need answer please

Question

A solution of iron (III) bromide at 3.5 M is diluted from its original volume of 7.52 L by adding 321.8 mL. What is the resulting concentration of this solution in moles per liter?

Answer 1

This question reads like it is from general chemistry, so there are two answers that the instructor may be looking for. The pedantic instructor will expect you to say that because of the osmolarity difference between the original solution and the diluted solution, the new volume will be some amount less than a simple addition between the two volumes. The regular instructor will ignore that and instead be looking for the following answer.

Because the concentration, in molar, is given in units of mol/L, multiplying it by the volume will just give the number of moles of solute:

mol/L x L = mol

We can use this fact, and the fact that diluting a solution does not involve chancing the number of moles of solute, to make an equation that will tell us the concentration of any solution that has been diluted:

mol1/L1 x L1 = mol1/L2 x L2

This is usually given as the "dilution equation" as:

C1 x V1 = C2 x V2

For this question, we know two of these variables from the question where we are told that C1 = 3.5M and V1 = 7.51L. We can also figure out the final volume by adding the volumes in the question so that 7.51L + 0.3218L = 7.8318L. Following the significant figure rules, this shortens to V2 = 7.83L (please note that I have converted the added water volume, 321.8mL = 0.3218L). So, we can plug these numbers in and rearrange the equation and then solve.

3.5M x 7.51L = C2 x 7.83L

3.5M x 7.51L / 7.83L = C2

3.357M = C2

With significant figure rules, this comes to.

C2 = 3.4M

So, the final concentration of the solution, after adding the water, is 3.4M iron (III) bromide.