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.
Solution
Molar mass of benzene (C6H6) = 6 x 12 + 6 x 1
= 78 gmol-1
Molar mass of toluene (C6H5CH3) = 7 x 12 + 8 x 1
= 92 g mol-1
Now, no. of moles present in 80 g of benzene = 80/78 mol = 1.026 mol
And, no. of moles present in 100 g of toluene = 100/92 mol = 1.087 mol
∴Mole fraction of benzene, `x_b = (1.026)/(1.026 + 1.087) = 0.486`
And, mole fraction of toluene, `x_1 = 1 - 0.486 = 0.514`
It is given that vapour pressure of pure benzene, `p_b^0 = 50.71 mm Hg`
And, vapour pressure of pure toluene, `p_1^0` = 32.06 mm Hg
Therefore, partial vapour pressure of benzene, `p_b = x_b xx p_b`
=0.486 x 50.71
= 24.645 mm Hg
And, partial vapour pressure of toluene, `p_t = x_t xx p_t`
= 0.514 x 32.06
= 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
