Question

A mixture of $HCOOH$ and $H_2{C}_2{O}_4$ is heated with conc. $H_{2}S{O}_4$. The gas produced is collected and on treating with $KOH$ solution, the volume of gas decreases by $\frac{1}{6}th$. Calculate the molar ratio of the two acids in original mixture.

• A. $4:1$
• B. $1:4$
• C. $3:1$
• D. $2:1$

Answer 1

The correct option is A $4:1$

$HCOOH\underset{conc}{\overset{H_{2}S{O}_4}{\to }}{H}_{2}O+CO$
$a$ mol $a$ mol
$H_2{C}_2{O}_4\underset{conc}{\overset{H_{2}S{O}_4}{\to }}{H}_{2}O+CO+C{O}_2$
$b$ mol $b$ mol $b$ mol
Total number of moles released as gases $=a+b+b=a+2b$.
The gas produced is collected and on treating with $KOH$ solution, the volume of gas decreases by $\frac{1}{6}th$.
Therefore, $b=\frac{1}{6}(a+2b)$ ( moles $\infty$ volume)
$\frac{a}{b}=4$
Hence, $a:b::4:1$.