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?

Explanation

Here, we are given that `w_(2)=0.5 g, w_(1)=39.0g, M_(1)="78 g mol"^(-1)`
`p^(@)="0.850 bar, "p_(s)="0.845 bar"`
Substituting these values in the formula, `(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(w_(2)//M_(2))/(w_(1)//M_(1))=(w_(2)M_(1))/(w_(1)M_(2))`, we get
`=("0.850 bar"-"0.845 bar")/("0.850 bar")=(0.5 gxx"78 g mol"^(-1))/(39.0g xxM_(2))`
`"or "M_(2)=(0.5xx78)/(39.0)xx(0.850)/(0.005)" g mol"^(-1)="170 g mol"^(-1)`