Question

At 80C, the vapour pressure of pure benzene is 753 mm Hg and of pure toluene 290 mm Hg. Calculate the composition of a liquid in mole per cent which at 80C is in equilibrium with the vapour containing 30 mole per cent of benzene.

Answer 1

$P_B^o=$ vapour pressure of pure Benzene $=753mmHg$

$P_1^o=$ vapour pressure of pure Tolvene $=290mmHg$

$x_B=$ mole frection of benzene above the liquid $=0.30$

Now, potential pressure of benzene above the liquid

$P_B=x_B\times P_{Total}$

$\Rightarrow 0.30=\frac{P_B}{P_{Total}}.....\left(1\right)$

Now, Also, potential pressure of benzene above the liquid using Roonth is law

$P_B=x_{Bliquid}\times P_B^o$

Similarly potential pressure of Tohve above the liquid

using Raonlt's law

$P_T=X_{Tliquid}\times P_T^o=(1-x_{Bliquid})P_T^o$

$\therefore P_{Total}=x_{Bliquid}\times P_B^o+(1-x_{Bliquid})P_T^o$

$=P_T^o+(P_B^o-P_T^o)x_{Bliquid}$

using $\left(1\right)$

$0.30=\frac{P_B}{P_T^O(P_B^o-P_T^o)x_{Bliquid}}=\frac{x_{Bliquid}{P}_B^o}{P_T^o+(P_B^o-P_T^o)x_{Bliquid}}$

Solving, we get

$x_{Bliquid}=\frac{0.30P_T^o}{0.70P_B^o-0.30P_T^o}=\frac{0.30\times 290}{(0.70\times 753-0.30\times 290)}$

$=0.142$

& $x_{Tliquid}=0.858$

$\therefore$ Mole percent benzene $=14\%$

Mole percent Tohene $=86\%$