An aqueous solution of a non-volatile, non-electrolyte solute (molecular mass = 150) boils at 373.26 — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Assuming pure water boils at 373 K, $\Delta T_b = 373.26 - 373 = 0.26 \text{ K}$. Molality $m = \Delta T_b / K_b = 0.26 / 0.52 = 0.5 \text{ mol/kg}$. Thus, 0.5 moles of solute are in 1000 g of water. Mass of solute $= 0.5 \times 150 = 75 \text{ g}$. Total mass $= 1000 + 75 = 1075 \text{ g}$. Mass % $= (75 / 1075) \times 100 = 6.976\% \approx 6.98\%$. Therefore, correct answer is C.