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(A) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{P^

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$29.1$

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(A) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is equal to the mole fraction of the solute: $\frac{P^{\circ} - P_s}{P^{\circ}} = X_{solute} = \frac{n_{solute}}{n_{solute} + n_{solvent}}$Given: $n_{solute} = 2 \ mol$,$n_{solvent} = 20 \ mol$,$P^{\circ} = 32 \ mm \ Hg$Substituting the values:$\frac{32 - P_s}{32} = \frac{2}{2 + 20} = \frac{2}{22} = \frac{1}{11}$$32 - P_s = 32 	imes \frac{1}{11} \approx 2.91 \ mm \ Hg$$P_s = 32 - 2.91 = 29.09 \ mm \ Hg \approx 29.1 \ mm \ Hg$

What is the vapour pressure of a solution when $2 \ mol$ of a non-volatile solute are dissolved in $20 \ mol$ of water (in $mm \ Hg$)? $(P_1^{\circ} = 32 \ mm \ Hg)$

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