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

Explanation

`"Given "P_(1)^(@)="0.850 bar, "w_(2)=0.5," "w_(1)=39.0," "m_(1)=78," "P_(1)="0.845 bar"`
`(p_(1)^(@)-p_(1))/(p_(1)^(@))=(w_(2))/(m_(2))xx(m_(1))/(w_(1))`
`=(0.850-0.845)/(0.850)=(0.5xx78)/(m_(2)xx39)`
`therefore M_(2)^(r)="170 g mol"^(-1)`