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. `61.2 kPa`
B. `36.1kPa`
C. `96.2kPa`
D. `144.5kPa`

Answer

1

Explanation

`P_(T)=X_("Heptane")P_("Heptane")^(@)+X_("Octane")^(@)P_("Octane")^(@)`
`=(0.25)/(0.557)xx75+(0.307)/(0.557)xx50`
`33.66+27.558=61.2kPa`