Freezing Point Depression and Boiling Point Elevation

Continuing with colligative properties, if the presences of a solute affects the vapor pressure of a solution, then it must also affect the boiling point.  Liquids "boil" when the vapor pressure of the liquid is equal to the atmospheric pressure on that liquid.  If a solute is added to a solvent, the vapor pressure is depressed, so the temperature must be increased further to increase the vapor pressure to the point that it is equal to the atmospheric pressure.  Boiling point elevation is determined by the equation:

ΔT_bp = k_bpe•m•i

Where m = molality = (mols solute) / (kg solvent)

i = van't Hoff factor

The van't Hoff factor, sometimes denoted as "n", describes the number of particles formed when a solute dissolved.  The presence of a solute also affect the freezing behaviour of a solvent by changing the energetics of the system.  For a solvent to freeze when solute is present, the average kinetic energy of the particles must be slowed even more (temperature must be lower) than for a pure solvent.  Freezing point depression uses essentially the same equation, although with different values for the constant.

ΔT_fp = k_fpd•m•i

For the problems we looked at in class, the answers are below.  The problem was "23.381g of “salt” dissolved in 500.00mL of water, what is the freezing/boiling point?":

Compound	m	i	ΔTfp	Tfp	ΔTbp	Tbp
KNO3	0.462522	2	1.72	-1.72	0.48	100.48
Na3PO4	0.331763	4	2.47	-2.47	0.69	100.69
Mg(ClO3)2	0.244565	3	1.36	-1.36	0.38	100.38
(NH4)2SO4	0.353884	3	1.97	-1.97	0.55	100.55
CaCl2	0.421340	3	2.35	-2.35	0.66	100.66