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Question

Question 15-16 is based on the following given information: A sample of ferrous oxide has actual formula Fe0.93O. In this sample what fraction of metal ions are Fe2+ ions?

A
0.93
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B
0.81
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C
0.79
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D
0.14
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Solution

The correct option is B 0.81

Let the formula of sample be

(Fe2+)x(Fe3+)yO.

On looking at the given formula of the compound

x + y = 0.93 ... (1)

Total positive charge on ferrous and ferric ions should balance the two

Units of negative charge on oxygen. Therefore,

2x + 3y = 2 ... (2)

x+(32)y=1 …(3)

On subtracting equation (1) from equation (3) we have

(y2) = 0.07 => y =0.14

On putting the value of y in equation (1) we get,

x + 0.14 = 0.93

x = 0.93 – 0.14

x = 0.79

Fraction of Fe2+ ions present in the sample =(0.790.93)=0.81


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