$$50\,c{m^3}$$  of $$0.04\,M\,{K_2}C{r_2}{O_7}$$    in acidic medium oxidizes a sample of $${H_2}S$$  gas to sulphur. Volume of $$0.03\,M\,KMn{O_4}$$    required to oxidize the same amount of $${H_2}S$$  gas to sulphur, in acidic medium is

Solution :
$${K_2}C{r_2}{O_7}$$   and $$KMn{O_4}$$   both are good oxidizing agents in acidic medium. The reactions are as follows :
$$\eqalign{ & C{r_2}{O_7}^{2 - } + 14{H^ + } + 6{e^ - } \to 2C{r^{3 + }} + 7{H_2}O \cr & Mn{O_4}^{2 - } + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O \cr & {\text{Normality of}}\,\,C{r_2}{O_7}^{2 - } \cr & = {\text{Molarity }} \times {\text{ no}}{\text{. of electrons involved}} \cr & = 0.04 \times 6 \cr & = 0.24\,N \cr & {\text{Normality of}}\,\,MnO_4^ - \cr & {\text{ = Molarity }} \times {\text{ no}}{\text{. of electrons involved}} \cr & = 0.03 \times 5 \cr & = 0.15\,N \cr & \mathop {{N_1}{V_1}}\limits_{\left( {{K_2}C{r_2}{O_7}} \right)} = \mathop {{N_2}{V_2}}\limits_{\left( {KMn{O_4}} \right)} \cr & \Rightarrow 0.24 \times 50 = 0.15 \times {V_2} \cr & \Rightarrow {V_2} = \frac{{0.24 \times 50}}{{0.15}} = 80\,c{m^3} \cr} $$