The vapour pressure of pure benzene at a certain temperature is `640 mm` of `Hg`. A non-volatile non-electrolyte solid weighing `2.175 g` added `39.0 g` of benzene. The vapour pressure of the solution is `600 mm` of `Hg`. What is the molecular weight of solid substance?

Options

A. `49.50`
B. `59.60`
C. `69.60`
D. `79.82`

Answer

C

Explanation

Given , vapour pressure of benzene,
`P^(@)=640` mm Hg
Vapour pressure of solution,
`P=600` mm Hg
Weight of solurte `w=2.175g`
Weighof benzene, `W=39.08g`
Molecular weight of benzene,
`M=78g`
Molecular weight of solute m=?
According to law,
`(P^(@)-p)/(p^(@))=(wxxM)/(mxxW)`
`(640-600)/(640) =(2.175xx78)/(mxx39.08)`
`(40)/(640)=(2.175xx78)/(mxx39.08)`
`m=(16xx2.175xx78)/(39.08)`
`m=69.60`