Experimentally it was found that a metal oxide has formula $M_{0.98} O .$ Metal M, present as $M^{2 +}$ and $M^{3 +}$ in its oxide. Fraction of the metal which exists as $M^{3 +}$ would be nearly
[Report your answer by rounding it upto nearset whole number]

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For a one mole of the oxide Moles of M = 0.98 , Moles of O 2 - = 1 Let moles of M 3 + = x Moles of M 2 + = 0.98 - x on balancing charge 0.98 - x × 2 + 3 x - 2 = 0 x = 0.04 % of M 3 + = 0.04 0.98 × 100 = 4.08 % ≈ 4 %

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