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 k Pa
B. 144.5 kPa
C. 72.0 kPa
D. 36.1 kPa

Answer

C

Explanation

`P_("total") = x_("hep") xx P_("hep")^(@) xx x_("oct") xx P_("oct")^(@)`
`n_("hep") = 25/100=0.25, n_("oct") = 35/114=0.307`
`x_("hep") = 0.25/(0.25 + 0.307)=0.45` `x_("oct")=1-0.45 = 0.55`
`P_("total")=0.45 xx 105 + 0.55 xx 45 kPa`
`P_("total") = 47.25 + 24.75 = 72 kPa`