A mixture of dihydgrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

A mixture of dihydgrogen and dioxygen at one bar pressure contains 20%

1.	A mixture of dihydgrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
Answer» <html><body><p></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/suppose-656311" style="font-weight:bold;" target="_blank" title="Click to know more about SUPPOSE">SUPPOSE</a>, Weight of mixture gas = 100 g<br/>`therefore` Weight of `H_(2)=20 g` and<br/>Weight of `O_(2)=80 g`<br/>`therefore` Mole of `n_(H_(2))=(20)/(2)=10` mol and<br/>Mole of `n_(O_(2))=(80)/(32)=2.5` mol<br/>Total mole of mixture gas `= (10+2.5)=12.5` mol<br/>Mole <a href="https://interviewquestions.tuteehub.com/tag/fraction-458259" style="font-weight:bold;" target="_blank" title="Click to know more about FRACTION">FRACTION</a> of `H_(2), chi_(H_(2))=(n_(H_(2)))/("Total Moles")=(10)/(12.5)`<br/>Mole fraction of `O_(2), chi_(O_(2))=(n_(O_(2)))/("Total moles")=(2.5)/(12.5)`<br/><a href="https://interviewquestions.tuteehub.com/tag/partial-596556" style="font-weight:bold;" target="_blank" title="Click to know more about PARTIAL">PARTIAL</a> Pressure = Mole fraction `xx` Total pressure<br/>`therefore p_(H_(2))=n_(H_(2))xx P=(10xx"1 bar")/(12.5)=(10xx"1 bar")/(12.5)=0.8` bar<br/>`therefore p_(O_(2))=n_(O_(2))xxP=(2.5xx"1 bar")/(12.5)=0.2` bar</body></html>