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

Answer

A

Explanation

`P_(T)=x_(H)*rho_(H)^(@)+x_(0)*rho_(0)^(@)`
`((25)/(100))/((25)/(100)+(35)/(114))`
`x_(H)=0.45,x_(0)=0.55`
`P_(T)=0.45xx105+0.55xx45`
`=72kPa`