Colligative Properties: Quick Revision

A brief note on Colligative Properties. Read it before a test. Try each question at the end before you open its answer.

What they are

• A colligative property depends on the number of solute particles, not on their kind.
• There are four: lowering of the vapour pressure, rise of the boiling point, fall of the freezing point and osmotic pressure.
• The results hold for dilute solutions of a non-volatile solute.

The formulas

• Molality: $m = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}$
• Boiling point: $\Delta T_b = K_b\,m$
• Freezing point: $\Delta T_f = K_f\,m$
• Molar mass: $M_2 = \dfrac{1000\,K\,w_2}{\Delta T\,w_1}$, with the masses in grams
• Osmotic pressure: $\pi = CRT$, with $C$ in mol/L and $T$ in kelvin
• Relative lowering of vapour pressure: $\dfrac{p_1^0 - p_1}{p_1^0} = x_2$
• Water: $K_b = 0.512\ \text{K kg/mol}$ and $K_f = 1.86\ \text{K kg/mol}$
• Gas constant: $R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}$, or $8.314\ \text{J K}^{-1}\text{mol}^{-1}$

Osmosis

• The solvent flows through a semipermeable membrane towards the more concentrated solution.
• Isotonic solutions have equal $\pi$. A cell swells in a hypotonic solution and shrinks in a hypertonic one.
• Reverse osmosis: a pressure greater than $\pi$ on the solution pushes solvent out of it.

The van’t Hoff factor

• $i = \dfrac{\text{observed effect}}{\text{calculated effect}} = \dfrac{\text{normal molar mass}}{\text{observed molar mass}}$
• Splitting into $\nu$ ions: $i = 1 + \alpha(\nu - 1)$, so $i > 1$.
• Joining in pairs: $i = 1 - \dfrac{\alpha}{2}$, so $i < 1$.
• Put $i$ in front of each formula: $\Delta T_b = i\,K_b\,m$, $\Delta T_f = i\,K_f\,m$ and $\pi = i\,C\,R\,T$.

A worked example

A solution has $0.36\ \text{mol}$ of sodium chloride in $1.20\ \text{kg}$ of water. Take $i = 1.9$. Find the fall of the freezing point.

1. Molality. $m = \dfrac{0.36\ \text{mol}}{1.20\ \text{kg}} = 0.30\ \text{mol/kg}$.
2. Fall. $\Delta T_f = i\,K_f\,m = 1.9 \times 1.86\ \text{K kg/mol} \times 0.30\ \text{mol/kg} = 1.06\ \text{K}$.
3. Freezing point. $-1.06\ ^\circ\text{C}$.

Practice with new numbers

A program to try

The program works out three colligative effects for one solution.

Mistakes to avoid

• Dividing by the mass of the solution. Molality uses the mass of the solvent.
• Adding $\Delta T_f$ to the freezing point. It falls.
• Using $^\circ\text{C}$ in $\pi = CRT$. Use kelvin.
• Forgetting $i$ for a salt. Count the ions.
• Using $i = 1 + \alpha(\nu - 1)$ for a solute that joins. For pairs, $i = 1 - \alpha/2$.

Check yourself

1. Which solution freezes lower: $0.10\ \text{mol/kg}$ glucose or $0.10\ \text{mol/kg}$ calcium chloride?

2. What is the osmotic pressure of $0.10\ \text{mol/L}$ of a non-electrolyte at $298\ \text{K}$?

3. Why does a solute that joins in pairs give a molar mass that is too large?

4. In reverse osmosis, on which side does the extra pressure go?