Question

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

Answer 1

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