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$500 ml$ sample of $H_{2} O_{2}$ was labelled as $" 11.2 V "$ $H_{2} O_{2}$ whereas on complete decomposition of the given sample it liberates $5 lit$ of $O_{2}$ at $1 atm$ & $273 K$ Calculate percentage error in labelling.

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12%

15%

10.7%

9.7%

answer is A.

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Detailed Solution

$2 H_{2} O_{2} \to 2 H_{2} O + O_{2} 0.4464 0.2232 M = 0.4464 \times 2 VS = 11.2 \times M VS = 11.2 \times 2 \times 0.4464 = 10 {VS}_{given} = 11.2 % error = \frac{1.2}{10} \times 100 = 12 %$

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