Question

The molar volume of liquid benzene (density$=0.877gm{L}^{-1}$) increase by a factor of $2750$ as it vaporizes at ${20}^{\circ }C$ and that of liquid toluene (density$=0.867gm{L}^{-1}$) increases by a factor of $7720$ at ${20}^{\circ }C$ solution has vapour pressure of $46.0torr$. Find the mole fraction of benzene in the vapour above the solution.

(write the value to the nearest integer)

Answer 1

$V_B=\frac{78}{0.877}\times 2750mL=244.583L$
$V_T=\frac{92}{0.867}\times 7720ml=819.192L$

$P_B=\frac{1\times 0.0821\times 293\times 760}{244.583}=74.74torr$

$P_T=\frac{1\times 0.0821\times 293\times 760}{819.192}=22.317$

$46=74.74X_B+23.317(1-X_B)$

$52.423X_B=23.683$

$X_B=0.451$

$Y_B=\frac{P_B^{\circ }\times X_B}{P_T}=\frac{74.74\times 0.451}{46}=0.732$