Question

Volume $V_1$ mL of 0.1 $M{K}_{2}C{r}_2{O}_7$ is needed for complete oxidation of 0.678 g $N_2{H}_4$ in acidic medium. The volume of 0.3 M $KMn{O}_4$ needed for same oxidation in acidic medium will be:

• A. $\frac{2}{5}{V}_1$
• B. $\frac{5}{2}{V}_1$
• C. $113V_1$
• D. None of the above

Answer 1

The correct option is A $\frac{2}{5}{V}_1$

For two reactants to be completely used to produce the product, the Gram equivalent of two reactants is equal.

Gram equivalent = Volume X Normality

Gram equivalent = Volume X n-factor X Molarity

Since, both $K_{2}C{r}_2{O}_7$ and $KMn{O}_4$ complete oxidation with 0.678 gm $N_2{H}_4$. Therefore, their Gram equivalent is equal.

Gram equivalent of $K_{2}C{r}_2{O}_7$ = Gram equivalent of $KMn{O}_4$

$V_1\times 6\times 0.1=V\times 5\times 0.3$

$\frac{2}{5}{V}_1=V$