What is the molarity of $11.2 V$ of $H_{2} O_{2}$ ?

1 M

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2 M

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5.6 M

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11.2 M

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Solution

The correct option is B $1 M$
1 L of $H_{2} O_{2} (a q)$ provides $11.2$ L of $O_{2}$ at STP.
Moles of $O_{2} = \frac{11.2}{22.4} = 0.5$
$n_{H_{2} O_{2}} r e q u i r e d = 0.5 \times 2$
$M_{H_{2} O_{2}} = \frac{n_{H_{2} O_{2}}}{V_{s o l u t i o n}} = 1 M$ .

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