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