ElectrochemistryHard
Question
Expermentally 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 :-
Options
A.7.01%
B.4.08%
C.6.05%
D.5.08%
Solution
Let no. of oxygen atom = 100
∴ no. of metal atom = 98
if no. of M+2 ion = x
then no. of M+3 ion = (98 - x)
Since compound is electrically neutral
x × (12) + (98 - x)x + 3 + 100 × (-2) = 0
2x + 98 × 3 - 3x - 200 = 0
294 - 200 - x = 0
x = 94
∴ No. of M+2 ion = 94
No. of M+3 ion = 98 - 94 = 4
∴ Fraction of metal which exists as M+3would be
∴ no. of metal atom = 98
if no. of M+2 ion = x
then no. of M+3 ion = (98 - x)
Since compound is electrically neutral
x × (12) + (98 - x)x + 3 + 100 × (-2) = 0
2x + 98 × 3 - 3x - 200 = 0
294 - 200 - x = 0
x = 94
∴ No. of M+2 ion = 94
No. of M+3 ion = 98 - 94 = 4
∴ Fraction of metal which exists as M+3would be
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