Question

A 1 g sample of ${H_2}{O_2}$ solution containing $x\% \,{H_2}{O_2}$ by mass requires $x\,c{m^3}$ of a $KMn{O_4}$ solution for complete oxidation under acidic condition. Calculate the normality of $KMn{O_4}$ solution.

Answer 1

Composition of a solution can be expressed based on its concentration. The terms dilute and concentrated provide a vague idea about the concentration of the solution and thus a quantitative representation is needed for expression of the concentration. The concentration of the solution depends on the amount of solute and solvent that are present in the solution.

Complete step by step answer:
As given in the sample , A 1 g sample of ${H_2}{O_2}$ solution contains $x\% \,{H_2}{O_2}$ by mass
This means that a 100 g sample has the mass of x gram of ${H_2}{O_2}$ . This can be represented as
1 g sample has $\dfrac{{1\,g \times xgm}}{{100gm}}{H_2}{O_2}$
Now according to the question the volume of $KMn{O_4}$ needed for the neutralization will be $x\,c{m^3}$
At equivalence,
Number of gram equivalent of ${H_2}{O_2}$ $=$ number of gram equivalent of $KMn{O_4}$
So,
${N_1}{V_1} = {N_2}{V_2}$
Where subscript 1 used for ${H_2}{O_2}$ and 2 used for $KMn{O_4}$
So putting the values in the equation we have
$\dfrac{x}{{100 \times 17}} = \dfrac{{{N_i}X}}{{1000}}$
Since the Normality = $\dfrac{{weight}}{{equivalent\,weight}}1000V\,inml$
And $equivalent\,weight\,of\,{H_2}{O_2} = \dfrac{{34}}{2} = 17$
So, ${N_2} = \dfrac{{10}}{{17}} = 0.588N$

So, the normality of $KMn{O_4}$ $=$ 0.55N

Note: There are other representations of the concentration of solution than normality.
While molality is represented by $m$ another representation, Molarity is represented by $M$.
Molality is represented by $m$ and can be calculated by the following formula
$Molality(m) = \dfrac{{{{Moles\, of \,solute}}}}{{{{Mass \,of\, solvent \,in \,kg}}}}$
Molarity of the solution can be represented in terms of the volume of the solution rather than the mass of the solvent as in the case of molality. It can be represented as
${{Molarity (M) = }}\dfrac{{{{Moles\; of\; solute}}}}{{{{Volume\; of \;solution\; in \;litre}}}}$