**An aqueous solution of NH**_{3}which is 20.00% NH_{3}by mass has a density of 0.9250 g/ml. What volume of NH_{3}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.
**Identify the unknown, including units.**In your problem, the unknown is the volume of NH_{3}gas. Units aren't specified, so we're free to choose any units we like. We'll try*liters of NH*._{3}gas? L NH

_{3}gas**List the given information.**You know the following:- The aqueous solution of NH
_{3}is 20.00% NH_{3}by mass, which means that there are 20.00 grams of NH_{3}per 100 grams of solution (20.00 g NH_{3}= 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.

- The aqueous solution of NH
**Connect the given information with the unknown.**Try to work backwards from the unknown. The only information given about the gaseous NH_{3}is that the gas is at STP. Assume that the gas is ideal. Then you can connect L NH_{3}gas with moles of NH_{3}, 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 NH _{3}1 mol NH _{3}

= 22.4 L NH_{3}L NH _{3}_{3}. If you had g NH_{3}, you could easily get moles NH_{3}using the molecular weight.? g NH _{3}17.0 g NH _{3}= 1 mol NH_{3}mol NH _{3}1 mol NH _{3}

= 22.4 L NH_{3}L NH _{3}_{3}. You know that 20.00 g NH_{3}= 100 g solution, so? g solution 100 g soln = 20.00 g NH _{3}g NH _{3}17.0 g NH _{3}= 1 mol NH_{3}mol NH _{3}1 mol NH _{3}

= 22.4 L NH_{3}L NH _{3}? mL soln 0.9250 g soln = 1 mL soln g solution 100 g soln = 20.00 g NH _{3}g NH _{3}17.0 g NH _{3}= 1 mol NH_{3}mol NH _{3}1 mol NH _{3}

= 22.4 L NH_{3}L NH _{3}2.500 L 1000 mL = 1 L mL soln 0.9250 g soln = 1 mL soln g solution 100 g soln = 20.00 g NH _{3}g NH _{3}17.0 g NH _{3}= 1 mol NH_{3}mol NH _{3}1 mol NH _{3}

= 22.4 L NH_{3}L NH _{3}**Do the math**. Use the roadmap above to set up a string of conversion factors so that all units but L NH_{3}gas cancel.**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.

General Chemistry Online! What volume of gas is needed to prepare a 20% NH_3_ solution?

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Last Revised 08/17/15.URL: http://antoine.frostburg.edu/chem/senese/101/gases/faq/print-preparing-gas-solutions.shtml