A 12.0 g sample of gas occupies 19.2 L at STP. What is the molecular weight of this gas?
To calculate the molecular weight of the gas, we can use the ideal gas law equation: PV = nRT
Where:
P = pressure (1 atm at STP)
V = volume (19.2 L)
n = number of moles of gas
R = ideal gas constant (0.0821 L.atm/mol.K)
T = temperature (273 K at STP)
First, let's calculate the number of moles of gas in the sample:
n = PV / RT
n = (1 atm * 19.2 L) / (0.0821 L.atm/mol.K * 273 K)
n = 0.709 moles
Next, let's calculate the molecular weight of the gas:
Molecular weight = mass / moles
Molecular weight = 12.0 g / 0.709 moles
Molecular weight ≈ 16.9 g/mol
Therefore, the molecular weight of this gas is approximately 16.9 g/mol.