What is vapour pressure of a solution when $2 \mathrm{~mol}$ of a non-volatile solute are dissolved in $20 \mathrm{~mol}$ of water? $\left(\mathrm{P}_1^{\circ}=32 \mathrm{~mm} \mathrm{Hg}\right)$

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{\mathrm{P}^{\circ}-\mathrm{P}_{\mathrm{s}}}{\mathrm{P}^{\circ}}=\mathrm{X}_{	ext {solute }}=\frac{\mathrm{n}_{	ext {solute }}}{\mathrm{n}_{	ext {solute }}+\mathrm{n}_{	ext {solvent }}} \ & \frac{32-\mathrm{P}_{\mathrm{s}}}{32}=\frac{2}{2+20} \ & 32-\mathrm{P}_{\mathrm{s}}=\frac{2}{22} 	imes 32=2.90 \ & \mathrm{P}_{\mathrm{s}}=32-2.90 \ & =29.1 \mathrm{~mm} \mathrm{Hg}\end{aligned}$

What is vapour pressure of a solution when $2 \mathrm{~mol}$ of a non-volatile solute are dissolved in $20…

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