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 ?

2.9 g of a gas at 95^(@)C occupied the same volume as 0.184 g of dihyd

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» SOLUTION :According to ideal gas equations, `pV=nRT=(wRT)/(M) "" therefore M=(wRT)/(pV)` `therefore (pV)/(R )=(WT)/(M)` where `(pV)/(R )=` constant of number for 1 gas `w=w_(1)=2.9` gram R = Gas constant `T=T_(1)=95^(@)C` `= (95+273)=368 K` p = p bar V = V L `M = M_(1)=M_(1)g mol^(-1)` For `H_(2)` gas : `w_(2)=0.184` gram R = Gas constant `T_(2)=(17+273)K=290 K` p = p bar V = V L `M=M_(2)=2.0 g mol^(-1)` `therefore (pV)/(R )=(w_(1)T_(1))/(M_(1))=(w_(2)T_(2))/(M_(2))=` constant `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))` `= 40 g mol^(-1)`