Question

For $$0.3g$$ of an oxalate salt was dissolved in $$100ml$$ solution required $$90ml$$ of $$N/20KMn{O_4}$$ for complete oxidation. The $$\% $$ of oxalate ion in salt is:-
(A) $$33\% $$
(B) $$66\% $$
(C) $$70\% $$
(D) $$40\% $$

Answer 1

Hint: Chemical name for $KMn{O_4}$ is Potassium Permanganate. It is a very good oxidising agent. It persists a purple colour and is used in titrations. Due to its colour, it acts as a self-indicator in titrations. Oxalic acid is an organic acid in which two carboxylic acid group are joined together and it has a chemical formula as ${(COOH)_2}.$

Complete answer:
We know that Potassium permanganate with chemical formula $KMn{O_4}$ is a very good oxidising agent and it is used in many chemical reactions. Oxalic acid is an organic acid in which two carboxylic acid group are joined together and it has a chemical formula as ${(COOH)_2}.$
Now, we will see the redox changes when potassium permanganate and oxalic acid reacts:
$M{n^{7 + }} + 5{e^ - } \to M{n^{2 + }}$
$C_2^{2 - } \to 2{C^{4 + }} + 2{e^ - }$
Therefore, we know that:
${M_{eq}}$ of oxalate ions is equal to the ${M_{eq}}$ of $KMn{O_4}$ .
Now, equivalent weight (E.W.) of any molecule or ion is calculated by dividing its molecular mass by the n-factor (the number of electrons that it loses or gains during the reaction).
$E.W. = \dfrac{{M.M.}}{{n - factor}}$
Equivalent weight (E) of oxalate ion is:
$\dfrac{{88}}{2} = 44$
n-factor is two because two electrons are involved in the reaction.
Now, equating the equivalents, we get:
$\dfrac{w}{{44}} \times 1000 = 90 \times \dfrac{1}{{20}}$
On solving the above equation, we get the weight of oxalate ions in $$0.3g$$ oxalate salt as:
$w = 0.198g$
Now, we will calculate the percentage of oxalate ions in $$0.3g$$ oxalate salt as:
$\% = \dfrac{{0.198 \times 100}}{{0.3}}$
$\% = 66\% $
Hence, the percentage of oxalate ions in $$0.3g$$ oxalate salt is $66\% .$
The correct option is (B) $$66\% $$ .

Note:
To solve these types of questions, we should have the knowledge about equivalent weights, number of moles and n-factor of any salt, acid or base. For some compounds (oxidising agents or reducing agents) n-factor is calculated as the number of electrons it is gaining or losing during the reaction.