Question

$20mL$ of a solution containing $0.2g$ of impure sample of $H_2{O}_2$ reacts with $0.316g$ of $KMn{O}_4$ (acidic). Calculate:
(i) Purity of $H_2{O}_2$.
(ii) Volume of dry $O_2$ evolved at ${27}^{\circ }C$ and $750mmP$.

Answer 1

1.The redox reaction is as follows:

$2{Mn{O}_4}^{-}+5H_2{O}_2\to 5O_2+2M{n}^{2+}$

The number of moles of $KMn{O}_4$ present $=\frac{0.316}{158}=2\times {10}^{-3}$ mol

The number of moles of pure $H_2{O}_2=\frac{5}{2}\times 2\times {10}^{-3}=5\times {10}^{-3}$

$0.2$ g of sample contains $0.17$ g of pure $H_2{O}_2$.

Therefore, percentage purity of $H_2{O}_2$ $=\frac{0.17}{0.2}\times 100=85\%$

2. The redox changes are as follows:

$M{n}^{7+}+5e⟶M{n}^{2+}$

${O_2}^{-}⟶O_2+2e$ $[E_{O_2}=\frac{M}{2}]$

Eq. of $O_2=$ Eq. of $KMn{O}_4$

$\frac{w}{\frac{32}{2}}=\frac{0.316\times 5}{158}$ $⟹w_{O_2}=0.16$ g

We know, $PV=nRT$.

Substituting the values, we get

$\frac{750}{760}\times V=\frac{0.16}{32}\times 0.0821\times 300$

$\therefore V_{O_2}=0.12479$ L $=124.79$ mL

So, the nearest integer is $125$ mL.