Question

Question 2 (4 points)

A 22.4 liter sample of gas at STP weighs 12.0 grams. What is the molecular weight of the gas?

a

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32.0 g/mol

16.0 g/mol

12.0 g/mol

29.4 g/mol

Answer 1

Molecular weight = 12.0 g/mol

The 3th option is correct

Step-by-step 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

Answer 2

12.0 g/mol

Step-by-step 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