Question

Concentrated aqueous ammonia is 28% NH3 by weight. Its density is 0.90g/ml. Calculate molarity and molality.

Answer 1

molar mass NH3 is 17 so
0.9 g/mL x 1000 mL x 0.28 x (1/17) = 14.8 mols/L and that is M.
m = molality = mols/kg solvent.
0.9 g/mL x 1000 mL = 900 g in the solution.
How much of that is ammonia? 900 x 0.28 = 252 g NH3.
How much is water? 900 g soln - 252 g NH3 = 648 g H2O
So you have 14.8 moles NH3/0.648 kg solvent = ? m

Answer 2

Hi thank you. You gave me better insight on this, but can you explain to me why the g/ml value was multiplied by 1000 ? I divided it to convert to liters. I'm also lost on how to find the amount of solution. I assumed 100g.

Answer 3

You multiply by 1000 for two reasons. First, the density is given in the problem as 0.900 grams/mL so that times 1000 mL gives you 900 grams. If you convert to L first then you must have the density in g/L so
0.900 g/mL x 1000 mL = 900 grams OR
900 g/L x 1 L = 900 grams. You do this and that answers your second question. That IS the mass of the solution. Here it is in detail
density = 0.900 g/mL so 1000 mL has a mass of 0.900 g/mL . x 1000 mL = 900 grams. Only 28% of that is NH3 so the mass of NH3 is
900 x 0.28 = 252 grams NH3. How many mols is that of NH3? That's grams/molar mass = mols = 252/17 = 14.8 mols. That is in 1 L solution. The definition of molarity is M = mols/L of solution. You have 14.8 mols/L of solution so that is 14.8 M.
NOTE: If you have other questions I suggest you post at the beginning. This is an old post and is getting lost. It's too far back in the pack to be noticed now.