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The ratio of number of oxygen atoms ( O ) in 16.0 ~g ozone ( O ₃ ), 28.0 ~g carbon monoxide ( CO ) and 16.0 oxygen ( O ₂ ) is (Atomic mass : C =12, O =16 and Avogadro's constant N _ A =6.0 10²³ ~mol ⁻¹ )

Options

  1. A3: 1: 2
  2. B1: 1: 2
  3. C3: 1: 1
  4. D1: 1: 1

Correct answer

D. 1: 1: 1

Step-by-step solution

16.0 ~g O ₃= 16 48 mole aligned & = 16 48 6.023 10²³ molecules & =3 16 48 6.023 10²³ atoms & =6.023 10²³ atoms & 28.0 ~g CO = 28 28 mole =1 mole & =1 6.023 10²³ molecules & =1 6.023 10²³ atoms & =6.023 10²³ atoms & 16.0 ~g O ₂= 16 16 mole =1 ~mole aligned =1 6.023 10²³ molecules =1 6.023 10²³ atoms =6.023 10²³ atoms Therefore, the ratio is 6.023 10²³: 6.023 10²³: 6.023 10²³ i.e. 1: 1: 1

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