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A mixture of dihydrogen and dioxygen at one bar pressure contains \(20 \%\) by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
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\(\because\) Moles \(=\frac{\text { given mass }}{\text { molar mass }}\)
\(\mathrm{n}_{\mathrm{H}_2}=20 / 2=10\) moles,
\(\mathrm{n}_{\mathrm{O}_2}=80 / 32=2.5\) moles
\(\mathrm{P}_{\mathrm{H}_2}=\frac{\mathrm{n}_{\mathrm{H}_2}}{\mathrm{n}_{\mathrm{H}_2}+\mathrm{n}_{\mathrm{O}_2}} \times \mathrm{P}_{\text {Total }}=\frac{10}{10+2.5} \times 1\) bar. \(=0.8 \mathrm{bar}\)
\(\mathrm{n}_{\mathrm{H}_2}=20 / 2=10\) moles,
\(\mathrm{n}_{\mathrm{O}_2}=80 / 32=2.5\) moles
\(\mathrm{P}_{\mathrm{H}_2}=\frac{\mathrm{n}_{\mathrm{H}_2}}{\mathrm{n}_{\mathrm{H}_2}+\mathrm{n}_{\mathrm{O}_2}} \times \mathrm{P}_{\text {Total }}=\frac{10}{10+2.5} \times 1\) bar. \(=0.8 \mathrm{bar}\)
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