volume of water in 97.0 mL of saturated ammonium hydroxide solution?

Question

I'm working in a lab with stock ammonium hydroxide solution that is 60.9% ammonium hydroxide by weight, which works about to be about 15.5M. What I want to know is how much water by volume is in 97.0 mL of this solution. I realize that the way to do this is with density, and the density of the solution is supposed to be .89 g/mL. But is that in g NH4OH per milliliter or in grams of solution per milliliter? Because if the former is true, then the calculation becomes much more difficult. Does anyone know how to do this?

Trevor H · Accepted Answer

The value of density = 0.89g/mL refers to the density of the ammonia solution. Ammonia solution is different to other solutions in that when you add the gas to the water, the density decreases.

You have 97.0mL at density = 0.89g/mL
Mass of solution = 97*0.89 = 86.33g

The ammonia solution = 15.5M
Mol ammonia in 97mL = 97/1000*15.5 = 1.5035 mol NH4OH 
Molar mass NH4OH = 35.0459 g/mol
1.5035mol = 1.5035*35.0459 = 52.69g NH4OH

The solution mass = 86.33g
Mass of NH4OH = 52.69g
Mass of water = 33.64g water in 97mL of the solution.

Edit the next day: I came back to this question because I think that a little further clarification is necessary: This is particularly so in consideration that the compound NH4OH does not exist. My original answer above followed the lines of the question.

We have determined that the 86.33g of solution contains 1.5035mol NH4OH. It is technically better to say that the solution contains 1.5035mol NH3 dissolved in water.
Molar mass NH3 = 17.03g/mol
1.5035mol NH3 = 1.5035*17.03 = 25.60g NH3

Now we have a more realistic situation:
 Mass of solution = 86.33g
Mass of NH3 dissolved = 25.60g
Mass of water in the solution = 60.73g

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volume of water in 97.0 mL of saturated ammonium hydroxide solution?

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