The correct option is A 32.19%, 67.81%
Let M be the average molecular mass of the mixture.
PV=wRTM
or, P=wRTM×V
or, P=dRTM
P=7376 atm, d=1.5 g L−1, M=?
T=303 K, R=0.0821 L atm K−1 mol−1
∴7376=1.5×0.0821×303M
M=38.85 g/mol
Suppose, the total moles = 100
Moles of CO=a
Moles 0f CO2=100−a
Mass of CO=28a
Mass of CO2=(100−a)44
Total mass =28a+(100−a)44
Molar mass of the mixture=28a+(100−a)44100
38.85=28a+(100−a)44100
a=32.19
Mole percent of CO=32.19%
Mole percent of CO2=(100−32.19)%=67.81%