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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. A2.25 g
  2. B4.5 g
  3. C0.44 g
  4. D1.125 g

Correct 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 = w₂ 1000 180 250 w₂= 180 250 20 1000 =2.25 ~g

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