Question

Half litre each of three samples of $H_{2}O_{2}$ labelled $10 vol, 15 vol, 20 vol$ are mixed and then diluted with $1700$ ml of water. Calculate relative strength of resultant $H_{2}O_{2}$ solution

Answer 1

Volume strength of $H_{2}O_{2} = 5.6 \times N$
$10 Vol. H_{2}O_{2} =\frac{10}{5.6} N H_{2}O_{2}$
$15 Vol. H_{2}O_{2} =\frac{15}{5.6} N H_{2}O_{2}$
$20 Vol. H_{2}O_{2} =\frac{20}{5.6} N H_{2}O_{2}$
Let $500 mL$ of each is mixed then total volume of mixture
$= 1500 mL$
Also this is diluted with $1700 \,ml$, so total volume becomes
$3200$ ml
$N \times 3200=\frac{10 \times500}{5.6} + \frac{15 \times500}{5.6} +\frac{20 \times500}{5.6}$
or $N_{H_2O_2}$ mixture$ = \frac{500 \times45}{5.6 \times3200} =1.255$
$\therefore $ Vol. strength of $H_{2}O_{2} =1.255 \times5.6 =7.03$