Heptane and octane form ideal solution. At `373K`, the vapour pressure of the two liquids are `105.2 kPa` and `46.8 kPa` respectively. What will be the vapour pressure, in bar, of a mixture of `26g` of heptane and `35g` of octane?

Explanation

Molar mass of heptane `(C_(7)H_(16))`
`=7xx12+16=100 " g mol"^(-1)`
Molar mass of octane `(C_(8)H_(18))`
`=8xx12+18=114 " g mol^(-1)`
Moles of heptane present in mixture
`=(26.0)/(100)=0.26 `mol
Moles of octane present in mixture
`(35.0)/(114)=0.307` mol
Mole fraction of heptane `x_(H)`
`=(0.26)/(0.26+0.307)=0.458`
Mole fraction of octane, `x_(0)=(1-0.458)=0.542`
Vapour pressure of heptane`=x_(H)xxP^(@)`
`=0.458xx105.2kPa=48.18 kPa`
Vapour pressure of octane `=x_(o)xxP^(@)`
`=0.542xx46.8 k Pa=25.36 kPa`
Vapour pressure of mixture
`=48.18+25.36=73.54k Pa`.