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(D) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by: $\frac{P^{0} - P_{s}}{P^{0}} = \frac{n_{A

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(D) According to Raoult's law for non-volatile solutes,the relative lowering of vapour pressure is given by: $\frac{P^{0} - P_{s}}{P^{0}} = \frac{n_{A}}{n_{A} + n_{solvent}}$.Given that the vapour pressure of the solution $P_{s}$ is half of the pure water $P^{0}$,so $P_{s} = \frac{P^{0}}{2}$.Substituting the values: $\frac{P^{0} - P^{0}/2}{P^{0}} = \frac{n_{A}}{n_{A} + n_{solvent}} \implies \frac{1}{2} = \frac{n_{A}}{n_{A} + n_{solvent}}$.This simplifies to $n_{A} + n_{solvent} = 2n_{A}$,which means $n_{A} = n_{solvent}$.The number of moles of water $(n_{solvent})$ is $\frac{100 \ g}{18 \ g/mol} = 5.55 \ mol$.Therefore,$n_{A} = 5.55 \ mol$.Rounding to the nearest integer,we get $6$.

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