This example problem demonstrates how to calculate freezing point depression.
31.65 g of sodium chloride is added to 220.0 mL of water at 34 °C. How will this affect the freezing point of the water?
Assume the sodium chloride completely dissociates in the water.
Given: density of water at 35 °C = 0.994 g/mL
water = 1.86 °C kg/mol
To find the temperature change elevation of a solvent by a solute, use the equation:
ΔT = iKf
ΔT = Change in temperature in °C
i = van 't Hoff factor
= molal freezing point depression constant or cryoscopic constant in °C kg/mol
m = molality of the solute in mol solute/kg solvent.
Calculate the molality of the NaCl
molality (m) of NaCl = moles of NaCl/kg water
From the periodic table
atomic mass Na = 22.99
atomic mass Cl = 35.45
moles of NaCl = 31.65 g x 1 mol/(22.99 + 35.45)
moles of NaCl = 31.65 g x 1 mol/58.44 g
moles of NaCl = 0.542 mol
kg water = density x volume
kg water = 0.994 g/mL x 220 mL x 1 kg/1000 g
kg water = 0.219 kg
= moles of NaCl/kg water
= 0.542 mol/0.219 kg
= 2.477 mol/kg
Determine the van 't Hoff factor
The van 't Hoff factor, i, is a constant associated with the amount of dissociation of the solute in the solvent. For substances which do not dissociate in water, such as sugar, i = 1. For solutes that completely dissociate into two ions, i = 2. For this example NaCl completely dissociates into the two ions, Na+
. Therefore, i = 2 for this example.
ΔT = iKf
ΔT = 2 x 1.86 °C kg/mol x 2.477 mol/kg
ΔT = 9.21 °C
Adding 31.65 g of NaCl to 220.0 mL of water will lower the freezing point by 9.21 °C.