The boiling point of a 2% (w/w) aqueous solution of a non-volatile, non-electrolyte solute is 0.102° — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Elevation in boiling point $\Delta T_b = 0.102^\circ C$. A 2% (w/w) solution contains 2 g of solute in 98 g of water. Molality $m = (2/M) / 0.098$. Using $\Delta T_b = K_b \times m$, $0.102 = 0.52 \times [ (2/M) / 0.098 ] = (1.04) / (0.098 \times M)$. Solving for M: $M = 1.04 / (0.102 \times 0.098) = 1.04 / 0.009996 \approx 104$ g/mol. Therefore, correct answer is D.