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
B
69.5
C
89.5
D
None of these

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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