How many grams of O2 must be placed in a 3.33 liter container in order to exert a pressure of 2.46 atm at 27°C?

10.656 g

Explanation:

Given data:

Mass of oxygen = ?

Volume of container = 3.33 L

Pressure of gas = 2.46 atm

Temperature = 27°C

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

27+273 = 300 K

Now we will put the values in formula.

n = PV/RT

n = 2.46 atm × 3.33 L /  0.0821 atm.L/ mol.K   × 300 K

n = 8.192 atm. L /24.63 atm.L/ mol

n = 0.333 mol

Mass in grams:

Mass = number of moles × molar mass

Mass = 0.333 mol × 32 g/mol

Mass = 10.656 g