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22 October, 08:25

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

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  1. 22 October, 08:50
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    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
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