(અણુભાર : $CHCl_3 = 119.5\, u, CH_2CI_2 = 85\,u$)
Where \(x_{A}\) and \(x_{B}\) are mole fraction of \(A\) and \(B\) in liquid phase.
\(P_{A}^{0}\) and \(P_{B}^{0}\) are vapour pressures of pure liquids.
Moles of \(C H C l_{3}=\frac{25.5}{119.5}=0.213\)
Moles of \(C H_{2} C l_{2}=\frac{40}{85}=0.471\)
Mole fraction \(X_{C H C l_{3}}=\frac{0.213}{0.471+0.213}=\frac{0.213}{0.684}=0.311\)
\(X_{C H_{2} C l_{2}}=\frac{0.471}{0.684}=0.689\)
vapour pressure \(=0.311 \times P_{C H C l_{3}}+0.689 P_{C H_{2} C l_{2}}\)
\(=0.311 \times 200+0.689 \times 41.5=90.7935 \;\mathrm{mm} \mathrm{Hg}\)