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

Options

A. 1g
B. 3g
C. 6g
D. 8g

Answer

B

Explanation

`(Delta p)/(p^(@))=(p^(@)-p_(s))/(p^(@))=(n_(2))/(n_(1))=(W_(2)M_(1))/(W_(1)M_(2))`
Since `Delta p` and `p^(@)` are same in two cases
`((W_(2)M_(1))/(W_(1)M_(2)))_("glucose")=((W_(2)M_(1))/(W_(1)M_(2)))_("urea")`
`(w_(2)xx18)/(30xx180)=(1xx18)/(50xx60)` or `w_(2) = 3 g`.