1
Question
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
(C) 5.08 percentage
(D) 7.01 percentage
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
(C) 5.08 percentage
(D) 7.01 percentage
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Solution
M0.98O
Let M2+ be x so that M3+ will be 0.98-x. Total charge on the compound must be zero so that
+2(x) + 3(0.98-x) - 2 = 0 (charge on oxygen atom = -2)
2x + 2.94 - 3x - 2 = 0
x = 0.94
Amount of M3+ present = 0.98 - 0.94 = 0.04
Fraction of M3+ present = 0.04/0.98
= 0.0408
so option A is correct
Hope you understand the query 😀
Let M2+ be x so that M3+ will be 0.98-x. Total charge on the compound must be zero so that
+2(x) + 3(0.98-x) - 2 = 0 (charge on oxygen atom = -2)
2x + 2.94 - 3x - 2 = 0
x = 0.94
Amount of M3+ present = 0.98 - 0.94 = 0.04
Fraction of M3+ present = 0.04/0.98
= 0.0408
so option A is correct
Hope you understand the query 😀
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