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Standard XII

Chemistry

Percentage Composition

32g of a samp...

Question

32g of a sample of ${F e S O}_{4} . {7 H}_{2} O$ 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 $M K M n O_{4}$ solution of complete oxidation. Calculate the weight % of $F e S O_{4} . {7 H}_{2} O$ in the sample.

34.75

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69.5

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89.5

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None of these

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Solution

The correct option is C 69.5
$M_{e q}$ of $F e S O_{4} .7 H_{2} O$ in $25 m L$ = $M_{e q}$ of $K M n O_{4}$
$= 20 \times 0.2 \times 5 = 80$
So, $M_{e q}$ of $F e S O_{4} .7 H_{2} O$ in 1000mL $= 20 \times 0.2 \times 5 \times \frac{1000}{25} = 80$
$⟹ \frac{W}{278} \times 1 \times 1000 = 80$
$⟹ W = 22.24$
So,% $f e S O_{4} .7 H_{2} O = \frac{22.4}{32} \times 100 = 69.5$

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32g of a sample of FeSO4.7H2O 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 KMnO4 solution of complete oxidation. Calculate the weight % of FeSO4.7H2O in the sample.

The correct option is C 69.5Meq of FeSO4.7H2O in 25mL=Meq of KMnO4=20×0.2×5=80So,Meq of FeSO4.7H2O in 1000mL=20×0.2×5×100025=80⟹W278×1×1000=80⟹W=22.24So,%feSO4.7H2O=22.432×100=69.5