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

Molar mass of benzene (C₆H₆) = 6 × 12 + 6 × 1

= 78 g mol⁻¹

Molar mass of toluene (C₆H₅CH₃) = 7 × 12 + 8 × 1

= 92 g mol⁻¹

Now, the number of moles present in 80 g of benzene = `80/78` mol

= 1.026 mol

And the number of moles present in 100 g of toluene = `100/92` mol

= 1.087 mol

∴ Mole fraction of benzene, χ_b = `(1.026)/(1.026 + 1.087)`

= 0.486

And the mole fraction of toluene, χ = 1 − 0.486

= 0.514

It is given that the vapour pressure of pure benzene, `p_b^0` = 50.71 mm Hg

And, vapour pressure of pure toluene, `p_t^0` = 32.06 mm Hg

Therefore, the partial vapour pressure of benzene, `p_b = chi_b xx p_b^0`

= 0.486 × 50.71

= 24.65 mm Hg

And, partial vapour pressure of toluene, `p_t = chi_t xx p_t^0`

= 0.514 × 32.06

= 16.48 mm Hg

As a result, the mole fraction of benzene in the vapour phase is as follows:

`p_b/(p_b + p_t)`

= `24.65/(24.65 + 16.48)`

= `24.65/41.13`

= 0.599

= 0.6