A 6% solution of urea is isotonic with

Options

A. 0.05 M solution of glucose
B. 6% solution of glucose
C. 25% solution of glucose
D. 1M solution of glucose

Answer

D

Explanation

When a solution of urea (6%) is isotonic with a solution of glucose
`pi_("glucose ")= C_("urea")`
`:.((w_(g)xx1000)/(m_(g)xx100))=((w_(u)xx1000)/(m_(u)xx100))`
where `w_(g)=` mass of glucose `=xg`
`m_(g)=` molecular mass o glucose `=180`
`w_(u)=` mass of urea =6g
`m_(u)=` molecular mass of urea =60
`:.(x xx 1000)/(180xx100)=(6xx1000)/(60xx100)impliesx=18g`
Thus 18 g of glucose are present in 100 ml of solution.
In other words 1M solution of glucose (18 g in 100 ml) is isotonic with 6% solution of urea.