The mass of glucose that should be dissolved in 50 g of water in order to produce the same lowering of vapour pressure as is produced by dissolving 1 g of urea in the same quantity of water is

Options

A. 1 g
B. 3 g
C. 6 g
D. 18 g

Answer

B

Explanation

`(p_(0)-p_(s))/(p_(s))=(w_(1)M_(2))/(w_(2)M_(1))`
To produce same lowering of vapour pressure `((p_(0)-p_(s))/(p_(s)))` term will be same for both the cases (solvent is same).
`therefore (w_(g)xx18)/(50xx180)=(w_(u)xx18)/(50xx60)`
(`w_(g)=` weight of glucose, `w_(u)=` weight of urea)
or, `(w_(g)xx18)/(50xx180)=(1xx18)/(50xx60)` or, `w_(g)=3`.