Question

5)A gaseous mixture composed of 16.0 g of O2 , 21.0 g of N2, and 16.0 g of He at 20.0°C is confined in a 25.0 L container.
What is the partial pressure of N2 in this mixture?
6)A gaseous mixture composed of 16.0 g of 2, 21.0 g of N2, and 16.0 g of He at 20.0°C is confined in a 25.0 L container.
What is the partial pressure of He in this mixture?
7)Assuming ideal gas behavior, what is the pressure in atm exerted by 1.57 mol Cl2(g) confined to a volume of 1.50 L at 273K?

Answer 1

5) moles of O2 = 16 g/32 gmol = 0.5 mol

moles of N2 = 21 g/28 g/mol = 0.75 mol

moles of He = 16 g/4 g/mol = 4 mol

total moles = 5.25 mol

Total pressure = nRT/V = 5.25 x 0.8205 x (273 + 20)/25 = 5.076 atm

Partial pressure of N2 = mole fraction of N2 x total pressure

                                    = (0.75/5.25) x 5.076 = 0.725 atm

6) Partial pressure of He = mole fraction of He x total pressure

                                        = (4/5.25) x 5.076

                                        = 3.867 atm

7) Pressure = nRT/V

n = 1.57 mol Cl2

R = gas constant

T = 273 K

V = 1.50 L

we get,

Pressure = 1.57 x 0.08205 x 273/1.50 = 23.445 atm