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Experimentally it was found that a metal oxide has formula M 0.98 O . Metal M, is present as M 2 + and M 3 + in its oxide. Fraction of the metal which exists as M 3 + would be:

Options

  1. A6 . 0 5 %
  2. B5 . 0 8 %
  3. C7 . 0 1 %
  4. D4 . 0 8 %

Correct answer

D. 4 . 0 8 %

Step-by-step solution

Consider one mole of the oxide, then Moles of M = 0.98; Moles of O 2- = 1 Let 'x' moles of M are in +3 oxidation state, then Moles of M 2+ = 0.98-x On doing charge balancing we get (0.98 - x)2 + 3x = 2 Or 1.96 - 2x + 3x - 2 = 0 Or x = 0.04 ∴ % of   M 3 + = 0 .04 0 .98 × 100 = 4 .08 %

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