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Molecular Mass

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Question

Calculate the percentage of iron in $K_{3} F e (C N)_{6}$ .

A

17

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B

34

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C

15

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D

1.7

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Solution

The correct option is A 17
Molar mass of $K_{3} [F e (C N)_{6}] = 329.24 g m$
% of iron in $K_{3} [F e (C N)_{6}] = 56 \times \frac{100}{329.24} = 17$ %

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