Calculate boiling point elevation of 0.5m glucose solution — colligative properties

Question

A 0.5 molal (0.5m) aqueous solution of glucose is prepared. Calculate the elevation in boiling point. Given: $K_b$ for water = 0.52 K·kg·mol⁻¹.

Solution — Step by Step

Boiling point elevation is a colligative property — it depends only on the number of solute particles, not their identity. The formula is:

\Delta T_b = K_b \times m

where $m$ is the molality of the solution.

Glucose ( $\text{C}_6\text{H}_{12}\text{O}_6$ ) is a non-electrolyte. It does not ionise in water, so the van’t Hoff factor $i = 1$ .

This means we use molality as given — no modification needed.

\Delta T_b = K_b \times m = 0.52 \times 0.5

\Delta T_b = 0.26 \text{ K}

The boiling point of pure water is 100°C. The solution boils at:

T_b = 100 + 0.26 = 100.26°C

The boiling point elevation is 0.26 K, and the solution boils at 100.26°C.

Why This Works

When glucose dissolves in water, its molecules occupy space among water molecules and reduce the number of water molecules at the surface. Fewer water molecules escape as vapour, so the vapour pressure drops.

To boil a liquid, we need its vapour pressure to equal atmospheric pressure. Since vapour pressure has decreased, we need a higher temperature to restore it — that extra temperature is $\Delta T_b$ .

The $K_b$ value (0.52 K·kg·mol⁻¹ for water) is a fixed property of the solvent. It tells us how much 1 mole of any non-dissociating solute raises the boiling point of 1 kg of water. Molality is used here (not molarity) because molality doesn’t change with temperature — a neat practical reason.

Alternative Method — Using Moles Directly

If you’re given grams instead of molality, here’s how to work backwards.

Say 90 g of glucose is dissolved in 1000 g of water. Molar mass of glucose = 180 g/mol.

\text{Moles of glucose} = \frac{90}{180} = 0.5 \text{ mol}

m = \frac{0.5 \text{ mol}}{1 \text{ kg}} = 0.5 \text{ mol/kg}

Same molality, same $\Delta T_b = 0.26$ K. The route is longer but identical in logic — useful when CBSE/JEE gives you grams.

In JEE Main, boiling point elevation questions often bundle in: “how many grams of glucose must be dissolved in 500 g of water to raise boiling point by X K?” — always rearrange $\Delta T_b = K_b \times m$ to solve for moles first, then convert to grams.

Common Mistake

Students often confuse molality and molarity here. The formula uses molality ( $m$ = moles per kg of solvent), not molarity ( $M$ = moles per litre of solution). If the question gives you concentration in mol/L, you cannot plug it directly into $\Delta T_b = K_b \times m$ . Convert first — or flag it in your solution to get method marks.

Also watch for electrolytes: if the question changes glucose to NaCl, you must multiply molality by $i = 2$ . Many students forget the $i$ factor and lose marks on what is otherwise a two-line calculation.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Moles Directly

Common Mistake

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