1.

2.9 g of a gas at 95^(@)C occupied the same volume as 0.184 g of dihydrogen at 17^(@)C, at the same pressure. What is the molar of the gas ?

Answer» <html><body><p></p><a href="https://interviewquestions.tuteehub.com/tag/solution-25781" style="font-weight:bold;" target="_blank" title="Click to know more about SOLUTION">SOLUTION</a> :According to ideal gas equations,<br/>`pV=nRT=(wRT)/(M) "" therefore M=(wRT)/(pV)`<br/>`therefore (pV)/(R )=(<a href="https://interviewquestions.tuteehub.com/tag/wt-1462289" style="font-weight:bold;" target="_blank" title="Click to know more about WT">WT</a>)/(M)` where `(pV)/(R )=` constant of number<br/>for <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> gas<br/>`w=w_(1)=2.9` gram<br/>R = Gas constant<br/>`T=T_(1)=95^(@)<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a>`<br/>`= (95+273)=<a href="https://interviewquestions.tuteehub.com/tag/368-1859650" style="font-weight:bold;" target="_blank" title="Click to know more about 368">368</a> K`<br/>p = p bar<br/>V = V L<br/>`M = M_(1)=M_(1)g mol^(-1)`<br/>For `H_(2)` gas :<br/>`w_(2)=0.184` gram<br/>R = Gas constant<br/>`T_(2)=(17+273)K=290 K`<br/>p = p bar<br/>V = V L<br/>`M=M_(2)=2.0 g mol^(-1)`<br/>`therefore (pV)/(R )=(w_(1)T_(1))/(M_(1))=(w_(2)T_(2))/(M_(2))=` constant<br/>`therefore M_(1)=(w_(1)T_(1))((M_(2))/(w_(2)T_(2)))=((2.9g)(368 K)(2g mol^(-1)))/((0.184 g)(290 K))`<br/>`= 40 g mol^(-1)`</body></html>


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