Question

Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at -14.5 C? The freezing point for pure water is 0.0 C and K_f is equal to 1.86 C/m

Answer 1

108.43 g.

Step-by-step explanation:

• Adding solute (KNO₃) to water causes depression of freezing point of water (collegative properties).

• We can predict the change in the freezing point (ΔTf) of water using the relation:

ΔTf = i.Kf.m,

where, ΔTf is the elevation in boiling water (ΔTf = 0.0°C - (- 14.5°C) = 14.5 °C).

i is van 't Hoff factor, The van 't Hoff factor is the ratio between the actual concentration of particles produced when the substance is dissolved and the concentration of a substance as calculated from its mass. For most non-electrolytes dissolved in water, the van 't Hoff factor is essentially 1.

i for KNO₃ = 2/1 = 2.

Kf is the molal freezing constant of water (Kf = 1.86 °C/m).

m is the molality of the solution.

∵ ΔTf = i.Kf.m,

∴ m = (ΔTf)/(i.Kf) = (14.5°C)/(2)(1.86 °C/m) = 3.9 m.

• molality (m) is the no. of moles of solute dissolved in 1.0 kg of solvent.

m = (no. of moles of KNO₃)/(mass of water (kg)) = (mass/molar mass of KNO₃)/(mass of water (kg)).

∴ 3.9 m = (mass of KNO₃ / 101.1 g/mol)/(0.275 kg).

∴ mass of KNO₃ = (3.9 m)(101.1 g/mol)(0.275 kg) = 108.43 g.