{mass of solute}}{\text{moles of solute}} = \frac{1.5}{0.0375} = 40 \text{ g/mol}

Final Answer: Molar mass =40= 40 g/mol.

Why This Works

Boiling-point elevation is a colligative property — it depends only on the number of solute particles, not their identity. So measuring ΔTb\Delta T_b directly gives us moles of solute (via molality), and then molar mass follows from the known mass.

The same logic underpins freezing-point depression (ΔTf=Kf⋅m\Delta T_f = K_f \cdot m) and osmotic pressure (π=MRT\pi = MRT). Pick whichever is easiest to measure for a given solute.

Alternative Method

Combine all steps into one formula:

M=1000⋅Kb⋅wsoluteΔTb⋅wsolvent (g)=1000×0.52×1.50.39×50=40 g/molM = \frac{1000 \cdot K_b \cdot w_{\text{solute}}}{\Delta T_b \cdot w_{\text{solvent (g)}}} = \frac{1000 \times 0.52 \times 1.5}{0.39 \times 50} = 40 \text{ g/mol}

Faster for MCQs once you trust the formula.

Forgetting to convert the solvent mass to kilograms is the most common error. Molality is mol/kg, not mol/g. A factor of 10001000 wrong gives a molar mass off by 1000×1000\times.

If the solute dissociates (like NaCl or CaCl2_2), use the van’t Hoff factor: ΔTb=i⋅Kb⋅m\Delta T_b = i \cdot K_b \cdot m. NEET often asks the same problem with an electrolyte to test whether you remember ii.

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