What is the density of ammonia (NH3) at 293 K and 0.913 atm?

Answer: The density of Ammonia is 0.648 g/l

Explanation:

Density = Mass / Volume

Mass of one mole of Ammonia (NH3) = 17.031g

Volume = ?

Using the ideal gas law we can determine the volume.

PV = nRT

P = 0.913 atm, V=?, n = 1, R = 0.08206 L. atm/K, and T = 293K

Make V the subject of the formular, we then have;

V = nRT / P = 1 mol x 0.08206 L. atm / K. mol x 293 / 0.913 atm

V = 24.04358 / 0.913 = 26.3L

Having gotten the value of Volume in this question, we then go back to solve for density.

Density = Mass / Volume

17.031g / 26.3L = 0.64756 ≈ 0.648 g/l