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

Answer

C

Explanation

`n_(h) = (25)/(100) = 0.25`
` n_(o) =(35)/(114) = 0.307`
` x_(h) = (0.25)/(114) = 0.307`
`x_(h) = (0.25)/(0.25+0.307) = 0.45`
`therefore" "x_(o) = 1 - 0.45 = 0.55`
` p = x_(x)p_(x)^(@) + x_(o)p_(o)^(@)`
` = 0.45 xx 105+ 0.55xx45`
` = 47.25 +24.75`
` = 72 k Pa`