WBJEE 2020 Chemistry Solved Paper

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