The strength of $11.2$ volume solution of $\mathrm{H}_{2} \mathrm{O}_{2}$ is : [Given that molar mass of $\mathrm{H}=1 \mathrm{~g} \mathrm{~mol}^{-1}$ and $\mathrm{O}=16 \mathrm{~g} \mathrm{~mol}^{-1}$ ]
Correct Option: , 2
$11.2 \mathrm{~V}$ strength of $\mathrm{H}_{2} \mathrm{O}_{2}$ means,
$11.2 \mathrm{~L}$ of $\mathrm{O}_{2}$ is liberated at STP.
$\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\frac{1}{2} \mathrm{O}_{2}$
$11.2 \mathrm{~L}$ of $\mathrm{O}_{2}$ at $\mathrm{STP}=0.5 \mathrm{~mol}$
$\therefore \quad$ No. of moles of $\mathrm{H}_{2} \mathrm{O}_{2}=1 \mathrm{~mol}$
i.e., $1 \mathrm{~L}$ of given $\mathrm{H}_{2} \mathrm{O}_{2}$ solution has 1 mole of $\mathrm{H}_{2} \mathrm{O}_{2}$ (i.e., 34 g)
Strength $=\frac{34}{1000} \times 100=3.4 \%$
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