The vapour pressure of pure benzene at certain temperature is 0.850 bars. A non-volatile, non-electrolyte solid weighing 0.5 grams when added to 39 grams of benzene (molar mass 78grams), vapour pressure of the solution becomes 0.845 bars. What is the molar mass of the solid substance? 

Explanation

Vapoure pressure of solvent benzene=`P^(0)=0.850` bar
Vapour pressure off solution=P=0.845 bar.
Lowering of vapour pressure=`P^(0)-P=0.850-0.845=0.005` bar
`(P^(0)-P)/(P^(0))=(0.005)/(0.850)=0.00588`
Mass of solute=`W_(2)=0.5g, " Mass of solvent benzene"W_(1)=39g`
Molar mass of benzene=78g `mol^(-1)`
Formula: `M_(2)=(W_(2))/(W_(1))xx(M_(1))/([(P^(0)-P)/(P^(0))])`
Substitution `M_(2)=(0.5)/(39)xx(78)/(0.0059)`
Answer `M_(2)=170g//mol`.