10. Solutions & Colligative Properties
Concentration describes how much solute sits in a solution; colligative properties describe how the solute changes the solvent's behaviour — and they depend only on how many particles are present, not what they are.
Molarity and Dilution
$$M = \frac{n_{\text{solute}}}{V_{\text{solution (L)}}} \qquad M_1V_1 = M_2V_2$$
Example: Dilute 50.0 mL of 2.0 M HCl to 250 mL:
$$M_2 = \frac{(2.0)(0.050)}{0.250} = 0.40\ \text{M}$$
Colligative Properties
- Freezing point depression: $\Delta T_f = iK_f m$
- Boiling point elevation: $\Delta T_b = iK_b m$
- Osmotic pressure: $\pi = iMRT$
$i$ is the van't Hoff factor — the number of particles a solute splits into (2 for NaCl, 3 for CaCl2).
Example: A 1.0 m NaCl solution ($i = 2$, $K_f = 1.86$ °C/m) freezes at $\Delta T_f = 2(1.86)(1.0) = 3.72$ °C below zero: $T_f = -3.72$ °C — which is why salt melts ice.
Example: Osmotic pressure of 0.10 M NaCl at 298 K: $\pi = (2)(0.10)(0.08206)(298) = 4.89$ atm.
Example: Osmotic pressure of 0.10 M NaCl at 298 K: $\pi = (2)(0.10)(0.08206)(298) = 4.89$ atm.
Practice Problems
- Dilute 100 mL of 6.0 M HCl to 500 mL. What is the new concentration?
- Find the boiling point of a 0.50 m CaCl2 solution (i = 3, Kb = 0.512 °C/m).
- Calculate the osmotic pressure of 0.20 M glucose at 25 °C.
- Why does spreading salt on icy roads melt the ice?
Answers: 1. 1.2 M 2. 100.77 °C 3. 4.89 atm 4. The dissolved salt lowers the freezing point of water below 0 °C