At 300 K an ideal solution is formed by mixing 460 g of toluene with 390 k benzene. If the vapour pressure of pure toluene and respectively the mole fraction of toluene in vapour phase is

Options

A. `0.196`
B. `0.588`
C. `0.294`
D. `0.444`

Answer

D

Explanation

Given `w_(a) = 460 g`(toluene)
Vapour pressure of pure toluene `(p_(A)) = 32 ` mm
Vapour pressure of pure benzene `(p_(B))` = 40 mm
Moles of benzene `n_(B) = (W_(B) )/( M_(B)) = (390)/( 78)`
Hence mole fraction of toluene
`X_(A)= ((460)/( 92))/((460)/(92) + (390)/(78))`
`P_(total ) = P^(@) A X A + p^(@) B X B`
`P_(total ) = 32 xx (((460)/(92))/((460)/(92)+ (390)/(78)))+ 40 (((390)/(78))/((390)/(78) + (460)/(92)))`
Mole fraction of toluene in vapour phase `(y_(1))` can be calculated as below :
`p^(@) T X T = Y_(T). T_(total) rArr Y_(T ) = (p^(@ ) T X T)/( P_(total ))`
where `X_(T)` is mole fraction of toluene .
`Y_(T) = (32xx (1)/(2))/(36) rArr Y_(T) = (16)/(36) = 0.444`