Question

A solution prepared by dissolving 1.25g of oil of winter green in 99.0g of benzene has a boiling point of 80.31 degree celsius. determine the molar mas of this compound

Answer 1

Answer

Explanation:

Mass of solute- (Wb)=1.25g

Mass of solvent- (Wa)= 99g

Elevation in boiling point- (∆Tb) = 8031 -90.10

Kb = 2.53℃ kg mol-1

Now,

∆Tb = Kb(Wb×100/Wa×Mb)

0.21 = (2.53×1.25×1000/99×Mb)

Mb = (2.53×1.25×1000/99×0.21)

Mb = 152.116g. (.....Ans.)