A gas mixture contains 1.25 g n2 and 0.85 g o2 in a 1.55 l c ontainer at 18 °c. calculate the mole fraction and partial pressure of each component in the gas mixture.

Answer: - mole fraction of nitrogen is 0.626 and mole fraction of oxygen is 0.374.

partial pressure of nitrogen is 0.687 atm and partial pressure of oxygen is 0.410 atm.

Solution: - Moles of nitrogen = 1.25 g x (1mol/28g) = 0.0446 mol

moles of oxygen = 0.85 g x (1mol/32g) = 0.0266 mol

Total moles of gases in the container = 0.0446 + 0.0266 = 0.0712

mole fraction of a gas = moles of gas/total moles of the gases

so, mole fraction of nitrogen = 0.0446/0.0712 = 0.626

mole fraction of oxygen = 0.0266/0.0712 = 0.374

Volume of the container is 1.55 L and the temperature is 18 degree C that is 18 + 273 = 291 K

From ideal gas law equation, PV = nRT

P = nRT/V

Partial pressure of nitrogen = (0.0446 x 0.0821 x 291) / 1.55 = 0.687 atm

and the partial pressure of oxygen = (0.0266 x 0.0821 x 291) / 1.55 = 0.410 atm