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. 96.2 kPa
B. 144.5 kPa
C. 72 kPa
D. 36.1 kPa

Answer

C

Explanation

Heptane A, `n_(A)=(25)/(100)=0.25`
Octane B, `n_(B)=(35)/(114)=0.31`
`x_(A)=(n_(A))/(n_(A)+n_(B))=(0.25)/(0.25+0.31)=0.45`
`x_(B)=0.55`
`p_(A)^(0)x_(A)+p_(B)^(0)x_(B)`
`=105xx0.45+45 xx0.55`
`=72 kPa`]