50 cm^{3} of 0.04 M K_{2}Cr_{2}O_{7} in ac | vedclass

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(A) The number of equivalents of the oxidizing agent must be equal to the number of equivalents of the reducing agent $(H_{2}S)$.Since the same amount

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$80$

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(A) The number of equivalents of the oxidizing agent must be equal to the number of equivalents of the reducing agent $(H_{2}S)$.Since the same amount of $H_{2}S$ is oxidized in both cases,the equivalents of $K_{2}Cr_{2}O_{7}$ must equal the equivalents of $KMnO_{4}$.For $K_{2}Cr_{2}O_{7}$ in acidic medium,the $n_{f} = 6$.For $KMnO_{4}$ in acidic medium,the $n_{f} = 5$.Using the formula: $n_{f1} 	imes M_{1} 	imes V_{1} = n_{f2} 	imes M_{2} 	imes V_{2}$.$6 	imes 0.04 	imes 50 = 5 	imes 0.03 	imes V_{KMnO_{4}}$.$12 = 0.15 	imes V_{KMnO_{4}}$.$V_{KMnO_{4}} = \frac{12}{0.15} = 80 \ cm^{3}$.

$50 \ cm^{3}$ of $0.04 \ M \ K_{2}Cr_{2}O_{7}$ in acidic medium oxidizes a sample of $H_{2}S$ gas to sulphur. The volume of $0.03 \ M \ KMnO_{4}$ required to oxidize the same amount of $H_{2}S$ gas to sulphur in acidic medium is: (in $cm^{3}$)

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