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Stoichiometric Calculations

25.0 g of F...

Question

$25.0$ g of $F e S O_{4} . 7 H_{2} O$ was dissolved in water containing dilute $H_{2} S O_{4}$ and the volume was made up to $1.0$ L. $25.0$ mL of this solution required $20$ mL of $\frac{N}{10} K M n O_{4}$ solution for complete oxidation. The percentage of $F e S O_{4} .7 H_{2} O$ in the acidic solution is :

A

78 %

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B

98 %

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C

89 %

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D

79 %

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Solution

The correct option is C $89 %$
Mili equivalents of $F e S O_{4} .7 H_{2} O$ in $25$ ml $=$ Milli equivalents of $K M n O_{4}$ used $= 2$ milli equivalents

Milli equivalents of $F e S O_{4} .7 H_{2} O$ in $1000$ ml $= 80$ milli equivalents.

Mass of $F e S O_{4} .7 H_{2} O$ in solution = $\frac{80}{1} \times 278 \times \frac{1}{1000} = 22.24$ g

$%$ of $F e S O_{4} .7 H_{2} O = \frac{22.24}{25} \times 100 = 88.89 = 89 %$

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