Question

200 ml of a 10 vol of $H_2{O}_2$ and 50 ml of a 20 vol of $H_2{O}_2$ are mixed. What is the volume strength of the mixture?

Answer 1

Given,

$200ml$ $10Vol.$ of $H_2{O}_2$ & $50ml$ $20Vol.$ of $H_2{O}_2$

The reaction involved is:

$2H_2{O}_2⟶2H_{2}O+O_2$

$\therefore 10Vol.$ $H_2{O}_2$ means

$1L$ of $H_2{O}_2$ would produce $10L$ of $O_2$

$\Rightarrow 10V$ $200ml$ of $H_2{O}_2$ would produce

$\frac{200\times 10}{1000}=2L$ of $O_2$

Again $20Vol.$ $H_2{O}_2$ means

$1L\left[1000ml\right]$ of $H_2{O}_2$ would produce $20L$ $O_2$

$\Rightarrow$ 20 Vol. 50ml of $H_2{O}_2$ would produce

$\frac{50\times 20}{1000}=1L$ of $O_2$

Thus total volume of oxygen in resulting solution

$=2+1=3L=3000ml$

Also, Total volume of the solution=$200ml+50ml=250ml$

$\Rightarrow 250ml$ of resulting solution produces $3000ml$ of $O_2$

So, $1000ml$ of resulting solution would produce

$\frac{1000}{250}\times 3000=12000ml$ of $O_2$ or $12L$ of $O_2$

$\therefore$ Strength of final solution=$12$ volume $H_2{O}_2$