A 5% solution (by mass) of cane sugar in water has a freezing point of 271 K. Calculate the freezing point of 5% glucose in water if the freezing point of pure water is 273.15 K.

Steps to Solve

1. Calculate the freezing point depression for the cane sugar solution

   We are given that the freezing point of a 5% cane sugar solution is -0.28 °C. The freezing point of pure water is 0 °C. The freezing point depression, $\Delta T_f$, is the difference between the freezing point of the pure solvent (water) and the solution. $$ \Delta T_f = T_{f, \text{pure solvent}} - T_{f, \text{solution}} $$ $$ \Delta T_f = 0 - (-0.28) = 0.28 \text{ }^\circ\text{C} $$

2. Determine the molality of the cane sugar solution

   We are given a 5% solution. This means 5 g of cane sugar in 100 g of solution. Assuming the density of the solution is approximately 1 g/mL, we have 5 g of cane sugar in 100 mL of solution, meaning 95 g of water. First, we need to find the number of moles of cane sugar ($C_{12}H_{22}O_{11}$). The molar mass of cane sugar is $12(12) + 22(1) + 11(16) = 144 + 22 + 176 = 342$ g/mol. $$ \text{moles of cane sugar} = \frac{5 \text{ g}}{342 \text{ g/mol}} = 0.0146 \text{ mol} $$ Now we can calculate the molality ($m$) of the solution. Molality is defined as moles of solute per kilogram of solvent. $$ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{0.0146 \text{ mol}}{0.095 \text{ kg}} = 0.154 \text{ mol/kg} $$

3. Calculate the freezing point depression constant, $K_f$

   The freezing point depression is related to the molality by the equation $\Delta T_f = K_f \cdot m$. We can solve for $K_f$ using the values we calculated for the cane sugar solution. $$ K_f = \frac{\Delta T_f}{m} = \frac{0.28 \text{ }^\circ\text{C}}{0.154 \text{ mol/kg}} = 1.82 \text{ }^\circ\text{C kg/mol} $$

4. Determine the molality of the 5% glucose solution

   Again, we have 5 g of glucose in 100 g of solution, meaning 95 g of water. We need to find the number of moles of glucose ($C_6H_{12}O_6$). The molar mass of glucose is $6(12) + 12(1) + 6(16) = 72 + 12 + 96 = 180$ g/mol. $$ \text{moles of glucose} = \frac{5 \text{ g}}{180 \text{ g/mol}} = 0.0278 \text{ mol} $$ The molality of the glucose solution is: $$ m = \frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{0.0278 \text{ mol}}{0.095 \text{ kg}} = 0.293 \text{ mol/kg} $$

5. Calculate the freezing point depression for the glucose solution

   Now we can use the $K_f$ value we found and the molality of the glucose solution to find the freezing point depression for the glucose solution. $$ \Delta T_f = K_f \cdot m = 1.82 \text{ }^\circ\text{C kg/mol} \cdot 0.293 \text{ mol/kg} = 0.533 \text{ }^\circ\text{C} $$

6. Calculate the freezing point of the glucose solution

   Finally, we can calculate the freezing point of the glucose solution using the freezing point depression. $$ T_{f, \text{solution}} = T_{f, \text{pure solvent}} - \Delta T_f = 0 - 0.533 = -0.533 \text{ }^\circ\text{C} $$