A) \[90\centerdot 67\]
B) \[91\centerdot 67\]
C) \[8\centerdot 33\]
D) \[9\centerdot 33\]
Correct Answer: B
Solution :
Let the amount of \[{{M}^{2+}}\] ions in the \[{{M}_{0.96}}{{O}_{1.00}}=x\] .The amount of \[{{M}^{3+}}\] ions will be = (0-96 - x). As the metal oxide is neutral, (+2)x + (+3) (0.96 -x) + (-2)(\[1\centerdot 00\]) = 0 or 2x + \[2\centerdot 88\] - 3x - \[2\centerdot 00\] =0 -1x + \[0\centerdot 88\] = 0 x = \[0\centerdot 88\] \[{{M}^{2+}}\] = \[0\centerdot 88\] \[{{M}^{3+}}\]= \[0\centerdot 96\] - \[0\centerdot 88\] = \[0\centerdot 08\] \[%\,\,of\,\,{{M}^{2+}}=\frac{0\centerdot 88}{0\centerdot 96}\times 100=91\centerdot 67%\]
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