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. `144.5 kPa`
B. `72.0 kPa`
C. `36.1 kPa`
D. `96.2 kPa`

Answer

B

Explanation

`P_(A) = P_(A)^(@) X_(A) = 105 xx (1//4)/(1//4 +0.307)`
`= 105 xx 0.449 = 47.13K Pa`
`P_(B) = P_(B)^(@) X_(B) = 45 xx 0.551 = 24.795`
`P_(tau) = P_(A) +P_(B) = 71.925 kPa`