How many grams of sucrose (molecular weight = 342) should be dissolved in 100 g water in order to pr — Solutions and Colligative Properties Chemistry Question
Question
How many grams of sucrose (molecular weight = 342) should be dissolved in 100 g water in order to produce a solution with a $104.76^\circ\text{C}$ difference between the freezing point and the boiling point temperature? ($K_f = 1.86, K_b = 0.52$)
💡 Solution & Explanation
The difference is $(100 + \Delta T_b) - (0 - \Delta T_f) = 100 + \Delta T_b + \Delta T_f = 104.76^\circ\text{C}$, so $\Delta T_b + \Delta T_f = 4.76^\circ\text{C}$. Since $\Delta T_b = K_b \cdot m$ and $\Delta T_f = K_f \cdot m$, we have $m(K_b + K_f) = 4.76 \Rightarrow m(0.52 + 1.86) = 4.76 \Rightarrow 2.38m = 4.76 \Rightarrow m = 2.0 \text{ mol/kg}$. Mass of sucrose $= m \times (\text{mass of water in kg}) \times M = 2.0 \times 0.1 \times 342 = 68.4 \text{ g}$. Therefore, correct answer is B.