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Sulphur dioxide and oxygen were allowed​ to diffuse through a porous partition. 20 dm 3 of SO 2 diffuses through the porous partition in 60 seconds. The volume of O 2 in dm 3 which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur = 32 u );

Options

  1. A28 . 2
  2. B10 . 0
  3. C14 . 1
  4. D7 . 09

Correct answer

C. 14 . 1

Step-by-step solution

As per Graham's law, V SO 2 / t SO 2 V O 2 / t O 2 = p SO 2 p O 2 M O 2 M SO 2 = M O 2 M SO 2 2 0 dm 3 / 6 0 sec V O 2 / 3 0 sec = 3 2 6 4 = 1 1 · 4 1 4 2 0 × 3 0 6 0 × V O 2 = 1 1 · 4 1 4 ⇒ V O 2 = 1 0 × 1 · 4 1 4 =   14 . 14   dm 3

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