What is the mol fraction of benzene in the vapour phase if equal mols of benzene and toluene are dissolved in the same solution? The pure vapor pressures of benzene and toluene are 22 torr and 75 torr respectively, at this temperature.
Answer:
\(\chi_{\text {benz }(g)}=0.773\)
The important thing to note is that the mole fraction in the vapor is asked for, not the mole fraction in the liquid (although that is readily found to be 0.50).
Recall Raoult’s law for one component in solution with a second:
\(P_{j}=\chi_{j(l)} P_{j}^{*}\)where:
- \(P_{j}\) is the vapor pressure above the solution of component \(j\) in solution.
- \(P_{j}^{*}\) is the pure vapor pressure in the absence of the solvent.
- \(\chi_{j(l)}\) is the mol fraction of \(j\) in the solution (liquid phase)
From Dalton’s law of partial pressures, we assume with ideal solutions that, upon plugging in (1) for each component in solution, we have:
\(P_{\text {tot }}=P_{i}+P_{j}\)
\(=\chi_{i(l)} P_{i}^{*}+\chi_{j(l)} P_{j}^{*} (2)\)
Let \(i\) be toluene and \(j\) be benzene. Then, we can get the total pressure (1 mm Hg = 1 torr), knowing that the solution is equimolar for both components:
