What volume of gas is needed to prepare a 20% NH₃ solution?

An aqueous solution of NH₃ which is 20.00% NH₃ by mass has a density of 0.9250 g/ml. What volume of NH₃ gas at STP would be required in the preparation of 2.500 L of solution?
Catherine

This problem is can be solved by breaking it into a chain of simpler problems. Follow these steps.

1. Identify the unknown, including units. In your problem, the unknown is the volume of NH₃ gas. Units aren't specified, so we're free to choose any units we like. We'll try liters of NH₃ gas.

   ? L NH₃ gas

2. List the given information. You know the following:
   • The aqueous solution of NH₃ is 20.00% NH₃ by mass, which means that there are 20.00 grams of NH₃ per 100 grams of solution (20.00 g NH₃ = 100 g solution).
   • The solution has a density of 0.9250 g/mL, which means that 0.9250 g solution = 1 mL solution.
   • 2.500 L of solution is being prepared.
   • The gas is at STP, meaning it has a temperature of 273.15 K and a pressure of 1 atm.
3. Connect the given information with the unknown. Try to work backwards from the unknown. The only information given about the gaseous NH₃ is that the gas is at STP. Assume that the gas is ideal. Then you can connect L NH₃ gas with moles of NH₃, using the temperature (273 K) and the pressure (1 atm) and the ideal gas law. (You can make the same connection using the fact that 1 mole of ideal gas at STP occupies 22.4 L):

   ?

   mol NH3

   1 mol NH3 = 22.4 L NH3

   L NH3

   The target is now moles of NH₃. If you had g NH₃, you could easily get moles NH₃ using the molecular weight.

   ?

   g NH3

   17.0 g NH3 = 1 mol NH3

   mol NH3

   1 mol NH3 = 22.4 L NH3

   L NH3

   The target is now g of NH₃. You know that 20.00 g NH₃ = 100 g solution, so

   ?

   g solution

   100 g soln = 20.00 g NH3

   g NH3

   17.0 g NH3 = 1 mol NH3

   mol NH3

   1 mol NH3 = 22.4 L NH3

   L NH3

   The target is now g of solution. From the given density, 0.9250 g solution = 1 mL solution, and

   ?

   mL soln

   0.9250 g soln = 1 mL soln

   g solution

   100 g soln = 20.00 g NH3

   g NH3

   17.0 g NH3 = 1 mol NH3

   mol NH3

   1 mol NH3 = 22.4 L NH3

   L NH3

   Finally, you know that there are 2.500 L solution, so you can easily get mL of solution:

   2.500 L

   1000 mL = 1 L

   mL soln

   0.9250 g soln = 1 mL soln

   g solution

   100 g soln = 20.00 g NH3

   g NH3

   17.0 g NH3 = 1 mol NH3

   mol NH3

   1 mol NH3 = 22.4 L NH3

   L NH3

4. Do the math. Use the roadmap above to set up a string of conversion factors so that all units but L NH₃ gas cancel.
5. Check the result Does the answer make sense? Check to see if you can work back towards the given information from your final answer, or try to devise another way to solve the problem.

Author: Fred Senese senese@antoine.frostburg.edu

Copyright © 1997-2001 by Fred Senese
Comments & questions to senese@antoine.frostburg.edu
Last Revised 02/23/18.URL: http://antoine.frostburg.edu/chem/senese/101-hidden/gases/faq/preparing-gas-solutions.shtml