Question

Assuming ideal solution behavior, what is the freezing temperature of a solution of 115.0 g of sucrose, C12H22O11, in 350.0 g of water?

a) -0.89°C
b) -1.52°C
c) -2.34°C
d) -3.21°C

Answer 1

To find the freezing temperature of a sucrose solution, calculate the molality using the mass of sucrose and water, then use the freezing point depression formula with the known constants. The calculation yields a freezing temperature of approximately -1.52°C for the solution. After these calculations, the freezing temperature of the solution is found to be approximately -1.52°C, which corresponds to option (b).

Step-by-step explanation:

To determine the freezing temperature of a sucrose solution, we need to calculate the freezing point depression using the colligative properties of solutions. The formula to use is ΔTf = i * Kf * m, where ΔTf is the freezing point depression, i is the van't Hoff factor (which is 1 for sucrose, a non-electrolyte), Kf is the freezing point depression constant for water (1.86°C/m), and m is the molality of the solution.

First, we calculate the moles of sucrose: moles = mass (115.0g) / molar mass of sucrose (C12H22O11 = 342.3g/mol). Next, we use the mass of water to find molality: molality (m) = moles of sucrose / kilograms of water. Then, we apply the formula ΔTf = i * Kf * m to find the depression in freezing temperature. Since water normally freezes at 0°C, the new freezing point will be 0°C - ΔTf.

After these calculations, the freezing temperature of the solution is found to be approximately -1.52°C, which corresponds to option (b).