Question

You have a 3.00-L container filled with N₂ (MM = 28.02 g/mol) at 298.15 K and 1.75 atm pressure connected to a 2.00-L container filled with Ar (MM = 39.95 g/mol) at 298.15 K and 2.15 atm pressure. A stopcock connecting the containers is opened and the gases are allowed to equilibrate between the two containers. What is the density of the final gas mixture? Assume ideal behavior. (Use R = 0.08206 L.atm/mol.K)

(HINT: What is the total mass, m, of the gaseous mixture? What is basic equation for density?)

g/L

Answer 1

Moles of N2 = PV / RT

= 1.75 * 3.00 / ( 0.08206 * 298.15 )

= 0.2146

Mass of N2 = mole * molar mass

= 0.2146 * 28

= 6.01 grams

Moles of Ar = 2.15 * 2.00 / ( 0.08206 * 298.15 )

= 0.1758

Mass of Ar = 0.1758 * 39.95

= 7.021 grams

Total mass = 6.01 + 7.021 = 13.031 grams

Total volume = 3.00 + 2.00 = 5.00 L

Density = mass / volume

= 13.031 / 5.00

= 2.6062 g/L

If you have any questions please comment