Question

The vapour pressure of a solvent at 283 K is 100 mm Hg. Calculate the vapour pressure of a dilute solution containing 1 mole of a strong electrolyte AB in 50 moles of the solvent at 283 K (assuming complete dissociation of solute AB).

Hint:

- The Van't Hoff factor \( i \) is critical in determining the effect of dissociation on the vapour pressure of a solution. For strong electrolytes, \( i \) represents the number of ions produced. - The relative lowering of vapour pressure is used to calculate the new vapour pressure of the solvent when a solute is added.

Answer 1

The formula for the relative lowering of vapour pressure is:

\[ \frac{p^\circ - p_s}{p^\circ} = i \times \chi \]

Where:

• \( p^\circ \) = vapour pressure of the pure solvent
• \( p_s \) = vapour pressure of the solution
• \( i \) = Van't Hoff factor (number of particles produced from dissociation; for a strong electrolyte AB, \( i = 2 \))
• \( \chi \) = mole fraction of the solute

For a dilute solution:

\[ \frac{100 - p_s}{100} = 2 \times \frac{n_{AB}}{n_{\text{solvent}}} \]

Where:

• \( n_{AB} = 1 \) mole (the amount of solute AB)
• \( n_{\text{solvent}} = 50 \) moles (the amount of solvent)

Substituting these values:

\[ \frac{100 - p_s}{100} = 2 \times \frac{1}{50} \]

Simplifying:

\[ \frac{100 - p_s}{100} = \frac{2}{50} \] \[ 100 - p_s = 4 \] \[ p_s = 96 \, \text{mm Hg} \]

Final Answer: The vapour pressure of the solution is 96 mm Hg.