Heptane and octane form ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane ?

Explanation

`"Molar mass of heptane "(C_(7)H_(16))="100 g mol"^(-1)" , Molar mass of octane "(C_(8)H_(18))="114 g mol"^(-1)`
`"26.0 g heptane"=("26.0 g")/("100 g mol"^(-1))="0.26 mol , 35.0 g octane"=("35.0 g")/("114 g mol"^(-1))="0.31 mol"`
`"x (heptane)"=("0.26 g")/(0.26+0.31)=0.456" , x (octane)"=1-0.456=0.544`
`"p (heptane)" = 0.456 xx105.2" kPa = 47.97 kPA, p (octane)"=0.544xx46.8" kPa = 25.46 kPa"`
`P_("Total")=47.97+25.46="73.43 kPa."`