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a
0.05 M solution of glucose
b
6% solution of glucose
c
25% solution of glucose
d
1 M solution of glucose
answer is D.
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Detailed Solution
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.