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
