ΔTb = Kb × m. Calculate BP elevation of 0.5m glucose in water.

Colligative Properties — Boiling Point Elevation

Question

A solution is prepared by dissolving 18 g of glucose (molecular mass = 180 g/mol) in 100 g of water. Calculate the elevation in boiling point. Given: $K_b$ for water = 0.52 K kg/mol.

Solution — Step by Step

Moles of glucose $= \dfrac{\text{mass}}{\text{molar mass}} = \dfrac{18}{180} = 0.1 \text{ mol}$

We always start here because molality is defined per kg of solvent, not per litre of solution — a distinction that trips up many students.

Mass of water $= 100 \text{ g} = 0.1 \text{ kg}$

The formula uses molality, so the solvent must be in kilograms. Don’t skip this conversion.

m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.1}{0.1} = 1 \text{ mol/kg}

Molality (m) is our concentration unit for colligative properties because unlike molarity, it doesn’t change with temperature — important when we’re talking about boiling points.

\Delta T_b = K_b \times m = 0.52 \times 1 = 0.52 \text{ K}

For non-electrolytes like glucose (which don’t dissociate), the van’t Hoff factor $i = 1$ , so we don’t need to multiply by $i$ here.

Normal boiling point of water $= 100°C$

New boiling point $= 100 + 0.52 =$ 100.52°C

Why This Works

Boiling point elevation is a colligative property — it depends only on the number of solute particles, not their identity. When glucose dissolves in water, its molecules occupy positions at the liquid surface, reducing the number of water molecules that can escape as vapour.

This lowers the vapour pressure of the solution (Raoult’s Law). To reach the external atmospheric pressure (760 mm Hg) and boil, the solution now needs a higher temperature. More solute particles → greater vapour pressure lowering → higher boiling point.

$K_b$ is the ebullioscopic constant — it tells you how many degrees the boiling point rises per molal concentration. For water, it’s 0.52 K kg/mol. For other solvents, NCERT gives different values — don’t mix them up.

For quick calculations in NEET/JEE: if molality is 0.5 m, $\Delta T_b = 0.52 \times 0.5 = 0.26$ K. The formula is linear, so doubling molality doubles the elevation. This mental scaling saves time on MCQs.

Alternative Method

Some questions give you mole fraction instead of mass. We can still reach molality:

If mole fraction of solute $x_2$ is given:

m = \frac{x_2 \times 1000}{(1 - x_2) \times M_1}

where $M_1$ is the molar mass of solvent in g/mol (18 for water).

This comes up in JEE Main when the question is designed to test whether you can convert between concentration units before applying the colligative property formula.

Common Mistake

Using molarity instead of molality. A very common error — students calculate concentration as mol/L and directly plug it into $\Delta T_b = K_b \times m$ . Molarity changes with temperature (volume expands on heating), while molality uses mass of solvent, which doesn’t change. The formula specifically requires molality. In this problem, using molarity would give a different (wrong) value because the solution volume is not 0.1 L.

Also watch out for the glucose molar mass. Glucose is $C_6H_{12}O_6$ , molar mass = 180 g/mol. Some students confuse it with fructose or write 182 — they’re the same formula but double-check the question always gives you the molar mass.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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