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The vapour pressure of benzene at 90^(@)...

The vapour pressure of benzene at `90^(@)C` is `1020` torr. A solution of `5g` of a solute in `58.5g` benzene has vapour pressure 990 torr. The molecular weight of the solute is?

A

78.2

B

178.2

C

200

D

220

Text Solution

Verified by Experts

The correct Answer is:
D
`(P^(@)-P_(s))/(P_(s))=(n_(B))/(n_(A))=(W_(B)//Mw_(B))/(W_(A)//Mw_(A))`
`(1020-990)/(990)=(5//x)/(58.5//18)=(5xx78)/(58.5x)`
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