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. 35.67 g
B. 356.7 g
C. 432.8 g
D. 502.7 g

Answer

B

Explanation

Relative lowering of vapour pressure is given by
`(P^(0)-P)/(P^(0))=(w//m)/(w//m+W//M)`
where , `P^(0)` =Vapour pressure of pure solvent
P=Vapour pressure of solution
w=mass of slute
m=molecular mass of solute
W=mass of solvent
M=molecular mass of solvent.
For dilute solution
`implies (121.8-120.2)/(121.8)=(15//m)/(250//78)implies m=356.265`