Find out the percentage of pure $M n O_{2}$ in the sample

48.88

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58.88

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38.88

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None of the above

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Solution

The correct option is B 48.88
We have, Meq of $M n O_{2} =$ Meq of oxalic acid added-Meq of oxalic acid left
$= (1 \times 50) - (0.1 \times 32 \times 10)$
Therefore, $\frac{w}{E} \times 1000 = 18 [i n 250 ㎖] = 18$
or, $w = \frac{(18 \times 86.9)}{2 \times 1000} [a s E = \frac{86.9}{2}]$
$w = 0.7821 ℊ$
Percentage of $M n O_{2} = \frac{0.7821}{1.6} \times 100$ %
$= 48.88$ %

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