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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Solution and Explanation
The formula for the relative lowering of vapour pressure is:
\[ \frac{p^\circ - p_s}{p^\circ} = i \times \chi \]Where:
- \( p^\circ \) = 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.
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