A 5% (by mass) solution of cane sugar in water has freezing point of 272.85 K. Calculate the freezin — Solutions and Colligative Properties Chemistry Question

💡 Solution & Explanation

For cane sugar ($M=342$), $\Delta T_{f1} = 273.15 - 272.85 = 0.30 \text{ K}$. Since mass % is the same, molality is inversely proportional to molar mass. $m_{\text{glucose}} / m_{\text{sucrose}} = 342 / 180 = 1.9$. Thus, $\Delta T_{f2}$ for glucose is $1.9 \times 0.30 = 0.57 \text{ K}$. The freezing point of the glucose solution is $273.15 - 0.57 = 272.58 \text{ K}$. Therefore, correct answer is B.