Question

What mass (in grams) of nitrogen must be added to 8.34 g of argon at 24.17oC, and in a 126.1 L contain for a final pressure of 895 mmHg?

Answer 1

Calculation of pressure exerted by Ar gas :

We know that PV = nRT

Where

T = Temperature = 24.17 ^oC = 24.17+273 = 297.17 K

P = pressure = ? atm

n = No . of moles = mass/molar mass = 8.34 g / 40(g/mol) = 0.2085 mol

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = volume of the container = 126.1L

Plug the values we get P = (nRT) / V

                                    = 0.0403 atm

                                    = 0.0403 x760 mmHg             since 1 atm = 760 mm Hg

                                    = 30.7 mm Hg

So the partial pressure exerted by Ar is p_Ar = 30.7 mm Hg

Given total pressure , P = 895 mm Hg

So the partial pressure exerted by N₂ is , p_N2 = P - p_Ar

                                                                    = 895 - 30.7

                                                                    = 864.3 mm Hg

                                                                     = 864.3/760 atm

                                                                    = 1.137 atm

Calculation of number of moles of Nitrogen :

We know that PV = nRT

Where

T = Temperature = 24.17 ^oC = 24.17+273 = 297.17 K

P = pressure = 1.137 atm

n = No . of moles = ?

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas = 126.1L

Plug the values we get n = (PV)/(RT)

                                     = 5.87 moles

Molar mass of N₂ = 2xAt.mass of N = 2x14 = 28 g/mol

So mass of N₂ added , m = number of moles x molar mass

                                     = 5.87 mol x 28 (g/mol)

                                     = 164.6 g

Therefore the mass of Nitrogen added is 164.6 g