Experimentally it was found that a metal oxide has a formula of M0.98 . 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 -----

(A) 4.08 percentage

(B) 6.05 percentage

(D) 7.01 percentage

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Experimentally it was found that a metal oxide has formula M0.98O. Metal M, present as M2+ and M3+ in its oxide. Fraction of the metal which exists as M3+ would be:

The correct option is B 4.08%4.08%Let M2+=x and M3+=y2x+3y=2.0x+y=0.982x+3y=2.02x+2y=1.96(−) (−) (−) ––––––––––––orM3+=4% [0.040.98×100]

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:

The correct option is B $4.08 %$
$4.08 % L e t M^{2 +} = x a n d M^{3 +} = y 2 x + 3 y = 2.0 x + y = 0.98 2 x + 3 y = 2.0 2 x + 2 y = 1.96 (-) (-) (-) - ---------- - o r M^{3 +} = 4 % [\frac{0.04}{0.98} \times 100]$