At T K,the vapour pressure of water is x kP | vedclass

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(D) The relative lowering of vapour pressure is given by Raoult's Law: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 + n_2}$.Here,$P^o = x \ kPa$ and $P_s$

Vedclass · Accepted Answer

$0.982$

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(D) The relative lowering of vapour pressure is given by Raoult's Law: $\frac{P^o - P_s}{P^o} = \frac{n_2}{n_1 + n_2}$.Here,$P^o = x \ kPa$ and $P_s$ is the vapour pressure of the solution.For a $1 \ m$ solution,there is $1 \ mol$ of solute in $1000 \ g$ of water.Moles of water $(n_1)$ = $\frac{1000 \ g}{18 \ g/mol} = 55.55 \ mol$.Moles of solute $(n_2)$ = $1 \ mol$.Substituting these values: $\frac{x - P_s}{x} = \frac{1}{55.55 + 1} = \frac{1}{56.55} \approx 0.01768$.$1 - \frac{P_s}{x} = 0.01768$.$\frac{P_s}{x} = 1 - 0.01768 = 0.98232$.$P_s \approx 0.982 x \ kPa$.

At $T \ K$,the vapour pressure of water is $x \ kPa$. What is the vapour pressure (in $kPa$) of a $1 \ m$ (molal) aqueous solution containing a non-volatile solute (in $x$)?

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