Question

1 mole of Benzene $(P_{Benzene}^{\circ }=42\text{mm Hg})$ and 2 mole of toluene $(P_{toluene}^{\circ }=36\text{mm Hg})$ will have

• A. Positive deviation from Raoult’s law
• B. Negative deviation from Raoult’s law
• C. Total vapour pressure $38\text{mm Hg}$
• D. Mole fraction of vapour of toluene above liquid mixture is $7/19$

Answer 1

The correct option is C Total vapour pressure $38\text{mm Hg}$

Benzene & Toluene form ideal solution
$P_{solution}=P_B^0{X}_B+P_T^0{X}_T$

$=\left[42(\frac{1}{3}+36\left(\frac{2}{3}\right))\right]$

$=14+24=38\text{mm Hg}$

Mole fraction of toluene in the vapour phase
$Y_T=\frac{P_T^0{X}_T}{P_T}=\frac{24}{38}=\frac{12}{19}$