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Q.
At $ 25{}^\circ C, $ at 5% aqueous solution of glucose (molecular weight $ =180\,g\,mo{{l}^{-1}} $ ) is isotonic with a 2% aqueous solution containing and unknown solute. What is the molecular weight of the unknown solute?
Since the two solutions are isotonic, they must have same concentrations in moles/litre. For glucose solution, concentration
$ =5\text{ }g/100\text{ }c{{m}^{3}} $ (given)
$ =50g/L $
$ \therefore $ $ \frac{50}{180}=\frac{20}{M} $
or $ M=72 $
( $ \because $ Molar mass of glucose $ =180\text{ }g\text{ }mo{{l}^{-1}} $ )
For unknown substance, concentration
$ =2\text{ }g/100\text{ }c{{m}^{3}} $ (given)
$ =20g/L=\frac{20}{M}mol/L $
$ \therefore $ $ \frac{50}{180}=\frac{20}{M} $ or $ M=72 $