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A vessel contains a mixture of equal masses of helium and oxygen at a pressure of 600 torr. Calculate the partial pressures of components in the mixture.

Answer» <html><body><p></p>Solution :Let the mass of each component gas is x. <br/> Total number of moles in the mixture = <br/> `(<a href="https://interviewquestions.tuteehub.com/tag/xg-747279" style="font-weight:bold;" target="_blank" title="Click to know more about XG">XG</a> " of " He)/(<a href="https://interviewquestions.tuteehub.com/tag/4g-318729" style="font-weight:bold;" target="_blank" title="Click to know more about 4G">4G</a> " of " mol^(-1)) + (xg " of " O_2)/(<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a> g " of " mol^(-1)) = (9x)/(32 g mol^(-1))` <br/> Partial pressure = pressure of the mixture x 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> <br/> Partial pressure of He `= 600 xx (x//4)/(9x//32) = 600 xx 8/9` <br/> `= 5333.33"<a href="https://interviewquestions.tuteehub.com/tag/torr-711044" style="font-weight:bold;" target="_blank" title="Click to know more about TORR">TORR</a>"` <br/>Partial pressure of `O_2 = 600 xx (x//32)/(9x//32)` <br/> `= 600 xx 1/9 = 66.67` torr <br/> (or) Partial pressure of `O_2 = 600 - 533.37 ` <br/> = 66.67 torr</body></html>


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