Question

A gas mixture containing N2 and O2 was kept inside a 2.00 L container at a temperature of 23.0°C and a total pressure of 1.00 atm. The partial pressure of oxygen was 0.722 atm. How many grams of nitrogen were present in the gas mixture?

Answer 1

You have all the information to use ideal gas law formula: PV = nRT

Answer 2

P total = Partial pressure of O2 + partial pressure N2
pressure N2 = 1-0.722 = 0.278 atm

PV = nRT

T = 273 + 23 = 296 K

n = (0.278)(2L)/(0.08206)(296)

n = ?

n * Mm = mass in grams