The volume strength (in $\mathrm{L}$ ) of $3 \mathrm{NH}_2 \mathrm{O}_2$ is approximately
The volume strength (in $\mathrm{L}$ ) of $3 \mathrm{NH}_2 \mathrm{O}_2$ is approximately
- $3$
- $8$
- $17$
- $9$
Solution
$3 \mathrm{~N}$ means 3 equivalent weight in $1 \mathrm{~L}$ solution,
$N=\frac{\text { Weight }}{\text { Equivalent weight }}$
Equivalent weight for $\mathrm{H}_2 \mathrm{O}_2=17 \mathrm{~g}$
$\therefore \quad 3=\frac{\text { Weight }}{17}$
Weight $=3 \times 17=51 \mathrm{~g}$
$\mathrm{H}_2 \mathrm{O}_2 \longrightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2$
We know that, $68 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}_2$ gives $22.4 \mathrm{~L}$ of oxygen.
$\therefore 51 \mathrm{~g}$ of $\mathrm{H}_2 \mathrm{O}_2$ will give $=22.4 \times \frac{51}{68}$ of oxygen
$=16.8 \approx 17$
Asked in: AP EAMCET 2021 (24 Aug Shift 2)
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