A `5%` solution of cane sugar is isotonic with `0.877%` solution of urea. Calculate the molecular mass of urea if the molecular mass of cane sugar is `342`.

Explanation

Let the molecular mass of urea be `m_(2)`
Molar concentration of sugar `= (w_(1))/(m_(1)xxV_(1))=(5)/(342xx0.1)`
and Molar concentration of urea `= (w_(2))/(m_(2)xxV_(2))=(0.877)/(m_(2)xx0.1)`
For isotonic solutions,
`(w_(1))/(m_(1)V_(1))=(w_(2))/(m_(2)V_(2))`
`(5)/(342xx0.1)=(0.877)/(m_(2)xx0.1)`
`m_(2)=(0.877xx342)/(5)=59.987`.