Question

Equal moles of benzene and toluene are mixed the vapour pressure of benzene and toluene in pure state are 700 and 600 mms Hg respectively.The mole fraction of benzene in vapour state is

Answer 1

Answer: The mole fraction of benzene in vapour state is 0.54

Explanation:-

According to Raoult's law, the vapor pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the vapor pressure of that component in the pure state.

$p_A=x_Ap_A^0$ and $p_B=x_BP_B^0$

where, x = mole fraction

$p^0$ = pressure in the pure state

According to Dalton's law, the total pressure is the sum of individual pressures.

$p_{total}=p_A+p_B$

$p_{total}=x_Ap_A^0+x_BP_B^0$

given $x_{benzene}=0.5$, $x_{toluene}=0.5$

$p_{benzene}^0=700 mmHg$

$p_{toluene}^0=600 mmHg$

$p_{total}=0.5\times700+0.5\times 600=650mmHg$

The mole fraction of benzene in vapour phase is given by:

$y_{benzene}=\frac{p_A}{P_{total}}$

$p_{benzene}=x_{benzene}\times p_{benzene}^0=0.5\times 700=350mmHg$

$y_{benzene}=\frac{350mmHg}{650mmHg}=0.54$