Asample of 1.58 g of an unknown gas has a volume of 0.682 l. the sample is kept at 60°c, and exerts a pressure of 1.38 atm. what is the gas?

The options given are as follows
a. Cl2
b. NO
c. NO2
d. SO2
e. C2H6

we can use the ideal gas law equation to find the number of moles of gas in the container
PV = nRT
where 
P - pressure - 1.38 atm x 101 325 Pa/atm = 139 829 Pa
V - volume - 0.682 x 10⁻³ m³
n - number of moles 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature - 60 °C + 273 = 333 K
Substituting these values in the equation 
139 829 Pa x 0.682 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 333 K
n = 0.0344 mol

mass present - 1.58 g
molar mass = mass present / number of moles
therefore molar mass - 1.58 g / 0.0344 mol = 46 g/mol 
the answer is NO₂
the molar mass of NO₂ is - 14 + (16x2) = 14 + 32 = 46
therefore unknown gas is NO₂