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).
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).
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- 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.
Updated On: Feb 25, 2025
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Solution and Explanation
The formula for the relative lowering of vapour pressure is:
\[
\frac{p^\circ - p_s}{p^\circ} = i \times \chi
\]
Where:
- \( p_0 \) is the vapour pressure of the pure solvent
- \( p_s \) is the vapour pressure of the solution
- \( i \) is the Van't Hoff factor (which represents the number of particles produced from dissociation; for a strong electrolyte AB, \( i = 2 \))
- \( \chi \) is the 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 in:
\[
\frac{100 - p_s}{100} = 2 \times \frac{1}{50}
\]
Simplifying:
\[
\frac{100 - p_s}{100} = \frac{2}{50}
\]
\[
100 - p_s = 4
\]
\[
p_s = 96 \, \text{mm Hg}
\]
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