For isotonic solutions [osmotic pressure must be equal] π1=π2 C1RT=C2RT C1=C2 180×1m=60×115( m is the mass of glucose) m=4180=45 g
NEET UG 2024 — Chemistry Physical Chemistry
Mass of glucose (C6H12O6) required to be dissolved to prepare one litre of its solution which is isotonic with 15 g L−1 solution of urea (NH3CONH2) is (Given: Molar mass in gmol−1C:12,H:1,O:16, N:14 )
Held on 30 Apr 2024 · Verified 9 Jul 2026.
55 g
15 g
30 g
45 g
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