Question

A mixture of $H_{2}S{O}_4$ and $H_2{C}_2{O}_4$ (oxalic acid) and some inert impurity weighing $3.822$ g was dissolved in water and the solution made up to $1$ litre. $10$ mL of this solution requires $3$ mL of $0.1$ N $NaOH$ for complete neutralization. In another experiment $100$ mL of the same solution in hot condition requires $4$ mL of $0.02$ M $KMn{O}_4$ solution for complete reaction. The wt. % of $H_{2}S{O}_4$ in the mixture was:

• A. $33.33$
• B. $50$
• C. $40$
• D. $80$

Answer 1

The correct option is A $33.33$

Let $x$ and $y$ are milli moles of $H_{2}S{O}_4$ and $H_2{C}_2{O}_4$ in given mixture when both reacted with base.
m.eq. of acid = m.eq. of base
$\Rightarrow (x+y)\times 2=3\times 0.1\times \frac{1000}{10}\Rightarrow x+y=15$
In second experiment only $KMn{O}_4$ reacted with $H_2{C}_2{O}_4$ in which $M{n}^{7+}$ converted into $M{n}^{2+}$ and $C_2{O}_4^{2-}$ converted into $C{O}_2$.
$\therefore$ milli equivalent of $H_2{C}_2{O}_4$
= milli equivalent of $KMn{O}_4$
$\Rightarrow y\times 2=4\times 0.02\times 5\times \frac{1000}{100}\Rightarrow y=2\therefore x=13\text{wt. of}{H}_{2}S{O}_4=13\times {10}^{-3}\times 98=1.274\text{g}\text{wt.}\%\text{of}{H}_{2}S{O}_4\text{in sample}=\frac{1.274}{3.822}\times 100=33.33\%$