Question

$0.56$ litre each of three samples of $H_2{O}_2$ labelled as $10vol,15vol$ and $20vol$ are mixed. The resultant normality of $H_2{O}_2$ solution is:

• A. $1.68\text{N}$
• B. $2.68\text{N}$
• C. $4.5\text{N}$
• D. $8.0\text{N}$

Answer 1

The correct option is B $2.68\text{N}$

$\text{Volume strength of}{H}_2{O}_2=5.6\times N$
$\therefore \text{Normality of 10 vol}=\frac{10}{5.6}N$
$\text{Normality of 20 vol}=\frac{20}{5.6}N$
$\text{Normality of 15 vol}=\frac{15}{5.6}N$
$\text{As 0.56 litre of}{H}_2{O}_2\text{is taken}$
$\therefore \text{Total equivalent of}{H}_2{O}_2=\frac{0.56\times 10}{5.6}+\frac{0.56\times 15}{5.6}+\frac{0.56\times 20}{5.6}=1+1.5+2.0=4.5N_{H_2{O}_2}=\frac{Eqv}{V}$
$N_{H_2{O}_2}=\frac{4.5}{3\times 0.56}=2.68\text{N}$