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 is decreased to 120.2 mm. The molecular weight of the solute is

Options

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

Answer

B

Explanation

As, `(P^@-P)/(P^@) = (w/m)/(w/m + W/M)`
`P^@` = vapour pressure of pure component
P= vapour pressure in solution
w= mass of solute, m= mol. wt, of solute
W = mass of solvent, M=mol. wt, of solvent
`implies (121.8-120.2)/(121.8) = (15/m)/(250/78) implies m = 356.7 g`