Question

Volume $V$ 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 $KMn{O}_2$ needed for same oxidation in acidic medium will be:

• A. $\frac{2}{5}{V}_1$
• B. $\frac{5}{2}{V}_1$
• C. $113$ $V_1$
• D. Can not be determined

Answer 1

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

According to the reaction

$2K_{2}C{r}_2{O}_7+2H_{2}S{O}_4+3N_2{H}_4\to 2K_{2}S{O}_4+3N_2+8H_{2}O+2C{r}_2{O}_3$

In acidic medium $C{r}^{+6}$ goes to $C{r}^{+3}$ state and hence there is a net gain of 3 electrons. As there are 2 moles of $Cr$ in $K_{2}C{r}_2{O}_7$

The equivalent moles of $K_{2}C{r}_2{O}_7=6$

Normality $N_1=0.1M\times 6=0.6N$

$5N_2{H}_4+4KMn{O}_4+6H_{2}S{O}_4\to 5N_2+4MnS{O}_4+2K_{2}S{O}_4+16H_{2}O$

In Acidic medium, $M{n}^{+7}$ goes to$M{n}^{+2}$ state and hence there is a net gain of 5 electrons

The equivalent moles of $KMn{O}_4=5$

Normality $N_2=0.3M\times 5=1.5N$

According to the law of equivalence

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

$0.6N\times V=1.5N\times V_1$

$V_1=\frac{0.6N}{1.5N}\times V$

$V_1=\frac{2}{5}V$

Hence, the correct option is $A$