To find the normality, we realize that HAc is a monoprotic acid, so the normality equals the molarity. So the solution is 0.904 M in HAc.

To calculate the molality of the solution, we find the number of moles of acetic acid per kilogram of solvent. Note that we divide by the mass of the solvent and not by the mass of the solution.

To calculate the mass percent of acetic acid in water we divide the mass of acetic acid, 14.1 g, by the total mass of solution, 264.1 g, and multiply by 100%. The solution is 5.34% acetic acid by mass.

The final concentration calculation is to find the mole fraction of acetic acid in the solution. To do so we find the number of moles of acetic acid, then divide that by the total number of moles in solution:

### Dilution

There are two common ways to prepare an aqueous solution. The first is to weigh out a known mass of solute, then add it to a given amount of solvent to achieve the desired concentration. The other method involves the dilution of a concentrated stock solution with more solvent to achieve a solution with a lower concentration than the original solution. To calculate the concentration of the diluted solution we will use the following formula: Figure %: The Relationship between Concentration and Volume for Dilution Problems

Suppose we wished to make 350 mL of a 0.15 M solution of sodium sulfate by diluting some 1.2 M sodium sulfate stock solution. To calculate the volume of stock solution necessary, we can solve the for v1:

There are several variants on the dilution problem such as asking for the volume of solution at a given concentration produced by diluting a known volume and concentration of a stock solution. All of these problems are readily solved by rearranging the concentration-volume equation then plugging in the known information.