Questions & Answers

Question

Answers

A.$3.15g$

B.$31.5g$

C.$6.3g$

D.$63g$

Answer
Verified

In the question, we have given that

Volume of oxalic acid$ = 50cc$

Normality of $KMn{O_4} = 0.20N$

Volume of $KMn{O_4} = 25cc$

The normality of a solution is defined as the measure of concentration equal to the gram equivalent weight per litre of solution.

$N = \dfrac{m}{M} \times \dfrac{1}{V}$

After substituting the values from the question we get,

$0.20 = \dfrac{m}{1} \times \dfrac{{1000}}{{25}}$

$m = 0.005g$$eq$

$0.005g$$eq$ of potassium permanganate neutralizes $0.005g$$eq$of oxalic acid.

The molecular formula of oxalic acid is ${C_2}{H_2}{O_4}.2{H_2}O$.

The equivalent mass of oxalic acid dihydrate is $\dfrac{{126}}{2} = 63g$$eq$

The mass of oxalic acid dihydrate is $0.05 \times 63 = 0.315g$

$0.315g$ of oxalic acid dihydrate is present in $50cc$ solution. Then $500cc$ solution contains,

$0.315 \times \dfrac{{500}}{{50}} = 3.15g$

The required mass of oxalic acid dihydrate present in $500cc$ is $3.15g$.