Total vapour pressure of mixture of 1 mole of volatile component A `(P_(A)^(@)=100 mm Hg)` and 3 mole of volatile component `B (P_(B)^(@)=80 mm Hg)` is `90 mm Hg`. For such case `:`

Options

A. there is positive deviation from Raoult's law
B. boiling point has been lowered
C. force of attraction between A and B is smaller than that between A and A or between B and B
D. all the above statements are correct

Answer

4

Explanation

`P_("ideal")=P_(A)^(@)x_(A)+P_(B)^(@)x_(B),`
`=100xx(1)/(4)+80xx(3)/(4)rArr85mmHg`
`P_("actual")=90mmHg,`
Actual `v.pr.` is greater than ideal solution `v.pr.` so `+ve` deviation from Raout's law.