Question

Calculate the density of oxygen, O 2 , under each of the following conditions: STP 1.00 atm and 35.0 ∘ C Express your answers numerically in grams per liter. Enter the density at STP first and separate your answers by a comma.

Answer 1

At STP gas 1 mole of gas occupy volume 22.414 Liter

thus, 1 mole of Oxygen atom gas occupy volume 22.414 Liter

molar mass of oxygen atom = 15.9994 gm/mol thus 1 mole of oxygen atom = 15.9994 gm

density = mass/volume = 15.9994/22.414 = 0.7138 gm/liter

density of oxygen atom at STP =  0.7138 gm/liter

At STP gas 1 mole of gas occupy volume 22.414 Liter

thus, 1 mole of O₂ molecule occupy volume 22.414 Liter

molar mass of O₂ = 31.9988 gm/mol thus 1 mole of oxygen molecule = 31.9988 gm

density = mass/volume = 31.9988/22.414 = 1.4276 gm/liter

density of O₂ molecule at STP =  0.7138 gm/liter

At STP 1 mole gas occupy volume 22.414 L then at 1 atm and 35⁰ C gas occupy volume can be calculated by using ideal gas equation

We know that PV = nRT

V = nRT/P

n =1 mole, T = 273.15 + 35 = 308.15 K, P= 1 atm, R = 0.08205 L atm mol^-1 K^-1 ( R = gas constant)

Substitute these value in above equation.

V = 1 x 0.08205 x 308.15/1 = 25.28 L

Volume of gas is 25.28 L

molar mass of oxygen atom = 15.9994 gm/mol thus 1 mole of oxygen atom = 15.9994 gm

density = mass/volume = 15.9994/25.28 = 0.63288 gm/liter

density of oxygen atom at 1 atm and 35⁰C =  0.0.63288 gm/liter

molar mass of O₂ = 31.9988 gm/mol thus 1 mole of oxygen molecule = 31.9988 gm

density = mass/volume = 31.9988/25.28 = 1.2657 gm/liter

density of O₂ molecule at 1 atm and 35⁰C = 1.2657 gm/liter