Some Basic Concepts of Chemistry - Result Question 94
####32. Hydrogen peroxide solution $(20 mL)$ reacts quantitatively with a solution of $KMnO _4(20 mL)$ acidified with dilute $H _2 SO _4$. The same volume of the $KMnO _4$ solution is just decolourised by $10 mL$ of $MnSO _4$ in neutral medium simultaneously forming a dark brown precipitate of hydrated $MnO _2$. The brown precipitate is dissolved in $10 mL$ of $0.2 M$ sodium oxalate under boiling condition in the presence of dilute $H _2 SO _4$. Write the balanced equations involved in the reactions and calculate the molarity of $H _2 O _2$.
(2001)
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Solution:
$\underline{\text { Weight of a compound in gram }(w)}=$ Number of moles $(n)$
Molar mass $(M)$
$$ \begin{aligned} & =\frac{\text { Number of molecules }(N)}{\text { Avogadro number }\left(N _A\right)} \ \Rightarrow \quad \frac{w\left(O _2\right)}{32} & =\frac{N\left(O _2\right)}{N _A} \end{aligned} $$
And
$$ \frac{w\left(N _2\right)}{28}=\frac{N\left(N _2\right)}{N _A} $$
Dividing Eq. (i) by Eq. (ii) gives
$$ \begin{aligned} \frac{N\left(O _2\right)}{N\left(N _2\right)} & =\frac{w\left(O _2\right)}{w\left(N _2\right)} \times \frac{28}{32} \ & =\frac{1}{4} \times \frac{28}{32}=\frac{7}{32} \end{aligned} $$
