Question

What are the freezing point and boiling point of water containing 1.00 mol of NaCl in 200. g of water? How do these values change if CaCl2 is used instead of NaCl?

Answer 1

Apply Colligative properties

This is a typical example of colligative properties.

Recall that a solute ( non volatile ) can make a depression/increase in the freezing/boiling point via:

dTf = -Kf*molality * i

dTb = Kb*molality * i

where:

Kf = freezing point constant for the SOLVENT; Kb = boiling point constant for the SOLVENT;

molality = moles of SOLUTE / kg of SOLVENT

i = vant hoff coefficient, typically the total ion/molecular concentration.

At the end:

Tf mix = Tf solvent - dTf

Tb mix = Tb solvent - dTb

for water:

Tb = 100 + 0.512*mol of solute / kg solvent * i

Tf = 0+ -1.86*mol of solute / kg solvent * i

then

a) NaCl = 2 ions,

Tb = 100 + 0.512*1/0.2*2 = 105.12°C

Tf = 0+ -1.86*1/0.2*2= -18.6°C

b) CaCl2 = 3 ions,

Tb = 100 + 0.512*1/0.2*3 = 107.68 °C

Tf = 0+ -1.86*1/0.2*3= -27.9°C