Answer: 39.86 g/mol
Explanation:
Assuming this gas is at a pressure of [tex]0.99 atm[/tex] and a temperature of [tex]37 \°C+273.15=310.15 K[/tex], we can solve this problem with the Ideal Gas Law equation:
[tex]P.V=n.R.T[/tex] (1)
Where:
[tex]P=0.99 atm[/tex] is the pressure of the gas
[tex]V[/tex] is the volume of the gas
[tex]n[/tex] the number of moles of gas
[tex]R=0.0821 \frac{atm.L}{mol.K}[/tex] is the gas constant
[tex]T=310.15 K[/tex] is the absolute temperature of the gas in Kelvin
On the other hand, we know the density [tex]\rho=1.55 g/L[/tex] is given by a relation between the mass [tex]m[/tex] and the volume [tex]V[/tex]:
[tex]\rho=\frac{m}{V}[/tex] (2)
Then, the mass is:
[tex]m=\rho V[/tex] (3)
Another important fact here is that the molar mass [tex]M_{mass}[/tex] of an element is given by a relation between the mass of that element and the number of moles [tex]n[/tex] it contains:
[tex]M_{mass}=\frac{m}{n}[/tex] (4)
Isolating [tex]n[/tex]:
[tex]n=\frac{m}{M_{mass}}[/tex] (5)
Substituting (5) in (1):
[tex]P.V=\frac{m}{M_{mass}}.R.T[/tex] (6)
Substituting (3) in (6):
[tex]P.V=\frac{\rho V}{M_{mass}}.R.T[/tex] (7)
Isolating [tex]M_{mass}[/tex]:
[tex]M_{mass}=\frac{\rho V}{PV}RT[/tex] (8)
Simplifying:
[tex]M_{mass}=\frac{\rho}{P}RT[/tex] (9)
Solving with the given data:
[tex]M_{mass}=\frac{1.55 g/L}{0.99 atm}(0.0821 \frac{atm.L}{mol.K})(310.15 K)[/tex] (10)
Finally:
[tex]M_{mass}=39.86 g/mol[/tex] This is the molar mass of the gas
