Freezing Point Depression

Sample Freezing Point Depression problem.....

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
Kf water = 1.86 °C kg/mol

Solution:
To find the temperature change elevation of a solvent by a solute, use the equation:

ΔT = iKfm

where
ΔT = Change in temperature in °C
i = van 't Hoff factor
Kf = molal freezing point depression constant or cryoscopic constant in °C kg/mol
m = molality of the solute in mol solute/kg solvent.

Step 1 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

mNaCl = moles of NaCl/kg water
mNaCl = 0.542 mol/0.219 kg
mNaCl = 2.477 mol/kg

Step 2  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+ and Cl-. Therefore, i = 2 for this example.

Step 3  Find ΔT

ΔT = iKfm

ΔT = 2 x 1.86 °C kg/mol x 2.477 mol/kg
ΔT = 9.21 °C

Answer:
Adding 31.65 g of NaCl to 220.0 mL of water will lower the freezing point by 9.21 °C.