Question

The gas described in parts a and b has a mass of 1.66 g. the sample is most likely which monatomic gas?

Answer 1

Mass of the gas m = 1.66 
The calculated temperature T = 273 + 20 = 293
 We have to calculate molar mass to determine the gas
 Molar Mass = mRT / PV
 M = (1.66 x 8.314 x 293) / (101.3 x 1000 x 0.001)
 M = 4043.76 / 101.3 = 39.92 g/mol
 So this gas has to be Argon Ar based on the molar mass.

Answer 2

Answer is: gas is argon (Ar).
Missing part of question find on internet:
part a: p(gas) = 1 atm, T(gas) = 20°C = 293,15 K.
part b: V(gas) = 1 L.
Ideal gas law: p·V = n·R·T or p·V = m/M·R·T.
R(universal gas constant) = 0,08206 L·atm/mol·K.
M = m·R·T÷p·V
M = 1,66 g·0,08206 L·atm/mol·K·293,15K÷(1 atm·1 L)
M = 39,9 g/mol (argon).