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.

Answer

`CHMi^(V)``_(benzen e)=0.60`

Explanation

`Mw` of benzene `(C_(6)H_(6))=78 g mol^(-1)`
`Mw` of toluen e `(C_(6)H_(5)CHM_(3))=92 g mol^(-1)`
`n_("benzene")=(80g)/(78g mol^(-1))=1.026mol`
`n_("toluence")=(100g)/(92g mol^(-1))=1.087mol`
`CHMi_(benzen e)=(1.026)/(1.026=1.087)=(1.026)/(2.113)=0.486`
`CHMi_("toluene")=1-0.486=0.514`
`p_("benzene")^(@)=50.71mm, p_("toluene")^(@)=32.06 mm`
Using Raould's law
`p_("benzen e")=CHMi_("benzene")xxp_("toluence")^(@)=0.486xx50.71mm`
`=24.65mm`
`p_("toluene")=CHMi_("toluene")xxp_("toluene")^(@)=0.514xx32.06mm`
`=16.48mm`
`P_("total")=p_("benzene")+p_("toluene")=24.65+16.48=41.13`
`CHMi_("benzene")^(V)=` mole fraction of benzen in vapour phase `=(p_("benzene"))/(P_("total"))=(24.65)/(41.13)=0.60`
`CHMi_("toluen e")^(V)=` mole fraction of toluene is vapour phase
`=1-0.60=0.40`.