Question

$V_1$ mL of $0.1M{K}_{2}C{r}_2{O}_7$ is needed for complete oxidation of $0.678$ g $N_2{H}_4$ in acidic medium. The volume of $0.3MKMn{O}_4$ 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 $\frac{2}{5}{V}_1$

$Cr$ changes from $+6$ to $+3$ on reduction.

Since, there are two $Cr$ atoms per molecule, net change $=2\times 3=6$.

$Mn$ changes from $+7$ to $+2$ on reduction.

Equivalent of $K_{2}C{r}_2{O}_7$= equivalent of $N_2{H}_4$

Also equivalent of $KMn{O}_4$= equivalent of $N_2{H}_4$

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

$N_1{V}_1=N_2{V}_2$

$0.1\times 6\times V_1=0.3\times 5\times V_2$

so $V_2=2/5V_1$