Given data, d=1.964gdm−3 T=273K p=76cmHg=1atm Where R=0.0821LatmK−1mol−1 R=82.1cc atm K−1mol−1 By using ideal gas equation, pV=nRT→(1) Where n=‌
m
M
,d=‌
m
V
Then equation(1) becomes pV=‌
m
M
RT ⇒M=‌
m
V
×‌
RT
p
⇒M=d×‌
RT
p
Then substituting the values, we get ⇒M=‌
1.964×10−3×82.1×273
1
⇒M=44 Then, a) MCH4=12+4=16 b) MC2H6=24+6=30 c) MCO2=12+32=44 d) MXe=131.29 Therefore the gas is CO2