Redox and Equivalent ConceptHard

Question

A definite mass of H2O2 is oxidized by excess of acidified KMnO4 and acidified K2Cr2O7 in separate experiments. Which of the following is/ are correct statements? (K = 39, Cr = 52, Mn = 55)

Options

A.Mass of K2Cr2O7 used up will be greater than that of KMnO4.
B.Moles of KMnO4 used up will be greater than that of K2Cr2O7.
C.Equal mass of oxygen gas is evolved in both the experiments.
D.If equal volumes of both the solutions are used for complete reaction, then the molarities of KMnO4 and K2Cr2O7 solutions are in 6: 5 ratio.

Solution

$n_{eq}H_{2}O_{2} = n_{eq}KMnO_{4} = n_{eq}K_{2}Cr_{2}O_{7} = n_{eq}O_{2}$

$\text{Or }\text{n}_{KMnO_{4}} \times 5 = n_{K_{2}Cr_{2}O_{7}} \times 6 $$$\Rightarrow n_{KMnO_{4}} > n_{K_{2}Cr_{2}O_{7}}$$

$\text{Or }\frac{W_{KMnO_{4}}}{158} \times 5 = \frac{W_{K_{2}Cr_{2}O_{7}}}{294} \times 6 $$${\Rightarrow W_{KMnO_{4}} < W_{K_{2}Cr_{2}O_{7}} }{\text{Or }\frac{V \times M_{KMnO_{4}}}{1000} \times 5 = \frac{V \times M_{K_{2}Cr_{2}O_{7}}}{1000} \times 6 }{\Rightarrow \frac{M_{KMnO_{4}}}{M_{K_{2}Cr_{2}O_{7}}} = \frac{6}{5}}$$

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