Question

Chemistry 2A Lab 13: Solution Calculations Introduction This lab involves sample calculations using various concentration units. This lab will work with the following types of concentration units: % mass, % volume, % mass/volume, and molarity. A solution is the sum total of the solute and solvent. % mass= mass solute mass solution -100 volumesolute .100 % volume- volume solution % mass/volume mass solute volumesolution -100 Molarity moles solute liters solution The dilution of a liquid requires the addition of more solvent. It should be noted when a problem refers to a solution and no solvent is stated, it is assumed to be water. To complete calculations using dilutions use the following convenient equation: CiV=C2V2 Where Ci is the initial concentration, C is the final concentration, V. is the initial volume and V, is the final volume. The density of water is 1.0 g/ml. For dilute aqueous (water is the solvent) solutions, it can be assumed the density is that of water. The general concept of the sum of the parts equals the whole applies to the addition of the mass of solutions, but not necessarily to solutions, why? If a solution has a specific mass, and more mass is added, the total mass of the solution is in fact equal to the mass of the solute plus the solvent. However, for the addition of liquids by volume measurements, if the density of the two solutions is different, then the addition of the two volumes will create a new solution with a different density. Its volume will not be additive. If the liquids to be added are dilute aqueous solutions we can assume the volume of solutions to be additive because the density of the two liquids is very close to that of water. A good rule of thumb is any solution that is 0.1 M or less in concentration is a dilute solution

Answer 1

1. % mass = mass of solute x 100% barium nitrate x108% mass of solution mass of barium nitrate mass of barium nitrate mass of water 25.09 - X100% 25-09+45:09 250 x 100% = (35.7% 7000 2 . 7. by mass = - x100 y. mass of solute mass of solution 14:09 x 100% = 135.0% / 40.09

1. 3. % mass/volume = mass of solute x 100% volume of solution 19.0 x100 % 200.0 9.50 % (0/0) 4. Yo volume = 25.0mL x 100% 100.0mL = 25.0% 1. 5. volume of solution = 250.0m mass of KBr that could be recovered = 50 x 250.0mL 1oooomL - 37.5 9] Mass of solute required = 3.50 9 x 6. -X 200 mL 100m2 - 67.009) 7. Mass of solution required = 100g x 25.09 12.809 119539