The relative lowering of vapour pressure produced by dissolving 71.5 g of a substance in 1000 g of water is 0.00713. The molecular mass of the substance will be:

Options

A. `18.0`
B. 342
C. 60
D. 180

Answer

D

Explanation

`(P^(@)-P_(s))/(P^(@))=((w)/(m))/((w)/(m)+(W)/(M))`
or `0.00713 = (71.5//m)/((71.5)/(m)+(1000)/(18))`