16.3. Colligative Properties http://www.ck12.org
- mass of H 2 O = 218 g = 0.218 kg
- mass of solute = 38.7 g
- Kf(H 2 O) =−1.86°C/m
Unknown
- molar mass of solute =? g/mol
Use the freeing point depression (∆Tf) to calculate the molality of the solution. Then use the molality equation to
calculate the moles of solute. Then divide the grams of solute by the moles to determine the molar mass.
Step 2: Solve.
m=
∆Tf
Kf=
− 5. 53 ◦C
− 1. 86 ◦C/m= 2. 97 m
mol solute =m×kg H 2 O = 2.97m×0.218 kg = 0.648 mol
38 .7 g
0 .648 mol
= 59 .7 g/molStep 3: Think about your result.
The molar mass of the unknown solute is 59.7 g/mol. Knowing the molar mass is an important step in determining
the identity of an unknown substance. A similar problem could be done with the change in boiling point.
Practice Problem- 26.80 g of an unknown alkali metal chloride (MCl) is dissolved in 100. g of water. The boiling point of the
solution is measured to be 103.7°C. Calculate the molar mass of the compound and determine the identity of
M. (Hint: Make sure to account for the fact that MCl is an electrolyte.)
Lesson Summary
- The colligative properties of a solution depend only on the number of dissolved particles and not on the
chemical nature of the solute. When a solute dissolves in a solvent, the vapor pressure decreases, the freezing
point decreases, and the boiling point increases. - Electrolytes are compounds that dissociate when they dissolve. Since more particles are produced, the effect
on the colligative properties is greater for electrolyte solutions than for nonelectrolyte solutions. - Freezing points and boiling points can be calculated for solutions of known molality by using the molal
freezing-point depression and boiling-point elevation constants. - The molar mass of an unknown solute can be determined from either its freezing point depression or boiling
point elevation.
Lesson Review Questions
Reviewing Concepts
- State the effect that dissolving a solute has on each of the following physical properties of a solvent.