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

 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

the answer

explanation:

the capillary tube was too close to the bottom of the beaker.

the ruler may have moved.

water got into the capillary tube.

the temperature was not allowed to equilibrate in the 2-4 minutes.

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