Concentrated aqueous ammonia is 28% NH3 by weight. Its density is 0.90g/ml. Calculate molarity and molality.
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
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.
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.
To calculate the molarity and molality of a solution of concentrated aqueous ammonia, we need to use the given information about the percentage composition and density.
1. Calculate the molarity (M):
Molarity (M) is defined as moles of solute per liter of solution. In this case, the solute is ammonia (NH3).
Step 1: Calculate the mass of NH3 in 100 g of the solution.
Mass of NH3 = 28% of 100 g
= 0.28 * 100 g = 28 g
Step 2: Calculate the number of moles of NH3 using its molar mass.
Molar mass of NH3 = (1 * 14.01 g/mol) + (3 * 1.01 g/mol) = 17.03 g/mol
Number of moles of NH3 = Mass of NH3 / Molar mass of NH3
= 28 g / 17.03 g/mol ≈ 1.64 mol
Step 3: Calculate the volume of the solution in liters.
Density = Mass / Volume
Volume = Mass / Density
Volume = 100 g / 0.90 g/ml
= 111.11 ml ≈ 0.111 L
Step 4: Calculate the molarity using the number of moles and volume.
Molarity (M) = Moles / Volume
Molarity = 1.64 mol / 0.111 L
Molarity ≈ 14.77 M
So, the molarity of the solution is approximately 14.77 M.
2. Calculate the molality (m):
Molality (m) is defined as moles of solute per kilogram of solvent. In this case, the solvent is water.
Step 1: Calculate the mass of water in the solution.
Mass of water = Total mass of solution - Mass of NH3
= 100 g - 28 g = 72 g
Step 2: Convert the mass of water from grams to kilograms.
Mass of water = 72 g / 1000 g/kg
Mass of water = 0.072 kg
Step 3: Calculate the number of moles of NH3 using its molar mass.
Number of moles of NH3 = 1.64 mol (from previous calculation)
Step 4: Calculate the molality using the number of moles and mass of water.
Molality (m) = Moles / Mass of solvent (in kg)
Molality = 1.64 mol / 0.072 kg
Molality ≈ 22.78 m
So, the molality of the solution is approximately 22.78 m.