Question

Volume $V-1$ mL of 0.1 $M{K}_{2}C{r}_2{O}_2$ 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. $\frac{201}{5}{V}_1$
• B. $\frac{5}{2}{V}_1$
• C. $113V_1$
• D. can't say

Answer 1

The correct option is A $\frac{201}{5}{V}_1$

We $K_2{Cr}_2{O}_7$ has 6 oxidation state change.

Hence moles oxidize by $K_2{Cr}_2{O}_7=V_1\times 0.1\times 6=0.6V_1\to \left(1\right)$

we ${KMnO}_4$ has 5 oxidation state changes moles oxidized by ${KMnO}_4=\left(0.3\right)\times V_2\times 5\to \left(2\right)$

we (1) & (2) are equal

$0.6V_1=5\times 0.3\times V_2$

$V_2=\frac{201}{5}$