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.

Explanation

Molar mass of `(C_(6)H_(5))`
`= 6xx12+6xx1`
`= 78 g mol^(-1)`
Molar mass of toluene `(C_(6)H_(5)CH_(3))`
`= 7xx12+8xx1`
`= 92 g mol^(-1)`
Now, no. of moles present in 80 g of benzene
`= (80)/(78)=1.026` mol
And, no. of moles present in 100 g of toluene
`= (100)/(92)=0.087` mol
Therefore, Mole fraction of benzene,
`x_(b)=(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 presure of pure benzene,
`p_(b)^(0)=50.71` mm Hg
Therefore, partial vapour presure of benzene,
`p_(b)=x_(b)xx p_(b)`
= 24.645 mm Hg
And, partial vapour pressure of toluene,
`p_(1)=x_(t)xx p_(t)`
`= 0.514xx32.06`
= 16.479 mm Hg
Hence, mole fraction of benzene in vapour phase is given by :
`=(p_(b))/(p_(b)+p_(t))`
`= (24.645)/(24.645+16.479)`
`= (24.645)/(41.124)`
= 0.599
= 0.6