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

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 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).