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NEET Chemistry Solutions 2024 NEET 2024 (Re-NEET)

NEET Chemistry Question (2024) — Solution

Question

The amount of glucose required to prepare 250 mL of M 20 aquepus solution is : (Molar mass of glucose : 180 ~g ~mol ^ -1 )

Options

  1. A. 2.25 g
  2. B. 4.5 g
  3. C. 0.44 g
  4. D. 1.125 g

Answer

A. 2.25 g

Step-by-step solution

Molarity, M = w _2 1000 M _2 ( V ) w _2= Amount of glucose Given molarity = M 20 1 20 = w_2 1000 180 250 w_2= 180 250 20 1000 =2.25 ~g

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