Question

A mixture of $H_{2}S{O}_4$ and $H_2{C}_2{O}_4$ (oxalic acid) and some inert impurity weighing $3.185g$ was dissolved in water and the solution was made up to $1litre.10mL$ of this solution required $3mL$ of $0.1N$ $NaOH$ for complete neutralization. In another experiment, $100mL$ of the same solution in hot conditions, required $4mL$ of $0.02MKMn{O}_4$ solution for complete reaction. The weight $\%$ of $H_{2}S{O}_4$ in the mixture was:

• A. $40$
• B. $50$
• C. $60$
• D. $80$

Answer 1

Answer: (A)

$40$

Let $x$ and $y$ be the milimoles of $H_{2}S{O}_4$ and $H_2{C}_2{O}_4$ in given mixture when both reacted with base.

Meq. of acids$=$Meq. of base

$(x+y)\times 2=3\times 0.1\times \frac{1000}{10}$

$x+y=15$

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 $KMn{O}_4$.

$y\times 2=4\times 0.02\times 5\times \frac{1000}{100}$

$y=2$

$\therefore x=13$

$Weightof{H}_{2}S{O}_4=13\times {10}^{-3}\times 98=1.274g$.

Weight $\%$ of $H_{2}S{O}_4$ in sample $=\frac{1.274}{3.185}\times 100=40\%$.

Hence, the correct option is $A$