Question

Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and toluene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase if 80 g of benzene is mixed with 100 g of toluene.

Answer 1

The correct answer is: 0.6
Molar mass of benzene $(C_6H_6)=6 \times 12+6 \times 1$
$=78gmol^{-1}$
Molar mass of toluene $(C_6H_5CH_3)=7 \times 12+8 \times 1$
$=92gmol^{-1}$
Now, no. of moles present in 80 g of benzene $=\frac{80}{78}mol=1.026mol$
And, no. of moles present in 100 g of toluene $=\frac{100}{92}mol=1.087mol$
∴Mole fraction of benzene, $x_b=\frac{1.026}{1.026+1.087}=0.486$
And, mole fraction of toluene, $x_t=1-0.486=0.514$
It is given that vapour pressure of pure benzene, $p^o_b=50.71mm\,Hg$
And, vapour pressure of pure toluene, $p^o_t=32.06mm\,Hg$
Therefore, partial vapour pressure of benzene, $p_b=x_b \times p_b$
$=0.486 \times 50.71$
=24.645mmHg
And, partial vapour pressure of toluene, $p_t=x_t \times p_t$
$=0.514 \times 32.06$
=16.479 mmHg
Hence, mole fraction of benzene in vapour phase is given by:
$\frac{p_b}{p_b+pt}$
$=\frac{24.645}{24.645+16.479}$
$=\frac{24.645}{41.124}$
=0.599
=0.6