sblock Elements 1 Question 5
5. 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 $\mathrm{O}=16 \mathrm{~g} \mathrm{~mol}^{-1}$ ]
(a) $1.7 $%
(b) $34 $%
(c) $13.6 $%
(d) $3.4 $%
(2019 Main, 8 April II)
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Answer:
Correct Answer: 5. (d)
Solution:
- 11.2 volume of $\mathrm{H}_2 \mathrm{O}_2$ means that $1 \mathrm{~mL}$ of this $\mathrm{H}_2 \mathrm{O}_2$ will give $11.2 \mathrm{~mL}$ of oxygen at STP.
$ \underset{2 \times 34 \mathrm{g}}{2 \mathrm{H}_2 \mathrm{O}_2(l)} \longrightarrow \underset{22.4 \mathrm{~L} \text { at STP }}{\mathrm{O}_2(g)}+2 \mathrm{H}_2 \mathrm{O}(l) $
$ 22.4 L of \mathrm{O} _ 2 \text {at STP is produced from} \mathrm{H}_2 \mathrm{O} _ 2=68 \mathrm{g} $
$\therefore 11.2 \mathrm{L}^2$ of $\mathrm{O}_2$ at STP is produced from
$\mathrm{H}_2 \mathrm{O}_2=\dfrac{68}{22.4} \times 11.2=34 \mathrm{~g} $
$ \therefore 34 \mathrm{g}$ of $\mathrm{H}_2 \mathrm{O}_2$ is present in 1000 g of solution
$ \therefore % w / w=\dfrac{34}{1000} \times 100=3.4 % $
BETA
