Heptane and octane form an ideal solution. At `373 K`, the vapour pressure of the two liquids are `105.0` kPa and `46.0` kPa, respectively. What will be the vapour pressure, of the mixture of `25 g` of heptane and `35 g ` of octane ?

Answer

`73.43kPa`

Explanation

`p_(hepten e)^(@)=105.2kPa,p_(" octane")^(@)=46.28kPa`
`Mw` of heptane `(C_(7)H_(16))=100`
`Mw` of octane `(C_(8)H_(18))=114`
Moles of heptane `=(26)/(100)=0.26`
Moles of octane `=(35)/(114)=0.31`
`CHMi_(hepta n e)=(0.26)/(0.26+0.1)=0.456`
` CHMi_(octa n e)=1-0.456=0.544`
`p_(hep ta n e)=p_(hepta n e)^(@)xxCHMi_(hep t an e)`
`=105.2xx0.456=47.97kPa`
`p_(oc t an e)=p_(ocatn e)^(@)xxCHMi_(octa n e)`
`=46.8kPaxx0.544=25.46kPa`
`p_(t otal)=47.97+25.46=73.43kPa`