A solution containing `10 g per dm^(3)` of urea (mol.wt. `= 60 g mol^(-1)`) is isotonic with a `5%` ( mass//vol.) of a non-volatile solute. The molecular mass of non-volatile solute is:

Options

A. `250 g mol^(-1)`
B. `300 g mol^(-1)`
C. `350g mol^(-1)`
D. `200 g mol^(-1)`

Answer

B

Explanation

10 g per `dm^(3)` of urea is isotonic with 5% solution of a non-volatile solute. Hence, between these solutions osomosis is not possible, so their molar concentrations are equal to each other. Thus, molar concentration of urea solution
`=(10g//dm^(3))/("Molecular weightof urea")=(10)/(60)M=(1)/(6)M`
Molar concentration of 5% non-volatile solute
`=(50g//dm^(3))/("Molecular weight of non-volatile solute")=(50)/(m)M`
Both solutions are isotonic to each other, therefore
`(1)/(6)=(50)/(m)`
or `m=50xx6=300 g mol^(-1)`