Heptane and octane form an ideal solution. At 373 K, the vapour pressure 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 25 g of octane ?

Explanation

Vapour pressure of heptane `(p_(1)^(0))=105.2` kPa
We know that,
Molar mass of heptane `(C_(7)H_(16))`
`= 7xx12+16xx1`
`= 100 g mol^(-1)`
Therefore, Number of moles of heptane
`= (26)/(100)=0.26` mol
Molar mass of octane `(C_(3)H_(18))`
`= 8xx12+18xx1`
= 114 g `mol^(-1)`
Therefore, Number of moles of octane
`= (35)/(114)=0.31` mol
Mole fractionn of heptane, `x_(1)`
`= (0.26)/(0.26+0.31)`
= 0.456
And, mole fraction of oxtane, `x_(2)`
`= 1-0.456`
= 0.544
Now, partial pressure of heptane,
`p_(1)=x_(1)p_(1)^(0)`
`= 0.456xx105.2`
= 47.97 kPa
Partial pressure of octane,
`p_(2)=x_(2)p_(2)^(0)`
`= 0.544xx46.8`
= 25.46 kPa
Hence, vapour pressure of solution,
`p_("total")=p_(1)+p_(2)`
`= 47.97+25.46`
= 73.43 kPa