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 fraction of pentane at vapour phase would be

Options

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

Answer

C

Explanation

`P_(M)=P_(C_(5)H_(12))^@.X_(C_(5)H_(14))^@+P_(C_(6)H_(14))^@.X_(C_(5)H_(14))`
Thus `P_(M)=440xx(1)/(40)+120xx(4)/(5)=184` lt brgt Now, `P_(C_(5)H_(12))=P_(C_(5)H_(12))^0. X_(C_(5)H_(14)(l))=P_(M).X_(C_(5)H_(12)(g))`
`:.440xx(1)/(5)=184xxX_(C_(5)H_(12)(g))`
`:. X_(C_(5)H_(12)(g))= 0.478`