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

Explanation

According to ideal gas equation
`PV=nRT`
First case: Let the molar mass of air `(M)= Mg mol^(-1))`
Mass of gas `(m)=2.9 g`
Number of moles `(m)= ("Mass")/("Molar mass")=2.9/M mol`
`T=273+95=368`
Pressure `=P, volume=V`
`PV=nRT`
`PV=2.9/MxxRxx368`
Second case: Mass of hydrogen `=0.184 g`
Number of moles of `H_(2)=0.184/2 mol`
`T=273+17=290`
`PV=0.184/2xxRxx290 (ii)`
From equations (i) and (ii), we get
`2.9/Mxx368=(0.184xx290)/2`
`M=(2.9xx368xx2)/(0.184xx290)=40 g mol^(-1)`