A solution has `1:4` mole ratio of pentane to hexane . The vapour pressure of pure hydrocarbons at `20^@C`are `440` mmHgfor pentane and `120`mmHg for hexane .The mole

Options

A. `0.549`
B. `0.200`
C. `0.786`
D. `0.478`

Answer

D

Explanation

Total vapour pressure of mixture = Vapour pressure of pentane in mixture + vapour pressure of hexane in mixture
Since, the ratio of pentane to hexane = 1 : 4
`therefore` Mole fraction of pentane `=(1)/(5)`
Mole fraction of hexane `=(4)/(5)`
= (mole fraction of pentane `xx` vapour presure of pentane) + (mole fraction of hexane `xx` vapour pressure of hexane)
`=((1)/(5)xx440+(4)/(5)xx120)=184 mm`
`because` Vapour pressure of pentane in mixture = Vapour pressure of mixture `xx` mole fraction of pentane in vapour phase
`88=184xx` mole fraction of pentane in vapour phase
`therefore` Mole fraction pentane in vapour phase
`=(88)/(184)=0.478`