2 moles each of ethylene glycol and glucose are dissolved in 500\, g of water. The boiling point of the resulting solution is: (Given: K b = 0.52\, K\,kg\,mol -1 )

Chemistry · Solutions and Colligative Properties

JEE Main 2025 — 3 April, Morning Shift — Question 1

$2$ moles each of ethylene glycol and glucose are dissolved in $500\,\mathrm{g}$ of water. The boiling point of the resulting solution is:

(Given: $K_b = 0.52\,\mathrm{K\,kg\,mol^{-1}}$ )

Both ethylene glycol and glucose are non-electrolytes, so $i=1$ for each.

Total moles of solute = $2+2=4\,\mathrm{mol}$

Mass of solvent = $500\,\mathrm{g}=0.5\,\mathrm{kg}$

Molality: $m=\frac{4}{0.5}=8\,\mathrm{mol\,kg^{-1}}$

Elevation in boiling point: $\Delta T_b = K_b \cdot m = 0.52 \times 8 = 4.16\,\mathrm{K}$

Boiling point of solution: $T_b = 373\,\mathrm{K} + 4.16\,\mathrm{K} = 377.16\,\mathrm{K} \approx 377.3\,\mathrm{K}$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.