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

Explanation

Moles of benzene `= (80)/(78) = 1.026 " "`(Molecular mass of `C_(6)H_(6) = 78`)
Moles of toluene ` = (100)/(92) = 1.087 " "` (Molecular mass of `C_(7)H_(8) = 92` )
Mole fraction of benzene,
`x_(b) = (1.026)/(1.023+1.087) = 0.486`
Mole fraction of toluene,
`x_(t) = 1 - 0.486 = 0.514`
`p_(b) = p_(b)^(@) xx x_(b)`
` = 50.71 xx 0.486 24.65 mm Hg`
`p_(t) = p_(t)^(@) xx x_(t)`
` =32.06 xx 0.514 = 16.48`
Total vapour pressure = 24.65+16.48.
= 41.13 mm Hg .
Mole fraction of benzene in vapour phase.
`y_(b) = (24.65)/(41.13) = 0.60`