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

If A = Heptane , B = octane
`P_(S)=P_(A)^(@)X_(A)+P_(B)^(@)X_(B)`
`X_(A)=(25//100)/((25)/(100)xx(35)/(114))=(0.25)/(0.25+0.30)=(0.25)/(0.55)=(5)/(11)`
Then `X_(B)=(6)/(11)`
`P_(S)=105xx(5)/(11)+45xx(6)/(11)=47.727+24.54~~72kPa`