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

Answer

C

Explanation

`P_(M) = P_(C_(5)H_(12))^(@).X_("L.P.of"C_(5)H_(12))+P_(C_(6)H_(14))^(@).X_("G.P.of"C_(6)H_(14))`
`= 440 xx (1)/(5) + 120 xx (4)/(5) = 184 mm`
Now `P_(C_(5)H_(12))^(@) . X_("L.P.of" C_(5)H_(12)) = P_(M) . X_("G.P.of"C_(6)H_(14))`
`440 xx (1)/(5) = 184 xx X_("G.P.of"C_(5)H_(12))`
`X_("G.P.of"C_(6)H_(12))=0.478`