The correct option is C 8.0 g
Graham's law of effusion,
r∝1√M
Where r= rate of effusion
M= molar mass of gas
Also,
r=Volume of gas effused(V)Time taken(t)
∴r=Vt
∴rArB=√MBMA
⇒VAtAVBtB=√MBMA
Where VA and VB are the volume of gas A and gas B.
tA and tB are time taken by gas for effusion.
According to question,
VA=VB=V
tA=2 min
tB=4 min
MB=32 g (Molar mass of O2)
∴V2V4=√32MA
⇒42=√32MA
⇒MA=8 g/mol