Question

Potassium tetraoxalate crystals $(KH{C}_2{O}_4.H_2{C}_2{O}_4.2H_{2}O)$ react with $KMn{O}_4$ (oxidising agent) as follows,
$4KMn{O}_4+5KH{C}_2{O}_4.H_2{C}_2{O}_4.2H_{2}O+13H_{2}S{O}_4\to 4MnS{O}_4+9KHS{O}_4+20C{O}_2+18H_{2}O$
The equivalent weight of potassium tetraoxalate crystals is:

• A. 84.76
• B. 63.5
• C. 127
• D. 158.75

Answer 1

The correct option is B 63.5

$4KMn{O}_4+5KH{C}_2{O}_4.H_2{C}_2{O}_4.2H_{2}O+13H_{2}S{O}_4\to 4MnS{O}_4+9KHS{O}_4+20C{O}_2+18H_{2}O$
$4KMn{O}_4\equiv 5KH{C}_2{O}_4.H_2{C}_2{O}_4.2H_{2}O$
$10\left[O\right]=5\times 254$
$10\times 16$ parts by wt. of oxygen $\equiv$ $1270$ parts by wt. of $KH{C}_2{O}_4.H_2{C}_2{O}_4.2H_{2}O$
8 parts by wt. of oxygen (1 equivalent) $\equiv \frac{1270}{20}$ $=$ $63.5$ parts by wt. of $KH{C}_2{O}_4.H_2{C}_2{O}_4.2H_{3}O$
Equivalent weight of $KH{C}_2{O}_4.H_2{C}_2{O}_4.2H_{3}O=63.5$