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$56 V, 500 m l H_{2} O_{2}$ solution is kept in an open container due to which some $H_{2} O_{2}$ is decomposed and evolves $8 g m O_{2}$ , simultaneously some $H_{2} O$ also vapourises. Due to all these changes, final volume is reduced by $20 %$ . Find final volume strength of $H_{2} O_{2} (a q)$ :
(Assume 1 mole of an ideal gas occupies $22.4 L$ at STP)

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$33.6 V$

$56 V$

$44.8 V$

$11.2 V$

answer is A.

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Detailed Solution

$H_{2} O_{2} \to H_{2} O + \frac{1}{2} O_{2}$

$\frac{56}{11.2} \times 0.5$

$= 2.5 moles$

$2.5 - 0.5$

$= 2.0 moles$

New V.S. $= 11.2 \times \frac{2}{0.8 \times 0.5}$

$= 56 V$

Hence, option A is correct.

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