The molal boiling point elevation constant of water is . When 0.1 mole of sugar is dissolved 200 g o — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

Molality $m = \text{moles} / \text{mass of solvent (kg)} = 0.1 \text{ mol} / 0.200 \text{ kg} = 0.5 \text{ m}$. Elevation $\Delta T_b = K_b \times m = 0.513 \times 0.5 = 0.2565^\circ\text{C}$. The new boiling point is $100^\circ\text{C} + 0.2565^\circ\text{C} = 100.2565^\circ\text{C}$. Therefore, correct answer is C.