The vapour pressure (at the standard boiling point of water) of an aqueous solution containing $28 \%$ by mass of a non-volatile normal solute (molecular mass $=28$ ) will be

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Aicharya · Accepted Answer

Moles of solute $=\frac{28}{28}=1$;
moles of water $=\frac{100-28}{18}=4$
V.P. of solution
$=\mathrm{P}_{\mathrm{H}_{2} \mathrm{O}}^{\mathrm{o}} \times \mathrm{X}_{\mathrm{H}_{2} \mathrm{O}}=760 \times \frac{4}{5}=608$ torr

The vapour pressure (at the standard boiling point of water) of an aqueous solution containing $28 \%$ by…

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