Colligative properties depend only on the number of solute particles in solution, not on their chemical identity. For a non-volatile solute there are four such properties, all linked by the lowering of vapour pressure.
1. Relative lowering of vapour pressure. From Raoult's law, $\frac{p^0-p}{p^0}=x_{solute}$. The relative lowering equals the mole fraction of the solute. For dilute solutions this gives $\frac{p^0-p}{p^0}=\frac{n_2}{n_1}=\frac{w_2 M_1}{w_1 M_2}$, a route to molar mass $M_2$.
2. Elevation of boiling point. Lowering the vapour pressure raises the boiling point. The elevation is $\Delta T_b=K_b\,m$, where $K_b$ is the ebullioscopic (molal boiling-point elevation) constant and $m$ is molality. For water $K_b=0.512\ \text{K kg mol}^{-1}$.
3. Depression of freezing point. Similarly the freezing point falls: $\Delta T_f=K_f\,m$, where $K_f$ is the cryoscopic constant. For water $K_f=1.86\ \text{K kg mol}^{-1}$. This underlies the use of ethylene glycol as antifreeze and salt for de-icing roads.
4. Osmotic pressure. When solvent flows through a semipermeable membrane from pure solvent into solution, the excess pressure needed to stop the flow is the osmotic pressure $\Pi$. For dilute solutions $\Pi=CRT$, where $C$ is molar concentration ($=n/V$), $R$ the gas constant and $T$ the absolute temperature. Osmotic pressure is the most accurate method for molar masses of macromolecules (proteins, polymers) because it gives measurable values even at low concentration. Solutions of equal osmotic pressure are isotonic.
Van't Hoff factor. Electrolytes dissociate (or some solutes associate), so the observed colligative effect differs from that calculated assuming no change. The van't Hoff factor is $i=\frac{\text{observed colligative property}}{\text{calculated (normal) value}}=\frac{\text{normal molar mass}}{\text{abnormal molar mass}}$. For dissociation $i>1$ (e.g. $\text{NaCl}\to2$ ions, $i\approx2$); for association $i<1$ (e.g. dimerised benzoic acid, $i\approx0.5$). The modified equations become $\Delta T_b=i K_b m$, $\Delta T_f=i K_f m$, and $\Pi=i CRT$. The degree of dissociation $\alpha=\frac{i-1}{n-1}$ where $n$ is the number of ions per formula unit.
The vapour pressure of pure water at 298 K is 23.8 mm Hg. On dissolving 18 g of glucose (M = 180) in 178.2 g of water, find the relative lowering of vapour pressure.
Solution- $n_2=18/180=0.10$ mol; $n_1=178.2/18=9.90$ mol.
- For dilute solution $\frac{p^0-p}{p^0}=\frac{n_2}{n_1+n_2}=\frac{0.10}{10.0}$.
- $\frac{p^0-p}{p^0}=0.010$.
Answer: Relative lowering $=0.010$.
Calculate the boiling point of a solution of 5.0 g of a non-volatile solute (M = 100) in 100 g of water. $K_b=0.512\ \text{K kg mol}^{-1}$.
Solution- $n_2=5.0/100=0.05$ mol; molality $m=0.05/0.100=0.5$ m.
- $\Delta T_b=K_b m=0.512\times0.5=0.256$ K.
- Boiling point $=373.15+0.256=373.41$ K $\approx100.26\ ^{\circ}\text{C}$.
Answer: $\Delta T_b=0.256$ K; b.p. $\approx100.26\ ^{\circ}\text{C}$.
45 g of ethylene glycol (M = 62) is dissolved in 600 g of water. Calculate the freezing point of the solution. $K_f=1.86\ \text{K kg mol}^{-1}$.
Solution- $n=45/62=0.726$ mol; $m=0.726/0.600=1.21$ m.
- $\Delta T_f=K_f m=1.86\times1.21=2.25$ K.
- Freezing point $=0-2.25=-2.25\ ^{\circ}\text{C}$.
Answer: Freezing point $\approx-2.25\ ^{\circ}\text{C}$.
200 mL of an aqueous solution contains 1.26 g of a protein. Its osmotic pressure is $2.57\times10^{-3}$ bar at 300 K. Calculate the molar mass of the protein. $R=0.083\ \text{L bar K}^{-1}\text{mol}^{-1}$.
Solution- $\Pi=CRT=\frac{n}{V}RT=\frac{w}{MV}RT \Rightarrow M=\frac{wRT}{\Pi V}$.
- $M=\frac{1.26\times0.083\times300}{2.57\times10^{-3}\times0.200}$.
- $M=\frac{31.37}{5.14\times10^{-4}}=6.10\times10^{4}\ \text{g mol}^{-1}$.
Answer: $M\approx6.10\times10^{4}\ \text{g mol}^{-1}$.
A $0.1$ m NaCl solution shows a freezing-point depression of 0.345 K. Given $K_f=1.86$, calculate the van't Hoff factor $i$ and the degree of dissociation.
Solution- Calculated (no dissociation): $\Delta T_f=K_f m=1.86\times0.1=0.186$ K.
- $i=\frac{\text{observed}}{\text{calculated}}=\frac{0.345}{0.186}=1.855$.
- For NaCl, $n=2$; $\alpha=\frac{i-1}{n-1}=\frac{0.855}{1}=0.855$.
Answer: $i\approx1.86$, degree of dissociation $\approx0.86$ (86%).
Benzoic acid dimerises in benzene. A solution of 0.50 mol benzoic acid in 1 kg benzene gives $\Delta T_f$ corresponding to an effective molality of 0.28 m. Find the van't Hoff factor and degree of association.
Solution- $i=\frac{\text{observed molality}}{\text{actual molality}}=\frac{0.28}{0.50}=0.56$.
- For association into dimers $n=2$; $\alpha=\frac{1-i}{1-\frac{1}{n}}=\frac{1-0.56}{1-0.5}=\frac{0.44}{0.5}$.
- $\alpha=0.88$.
Answer: $i=0.56$; degree of association $\approx0.88$ (88%).