Question

The vapour pressure of benzene at `90^(@)C` is `1020` torr. A solution of `5 g` of a solute in `58.5 g` benzene has vapour pressure `990` torr. The molecualr weight of the solute is:
A. 78.2
B. 178.2
C. 206.2
D. 220

Answer 1

Correct Answer - D
`(P_(1)^(o)-p)/p_(1)^(o)=(W_(2)//M_(2))/(W_(1)//M_(1))=W_(2)/M_(2) xx M_(1)/W_(1)`
`(1020-990)/990 = (5 xx 78)/(M_(2) xx 58.5)`
`:. M_(2)=(5 xx 78 xx 990)/(30 xx 58.5)=220`