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Question

If a piece of iron gains 10% of its weight due to partial rushing into Fe2O3 the percentage of total iron that has rusted is:

A
23
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B
13
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C
23.3
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D
90
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Solution

The correct option is D 90
Let mass of piece of iron be m gram
After resulting, iron obtains= m×10100=0.1m
Number of moles before rusting= m56g
Number of moles obtained after rusting= 0.1m56
So, amount of mole iron that has rusted is
m560.1m56=0.9m56
% of iron that has rusted= 0.9m56m56×100=90 %

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