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 ?

Explanation

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)`