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A neon - dioxygen mixture contains 70.6 g dioxygen and 167.5 g neon. If pressure of the mixture of gases in the cylinder is 25 bar. What isd the partial pressure of dioxygen and neon in the mixture ?

Answer» <html><body><p></p>Solution :Calculation of moles of Dioxygen gas,<br/> Where, molecular mass of `O_(2)=<a href="https://interviewquestions.tuteehub.com/tag/32g-307750" style="font-weight:bold;" target="_blank" title="Click to know more about 32G">32G</a> <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>^(-1)`<br/>mole `=("Mass")/("Molecular Mass")=(70.6 g)/(32g mol^(-1))`<br/>`therefore n(O_(2))=2.206 ~~2.21` mol<br/>Caslculation of mole <a href="https://interviewquestions.tuteehub.com/tag/number-582134" style="font-weight:bold;" target="_blank" title="Click to know more about NUMBER">NUMBER</a> of Neon gas :<br/>Mole = n (Ne) `= ("Weight")/("Molar Mass")`<br/>`= (167.5 g)/(20 g mol^(-1))=8.375` mol<br/>Total moles of mixture `= n(O_(2))+n (Ne)`<br/>`= (2.21+8.375)` mol<br/>= 10.585 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 `O_(2)chi (O_(2))=("Moles of "O_(2))/("Total Moles")`<br/>`= (2.21 mol)/(10.585 mol)=0.2087`<br/>Partial <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> of `O_(2)p(O_(2))`<br/>`= chi (O_(2))xx` total Pressure (P)<br/> `=(2.21mol)/(10.585 mol)xx25` bar<br/>`= 5.2196` bar `~~ 5.22` bar<br/>Partial Pressure of `N_(2)` gas `(P_(N_(2)))`<br/>`= chi_(Ne)xx` total Pressure,<br/>`= 25xx((8.375)/(10.585))`<br/>= 19.7803 bar<br/>OR<br/>`rho_(Ne)=P_("total")-rho_(O_(2))`<br/>`= (25.5.2196)` bar = 19.7804 bar</body></html>


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