Experimentally it was found that a metal oxide has formula M_0.98O. Metal M, is present as M^(2+) and M^(3+) in its oxide. Fraction of the metal which…

Question

Experimentally it was found that a metal oxide has formula M_0.98O. Metal M, is present as M^(2+) and M^(3+) in its oxide. Fraction of the metal which exists as M^(3+) would be:

MARKS Content Team · Accepted Answer

Correct answer: (D) 4.08%. Consider one mole of the oxide, then Moles of M = 0.98; Moles of O2- = 1 Let 'x' moles of M are in +3 oxidation state, then Moles of M2+ = 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%

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:

Solution

Relevant Questions

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

Solution

Relevant Questions

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: