On mixing, heptane and octane form an ideal solution. At `373K` the vapour pressure of the two liquid components (heptane and octane) are `105 kPa` and `kPa` respectively. Vapour pressure of the solution obtained by mixing `25.0` of heptane and `35g` of octane will be (molar mass of heptane `= 100 g mol^(-1)` and of octane `= 114 g mol^(-1))`:-

Options

A. `72.0kPa`
B. `36.1kPa`
C. `96.2kPa`
D. `144.kPa`

Answer

A

Explanation

`p_(T)=x_(H)*p_(H)^(@)+x_(o)*p_(o)^(@)`
`x_(H)((25)/(100))/(25/100+35/114)=0.45`
`:.x_(o)=1-0.45=0.55`
`p_(T)=0.45xx105+0.55xx45=72kpa`