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

Answer

3

Explanation

Two solution are said to be isotonic if they have same molarity (or molar concentration)
Molarity = no. of moles of solute per litre of solution
`=(("mass of solute per")/("litre "("or "dm^(3))" of solution"))/("molar mass of solute")`
`"Molarity "(urea) =(10 g per dm^(3))/(60 g mol^(-1))=1/6=M`
`5%` solution of nonvolatile solute contains `5g` of solute per `100 mL` of solutions. Thus `1000 mL ("or 1L or 1 "dm^(3))` of solution will hold `50 g` of solute. Thus
`"Molarity" ("non volatile solute")=(50" g per dm"^(3))/("mm"_(solute))`
where `mm` solute stands for molar mass of solute.
As both the solutions are isotonic to each other we have Molarity (urea) `="Molarity "("nonvolatile solute")`
`1/6M=50/(mm_("solute"))M`
or `mm_("solute")=(50xx6) g mol^(-1)`
`=300" g mol"^(-1)`