Calculate the volume in $m L$ hydrogen peroxide labelled 10 volume required to liberate 600 $m L$ of oxygen at $27^{\circ} C$ and 760 $m m$ .

V = 27.3 m L

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V = 54.6 m L

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V = 109.2 m L

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None of these

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Solution

The correct option is B $V = 54.6 m L$
$10 V o f H_{2} O_{2}$
$1 V o f H_{2} O g i v e s 10 V o f O_{2} a t S T P$ .
$1 m L o f 10 V H_{2} O_{2} g i v e s 10 m l o f O_{2}$ .
$V o l u m e o f O_{2} r e q u i r e d a t N T P$
$V_{1} = 600$
$P_{1} = 760 m m$
$V_{2} = ? P_{2}$
$= 760 m m$
$T_{1} = 300 K$
$T_{2} = 273 K$
$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$
$V_{2} = \frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}} = 600 \times \frac{273}{300} = 546 m L$
$10 V o f H_{2} O_{2} p r o d u c e s 10 m L o f O_{2} a t S T P f o r 1 m L o f i t s v o l u m e$ .
$H e n c e, f o r 546 m L o f O_{2} a t N T P, t h e v o l u m e o f 10 V o f H_{2} O_{2} r e q u i r e d = \frac{546 \times 1}{10} = 54.6 m L$

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Similar questions

Volume occupied by a gas at one atmospheric pressure and 0 $^{\circ}$ C is V ml. Its volume at 273 K will be

Time (minutes)	$0$	$10$	$20$	$30$
$V (m l)$	$46.1$	$29.8$	$19.6$	$12.3$

V

m L

34 g

1120 m L

H_{2} O_{2}

100

15 V

H_{2} O_{2}

27 C

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