Question

Vapour pressure of benzene at 30°C is 121.8 mm.
When 15 g of a non-volatile solute is dissolved
in 250 g of benzene its vapour pressure decreased
to 120.2 mm. The molecular weight of the solute
(Mo. wt. of solvent = 78)
(a) 356.2 (b) 456.8
(c) 530.1 (d) 656.7

Answer 1

Answer:

Explanation:

Given:

Vapour pressure of pure benzene = 121.8 mm

Vapour pressure of solution = 120.2 mm

Mass of non volatile solute = 15 g

Mass of benzene = 250 g

Calculation of molecular weight of solute:

Using the relation derived from relative lowering in vapour pressure,

MB = wBMAwA×(p°A−pAp°A)Where wA and wB are weights of solvent and solute respectively and MA and MB are their corresponding molar mass.p°A = Vapour pressure of pure benzenepA = Vapour pressure of solutionMB = 15×78250×(121.8−120.2121.8) MB = 1170×121.8250×1.6 = 356.2 g/mol