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\( 50 \mathrm{~cm}^{3} \) of \( 0.04 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} \) in acidic medium oxidizes a sample of \( \mathrm{H}_{2} \mathrm{~S} \) gas to sulphur.
Volume of \( 0.03 \mathrm{M} \mathrm{KMnO}_{4} \) required to oxidize the same amount of \( \mathrm{H}_{2} \mathrm{~S} \) gas to sulphur, in
acidic medium is
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Volume of \( 0.03 \mathrm{M} \mathrm{KMnO}_{4} \) required to oxidize the same amount of \( \mathrm{H}_{2} \mathrm{~S} \) gas to sulphur, in
acidic medium is
Solution:
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Verified Answer
The correct answer is:
\( 80 \mathrm{~cm}^{3} \)
Equivalents of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}=\) Equivalents of \(\mathrm{H}_{2} \mathrm{~S}=\) Equivalents of \(\mathrm{KMnO}_{4}\)
\(n_{f}\) of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}=6\)
\(n_{f}\) of \(\mathrm{KMnO}_{4}=5\)
Equivalents \(=n_{f} \times\) Molarity \(\times\) Volume
\(6 \times 0.04 \times 50=5 \times 0.03 \times V_{K M n O_{4}}\)
\(V_{K M n O_{4}}=80 \mathrm{~cm}^{3}\)
\(n_{f}\) of \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}=6\)
\(n_{f}\) of \(\mathrm{KMnO}_{4}=5\)
Equivalents \(=n_{f} \times\) Molarity \(\times\) Volume
\(6 \times 0.04 \times 50=5 \times 0.03 \times V_{K M n O_{4}}\)
\(V_{K M n O_{4}}=80 \mathrm{~cm}^{3}\)
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