How many grams of pure sucrose (Molar mass = ) should be dissolved in exactly of water to produce a — Solutions and Colligative Properties Chemistry Question
Question
How many grams of pure sucrose (Molar mass = $342\text{ g mol}^{-1}$) should be dissolved in exactly $100\text{ g}$ of water to produce a theoretical difference of precisely $105^\circ\text{C}$ between the boiling point and freezing point of the solution at $1\text{ atm}$ pressure? (Assume $K_f = 2.0\text{ K kg mol}^{-1}$ and $K_b = 0.5\text{ K kg mol}^{-1}$ for water. Round your final answer to one decimal place).
💡 Solution & Explanation
The difference between boiling and freezing points is $\Delta T = T_b - T_f$. Since $T_b = 100 + \Delta T_b$ and $T_f = 0 - \Delta T_f$, $\Delta T = 100 + \Delta T_b + \Delta T_f = 105^\circ\text{C}$. Therefore, $\Delta T_b + \Delta T_f = 5$. Substituting the formulas: $(K_b \times m) + (K_f \times m) = 5 \implies (0.5 + 2.0)m = 5 \implies m = 5 / 2.5 = 2.0\text{ mol kg}^{-1}$. Moles of sucrose required for $100\text{ g}$ ($0.1\text{ kg}$) of water = $2.0 \times 0.1 = 0.2\text{ moles}$. Mass of sucrose = $0.2 \text{ mol} \times 342 \text{ g mol}^{-1} = 68.4\text{ g}$.