Q.

When a solution of urea (6%) is isotonic with a solution of glucose                     πgluocse =πurea,Cglucose=Curea ∴               wg×1000mg×100=wu×1000mu×100 where, wg= mass of glucose=x g               mg=molecular mass of glucose=180 g mol-1                mu=molecular mass of urea=60 g mol-1 ∴             x×1000180×100=6×100060×100⇒x=18 gThus, 18 g of glucose is present in 100 mL of solution. In other words, 1 M solution of glucose ( 18 g in 100 mL) is isotonic with 6% solution of urea.