A sample of an unknown compound is vaporized at 190.°C. The gas produced has a volume of 2160.mL at a pressure of 1.00atm, and it weighs 2.87g. Assuming the gas behaves as an ideal gas under these conditions, calculate the molar mass of the compound. Be sure your answer has the correct number of significant digits.

Molar mass = 50.4 g/mol

Explanation:

Pressure . Volume = number of moles (n) . R . T°K

This is the Ideal Gases Law where R is the Ideal Gases Constant and T°, Absolute Temperature.

We convert the T°C to T°K → 190°C + 273 = 463 K

R = 0.082 L.atm /mol.K

We need to convert the volume from mL to L → 2160 mL . 1L / 1000 mL = 2.16 L; now we replace:

1 atm . 2.16L = n . 0.082 L.atm /mol.K . 463K

(1 atm . 2.16L) / (0.082 L.atm /mol.K . 463K) = n → 0.569 moles

These moles refers to 2.87 g so let's find out the molar mass:

Molar mass (g/mol) = 2.87 g / 0.0569 mol =  50.4 g/mol

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.