Question

At a given temperature, you have a mixture of benzene ( vapour pressure of pure benzene = 745 torr ) and toulenen (vapor pressure of pure toluene = 290 torr ). The mole fraction of benzene in the solution is 0.590. Assuming ideal behavior, calculate the mole fraction of toluene in the vapor above the solution.

Answer 1

Answer: 0.22

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.

$p_1=x_1p_1^0$ and $p_2=x_2P_2^0$

where, x = mole fraction  in solution

$p^0$ = pressure in the pure state

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

$p_{total}=p_1+p_2$$p_{total}=x_Ap_A^0+x_BP_B^0$

$x_{benzene}=0.590$,  

$x_{toluene}=1-x_{benzene}=1-0.590=0.41$,  

$p_{benzene}^0=745torr$

$p_{toluene}^0=290torr$

$p_{total}=0.590\times 745+0.41\times 290=558.45torr$

$y_{benzene}$ = mole fraction of benzene in vapor phase

$y_{benzene}=\frac{p_{benzene}}{p_{total}}=\frac{0.590\times 745}{558.45}=\frac{439.55}{558.45}=0.78$

$y_{toluene}$ = mole fraction of toluene in vapor phase

$y_{toluene}=1-y_{benzene}=1-0.78=0.22$

Thus the mole fraction of toluene in the vapor above the solution is 0.22