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Question

32g of a sample of $${ FeSO }_{ 4 }.{ 7H }_{ 2 }O$$ where the dissolved in dilute Sulphuric acid and water and it's volume was made up to 1 litre, 25 mL of this solution required 20 mL of 0.02 $$M\ KMn{ O }_{ 4 }$$ solution of complete oxidation. Calculate the weight % of $$FeS{ O }_{ 4 }.{ 7H }_{ 2 }O$$ in the sample. 


A
34.75
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B
69.5
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C
89.5
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D
None of these
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Solution

The correct option is C 69.5
$$M_{eq}$$ of $$FeSO_4.7H_2O$$ in $$25mL$$=$$M_{eq}$$ of $$KMnO_4$$
$$=20\times 0.2 \times 5=80$$
So,$$M_{eq}$$ of $$FeSO_4.7H_2O$$  in 1000mL$$=20\times 0.2 \times 5\times\dfrac{1000}{25}=80$$
$$\implies \dfrac{W}{278}\times 1\times 1000=80$$
$$\implies W=22.24$$
So,%$$feSO_4.7H_2O=\dfrac{22.4}{32}\times 100=69.5$$

Chemistry

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