Answer :
Answer:
12.0 g/mol
Explanation:
We have a 22.4 liter sample of gas at STP (273.15 K, 1.00 atm). We can find the moles of gas using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.00 atm × 22.4 L / 0.0821 atm.L/mol.K × 273.15 K
n = 1.00 mol
1.00 mol of gas weighs 12.0 grams. The molar mass of the gas is:
12.0 g / 1.00 mol = 12.0 g/mol
Answer:
Molecular weight = 12.0 g/mol
The 3th option is correct
Explanation:
Step 1: Data given
Volume of a gas = 22.4 L
STP = 1 atm and 273 K
Mass of the gas = 12.0 grams
Step 2: Calculate moles of the gas
p*V = n*R*T
⇒with p = the pressure of the gas = 1.0 atm
⇒with V = the volume of the gas = 22.4 L
⇒with n = the number of moles = TO BE DETERMINED
⇒with R = the gas constant = 0.08206 L*Atm/mol*K
⇒with T = the temperature = 273
n = (p*V)/ (R*T)
n = (1* 22.4)/ (0.08206 * 273)
n = 1.00 moles
Step 3: Calculate molecular weight of the gas
Molecular weight = mass of the gas / moles of gas
Molecular weight = 12.0 grams / 1.00 moles
Molecular weight = 12.0 g/mol
The 3th option is correct