Question

Benzene and toluene forms nearly ideal solution.At 313 K the vapour pressure of pure benzene is 150 mm Hg and pure toluene is 50 mm If.Calculate the vapour pressure of a mixture of these two containing their equal masses at 313 K

Answer 1

Answer: The vapour pressure of a mixture of these two containing their equal masses at 313 K is 104 mm Hg.

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

$p^0$ = pressure in the pure state

$x_{benzene}=\frac{\text {no of moles of benzene}}{\text {total no of moles}}=\frac{\frac{x}{78}}{\frac{x}{78}+\frac{x}{92}}=0.54$

$x_{toluene}=\frac{\text {no of moles of toluene}}{\text {total no of moles}}=\frac{\frac{x}{92}}{\frac{x}{78}+\frac{x}{92}}=0.46$

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

$p_{total}=p_1+p_2$

$p_{total}=x_{benzene}p_{benzene}^0+x_{toluene}P_{toluene}^0$

$p_{benzene}^0=150 mmHg$

$p_{toluene}^0=50 mmHg$

$p_{total}=0.54\times 150+0.46\times 50=104mmHg$