I need help doing these ..here is the introduction and 2nd paper the questions C

Question

I need help doing these ..here is the introduction and 2nd paper the questions

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 volume solute 100 volume solution % volume = mass solute volume solution % mass/volume 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: Where Ci is the initial concentration, C2 is the final concentration, Vi is the initial volume and V2 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.

Explanation / Answer

Ans 1 :

% mass = (mass of solute / mass of solution) x 100

mass of solution = mass of solute + mass of solvent

= 25.0 + 45.0

= 70.0 g

So the % mass of barium nitrate in the solution =

= ( 25.0 / 70.0) x 100

= 35.7 %

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