Flask Q (volume = 1.00 x 103 cm3 ) is filled with ammonia (NH3) at 102 kPa and 300 K. The tap is closed and there is a vacuum in flask P. (Gas constant R = 8.31 J K−1 mol−1 ) Calculate the mass of ammonia

This question can be solved using the ideal gas equation, given by: PV=nRT 

Rearranging the ideal gas equation gives an expression for the number of moles as

n=PV/RT 

Subbing the values given in the question, and being careful to change the units to SI (ie, kPa to Pa and cm^3 to m^3) we obtain the following value:

n=[(102.10^3)(1x10^-3)]/[(8.31)(300)]

 = 0.0409 moles of Ammonia

with the moles calculated, we can then obtain the mass of Ammonia using the following equation:

n = mass/RMM

rearranging gives:

mass = n*RMM

Subbing in the value of n for Ammonia, and the RMM of Ammonia (17 g/mol) we obtain the following value:

mass = 0.0409*17

         = 0.696 g 

Hence the mass of Ammonia in Flask Q is 0.696 grams.