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?
Given:
- Mass of sucrose (C12H22O11) = 115.0 g
- Mass of water (solvent) = 350.0 g = 0.3500 kg
- Molar mass of sucrose = 342.30 g/mol
- Freezing point depression constant (K_f) for water = 1.86 °C·kg/mol
- Freezing point of pure water = 0 °C

Step 1: Calculate the moles of sucrose:
- Moles of sucrose = mass of sucrose / molar mass of sucrose = 115.0 g / 342.30 g/mol = 0.336 mol

Step 2: Calculate the molality (m) of the solution:
- m = moles of solute / mass of solvent (kg) = 0.336 mol / 0.3500 kg = 0.96 mol/kg

Step 3: Calculate the freezing point depression (ΔT_f):
- ΔT_f = K_f * m = 1.86 °C·kg/mol * 0.96 mol/kg = 1.78 °C

Step 4: Calculate the freezing point of the solution:
- Freezing point of the solution = Freezing point of pure water - ΔT_f = 0 °C - 1.78 °C = -1.78 °C

Final Answer:
- Freezing point of the solution = -1.78 °C

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