Molarity, M=M2×(V)w2×1000 w2= Amount of glucose Given molarity =20M 201=180×250w2×1000 w2=20×1000180×250 =2.25 g
The amount of glucose required to prepare 250 mL of 20M aquepus solution is : (Molar mass of glucose : 180 g mol−1 )
Held on 30 Apr 2024 · Verified 9 Jul 2026.
2.25 g
4.5 g
0.44 g
1.125 g
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