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 is (mol. Weight of solvent = 78)

Options

A. 356.2
B. 456.8
C. 530,1
D. 656.7

Answer

A

Explanation

Given vapour pressure of pure solvent `(P^@) = 121.8 mm Hg`, Weight of solute (w) = 15 g
Weight of solvent (W) = 250 g, Vapour pressure of solution (P) = 120.2 mm Hg and Molecular weight of solvent(M) = 78
From Raoult.s law
`=(P^@-P)/P^@ = w/m xx M/W`
`(121.8 -120.2)/(121.8) = 15/m xx 78/250`
or `m = (15 xx 78)/(250)xx(121.8)/1.6 = 356.2`