Answer:
[tex]M=39.1g/mol[/tex]
[tex]\rho=1.70g/L[/tex]
Explanation:
Hello,
In this case, one uses the ideal gas equation:
[tex]PV=nRT[/tex]
By solving for moles in terms of mass and molar mass one obtains:
[tex]PV=\frac{m}{M}RT\\M=\frac{mRT}{PV} =\frac{3.15g*0.082 \frac{atm*L}{mol*K} *(35+273)K}{1.10atm*1.85L} \\M=39.1g/mol[/tex]
On the other hand, the density is easily computed as shown below:
[tex]\rho=\frac{m}{V}=\frac{3.15g}{1.85L}=1.70g/L[/tex]
Best regards.
