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Percentage Composition

40 g BaMnO4 2...

Question

$40 g B a (M n O_{4})_{2}$ (mol. wt. $= 375$ g/mol) sample containing some inert impurities in acidic medium is completely reacted with $125$ mL of $33.6 V$ of $H_{2} O_{2}$ . What is the percentage purity of the sample?

28.12 %

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70.31 %

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85 %

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None of the above

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Solution

The correct option is B $70.31 %$
Mass of $H_{2} O_{2}$ in $125$ ml solution $= \frac{17}{5600} \times 33.6 \times 125$ g
So, number of equivalents of $H_{2} O_{2} = 33.6 \times \frac{125}{5600}$
As number of equivalents of both are same,
$\frac{x}{\frac{375}{10}} = 33.6 \times \frac{125}{5600}$
or, $x = 28.125$
So, percentage purity of the sample $= \frac{28.125}{40} \times 100 = 70.31 %$

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