# To calculate total vapour pressure of a solution

Question: 1 mole each of A and B were mixed to obtain an ideal solution of which one mole went to the vapour phase at equilibrium. If $$P°_\mathrm{A} = \pu{100 torr}$$ and $$P°_\mathrm{B} = \pu{900 torr}$$, then $$P_\mathrm{S}$$ would be?

For the above question, I first calculated vapour pressure of A and B which came out to be $$\pu{50 torr}$$ and $$\pu{450 torr}$$ respectively (using $$P°_\mathrm{A} × X_\mathrm{A}$$ where $$X_\mathrm{A}$$ is mole fraction of A) and then I calculated vapour pressure for A and B in vapour phase which came out to be $$0.1$$ and $$0.9$$ respectively, so the amount remaining in the liquid phase will be $$\pu{0.9 mole}$$ for A and $$\pu{0.1 mole}$$ for B, according to which the $$P_\mathrm{S}$$ should be $$\pu{180 torr}$$, but the answer provided is $$\pu{300 torr}$$, am I missing something? (also see the image below for better understanding of my approach).

Any help is appreciated! Thanks in advance!

• Could you please type out what you have written instead of posting an image? This is because images are not searchable, and harder to read. – Shoubhik R Maiti Apr 19 at 9:19
• I'm really sorry for that, actually i am pretty bad with message formatting, and i was running out of time so i decided to add an image instead, i'll redo that with a pen and replace the image as i get time. Thanks for pointing out. – kanishk6103 Apr 19 at 16:04
• Images are discouraged. Please use mathjax/mhchem, see Notation basics and How can I format math/chemistry expressions on Chemistry SE. See also Math SE MathJax tutorial. – Buck Thorn Apr 20 at 6:55

The starting point to solve the problem is Raoult's law. For each component we can write that the vapour pressure (related by Dalton's law to the total pressure) is equal to the product of the mole fraction in the solution and the vapour pressure of the pure component:

$$y_iP_S = \chi _i P^{\circ} _i$$

You can exploit the fact that the molar amounts in the problem are particularly tidy:

\begin{align} n_{1l} + n_{1g} &=1 \\ n_{2l} + n_{2g} &=1 \\ n_{1g} + n_{2g} &=1 \end{align}

So that for the first component: \begin{align} y_1 &= n_{1g} \\ \chi _1 &= n_{1l} \\ \rightarrow n_{1g}P_S &= n_{1l}P^{\circ}_1 \\ \rightarrow n_{1g}P_S &= (1-n_{1g})P^{\circ}_1 \\ \rightarrow P_S &= \frac{(1-n_{1g})P^{\circ}_1}{n_{1g}} = \frac{n_{2g}P^{\circ}_1}{(1-n_{2g})} \tag{1}\label{eq:firstcomponent} \end{align}

For the second component we can similarly write \begin{align} y_2 &= n_{2g} \\ \chi_2 &= n_{2l} \\ \rightarrow n_{2g}P_S &= n_{2l}P^{\circ}_2 \\ \rightarrow n_{2g}P_S &=(1-n_{2g})P^{\circ}_2 \\ \rightarrow P_S &=\frac{(1-n_{2g})P^{\circ}_2}{n_{2g}} \tag{2}\label{eq:secondcomponent} \end{align}

Equating the expressions for $$P_S$$ in \eqref{eq:firstcomponent} and \eqref{eq:secondcomponent} and rearranging terms we can write a quadratic equation in $$n_{2g}$$ whose solution can be inserted back into the expression for $$P_S$$.