Question

A sample of glucose ( C6H12O6 ) of mass 8.44 grams is dissolved in 2.11 kg water. What is the freezing point of this solution? The freezing point depression constant, Kf , for water is 1.86 °C/mol. (Round your answer to the nearest thousandth)

Answer 1

Answer:

- 0.0413°C ≅ - 0.041°C (nearest thousands).

Explanation:

• Adding solute to water causes the depression of the freezing point.

• We have the relation:

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).