Answer:
- 0.0413°C ≅ - 0.041°C (nearest thousands).
Explanation:
- Adding solute to water causes the depression of the freezing point.
ΔTf = Kf.m,
Where,
ΔTf is the change in the freezing point.
Kf is the freezing point depression constant (Kf = 1.86 °C/m).
m is the molality of the solution.
Molality is the no. of moles of solute per kg of the solution.
- no. of moles of solute (glucose) = mass/molar mass = (8.44 g)/(180.156 g/mol) = 0.04685 mol.
∴ molality (m) = no. of moles of solute/kg of solvent = (0.04685 mol)/(2.11 kg) = 0.0222 m.
∴ ΔTf = Kf.m = (1.86 °C/m)(0.0222 m) = 0.0413°C.
∴ The freezing point of the solution = the freezing point of water - ΔTf = 0.0°C - 0.0413°C = - 0.0413°C ≅ - 0.041°C (nearest thousands).
