The vapour pressure of pure benzene at a certain temperature is `0.850` bar. A non-volatile, non-electrolyte solid weighting `0.5 g` when added to `39.0 g` of benzene (molar mass `78 g mol^(-1)`). The vapour pressure of the solution then is `0.845` bar. What is the molar mass of the solid substance?

Options

A. 58
B. 180
C. 170
D. 145

Answer

C

Explanation

`P_(A)^(@)=0.850" bar, " P_(S)=0.845` bar
`w=0.5 g, " " m = ?`
weight of solvent (benzene) = 39.0 g
and molecular weight of benzene = 78 g
As we knowm `(P_(A)^(@)-P_(S))/(P_(A)^(@))=x_(B)=(n_(B))/(n_(A))`
`(0.850-0.845)/(0.850)=((w_(B))/(M_(B)))/((w_(A))/(M_(A)))=((0.5)/(m))/((39)/(78))`
On solving, we get
molecular mass of solid structure (m) = 170 g.