A `4%` solution of sucrose `C_(12)H_(22)O_(11)` is isotonic with `3%` solution of an unknown organic substance. Calculate the molecular mass of the unknown substance.

Explanation

Since the two solutions are isotonic, they must have same concentrations in moles/litre.
For sucrose solution, we gave`" "` Concentration `=4g//100cm^(3)" (Given) = 40 g/litre "=(40)/(342)" moles/litre"`
`" "(because" Molar mass of sucrose, "C_(12)H_(22)O_(11)="342 g mol"^(-1))`
For unknown substance, suppose M is the mol. mass. Then
Concentration = `3g//100cm^(3)" (Given) = 30 g/litre"=(30)/(M) "moles/litre"`
Thus, we have `(30)/(M)=(40)/(342) or M=(30xx342)/(40)="256.5 g mol"^(-1)`