Question

3.4 g sample of H$_2$O$_2$ solution containing x% H$_2$O$_2$ by weight requires x ml of KMnO$_4$ solution for complete oxidation under acidic condition. The normality of KMnO$_4$ solution is:
A) 1N
B) 2N
C) 3N
D) 0.5N

Answer 1

Hint: The normality of a solution is defined as the number of gram, or mole equivalents of solute present in one litre of a solution. By using this definition we can determine the normality of a solution.

Complete step by step solution:
First, write the chemical reaction of KMnO$_4$, and H$_2$O$_2$ i.e.
H$_2$O$_2$ + KMnO$_4$ $\rightarrow$ 2KOH + 2MnO$_2$ + 2O$_2$
Now, according to the question;
Let us say that 100 g of H$_2$O$_2$ sample solution contains x g of H$_2$O$_2$.
Then 3.4 g of solution contains = $\dfrac{x}{100}$ $\times$ 3.4 (3.4 g is given)
So, we can say that weight of H$_2$O$_2$ = $\dfrac{3.4x}{100}$
 Thus, the equivalent weight of H$_2$O$_2$ = $\dfrac{3.4x}{100}$ $\times$ $\dfrac{1}{17}$
Now, the milliequivalents of H$_2$O$_2$ = $\dfrac{3.4x}{1700}$ $\times$ 1000= 2x.
Now, we can calculate the milliequivalents of KMnO$_4$ = x $\times$ N
Here, N represents the molarity of a solution.
So, $x \times N = 2x$
Now, we have N = 2

Hence, the normality of KMn$O_4$ solution is 2 N. The correct option is (B).

Additional information: Generally, normality is used in the place of molarity, because sometimes 1 mole of acid is unable to neutralise 1 mole of base. The SI unit of normality is mol/L.

Note: Don’t get confused between the equivalent weight, and the milliequivalent weight. The equivalent weight of an element is the gram atomic weight divided by its valence, and when we divide the equivalent by 1000, it leads to the milliequivalent. In the question, we have calculated by the given value of the sample, so here we multiplied by 1000 to attain the milliequivalent weight.