Question

What is the colligative property? Write an expression for the determination of molar mass of solute using relative lowering in vapour pressure.

Hint:

Use Raoult's law: relative lowering of vapour pressure equals mole fraction of solute; rearrange to get \( M_2 \).

Answer 1

Step 1: Definition of colligative property.
Colligative properties are those properties of a dilute solution which depend only on the number of solute particles present and not on the chemical nature of the solute. The four colligative properties are: relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, and osmotic pressure.

Step 2: Raoult's law for relative lowering of vapour pressure.
When a non-volatile solute is added to a solvent, the vapour pressure of the solvent falls. According to Raoult's law, the relative lowering of vapour pressure equals the mole fraction of the solute:
\[ \frac{p^{\circ} - p}{p^{\circ}} = x_2 = \frac{n_2}{n_1 + n_2} \]
where \( p^{\circ} \) is the vapour pressure of pure solvent, \( p \) is that of the solution, and \( n_1, n_2 \) are the moles of solvent and solute.

Step 3: Dilute solution approximation.
For a dilute solution \( n_2 \) is very small compared with \( n_1 \), so \( n_1 + n_2 \approx n_1 \):
\[ \frac{p^{\circ} - p}{p^{\circ}} = \frac{n_2}{n_1} = \frac{w_2 / M_2}{w_1 / M_1} \]
where \( w_1, w_2 \) are masses and \( M_1, M_2 \) are molar masses of solvent and solute.

Step 4: Expression for molar mass of solute.
Rearranging for \( M_2 \):
\[\boxed{ M_2 = \frac{w_2 \times M_1 \times p^{\circ}}{w_1 \,(p^{\circ} - p)} }\]