solvent constant

B. Boiling Point Elevation, Chang p. 520, Eq 12.6
T = Kb m
T = i Kb m
Calculating Change in Boiling Point
The change in boiling point is  T in the formula above. The key to doing these
problems is the calculation of molality and choosing the right constant (Kb) from
the table - [Table 12.2 p. 521] (Always choose the constant based on the solvent
and be sure to choose the boiling point constant for boiling point problems and
freezing point constant for freezing point problems.
Example 6:
What is the change in boiling point if 3.69 g of K2SO4 was added to 100.0 mL of
T = Kb m
Kb = boiling point constant for water = 0.52 °C/m
moles K2SO4
kg water (solvent only)
moles K2SO4= 3.69 g x 1 mole
= 0.0212 mole
174.20 g
kg water = 100.0 mL x 1.00 g x 1 kg
= 0.100 kg
1 mL
1000 g
m K2SO4 = 0.0212 mol
= 0.212 m
0.100 kg
Remember that you need the molality of particles and when K2SO4 dissolves in
water it dissociates into two K+ ions and one SO4 2– ion. Therefore, i = 3 and
molality of particles = i m = 3 (0.212 m) = 0.636 m
T = i m( 0.52 °C /m); T = 0.636 m (0.52 °C /m) T = 0.33 °C
The answer has two significant figures because the Kb only has two.
Calculating the Actual Boiling Point
Problems involving actual boiling point are different because they ask for the
boiling point of the solution rather than for the change in boiling point of the
solvent. The process is initially the same. You must still determine the change in
boiling point, but there is an extra step. The extra step is to add the
change in boiling point, T, to the boiling point of the pure solvent. It will help you
to remember to add if you remember the name of this colligative property is
boiling point elevation.
Example 7:
What is the boiling point of a solution made by mixing 13.63 g ethanol and
540.00 g water?
The sneaky part of this question has to do with both ethanol and water being
listed as solvents. Remember that the solvent is the substance in greater
quantity. In this case, it is obviously water.
solute = 13.63 g ethanol (C2H5OH)
solvent = 540.00 g water (H2O)
T = Kb m
The value of Kb is the value of the solvent (for water) = 0.52 °C/m
The value of m is the value of particles of solute. Since ethanol is a
nonelectrolyte, i = 1. The molality of ethanol is the same as the molality of
m = mole solute =
mole ethanol
kg solvent
kg water
mol ethanol = 13.63 g C2H5OH x 1 mole = 0.2959 mol
46.07 g
kg water = 540.00 g x 1 kg
= 0.54000 kg
1000 g
m = 0.2959 mol = 0.5479 m
0.54000 kg
T = Kb m [ Kb = 0.52 °C/ m]; T = 0.5479 m (0.52 °C/ m ) = 0.28 °C
Don't forget (and it is the most common mistake) that this does not answer the
boiling point (bp) solution = bp solvent + T = 100 °C + 0.28 °C = 100.28 °C
Recall that the 100°C was defined as the boiling point of water; therefore it is an
exact number with infinite significant figures. The significant figure rule for
addition is to use the value with the fewest decimal places.