For this experiment you will need to perform a serial dilution of CO(NO3)2 solutions, meaning that you will begin with a stock solution, dilute it to make a new solution, and then use that new solution as the stock solution for the next dilution. You will start with a 0.25 M CO(NO3)2 solution. Using the values below, calculate the volume of solution and water needed at each step of the dilution. Concentration of original solution mL of original solution required mL of water required Concentration of new solution
0.25M 0.1M
0.1M 0.05M
0.05M 0.01M

See answer below

Explanation:

In this case, I will put the original photo of this exercise, because we are missing one data. The first picture is the original exercise.

Now, according to this, we need to make a serial dilution of CO(NO₃)₂. We don't know the volume of this solution, but we do know the total volume of the preparing solution (In the picture states that the total volume will be 10 mL).

So, we know the final volume of the solutions to be prepared, so, le'ts use the expression that will help us to solve this:

C₁V₁ = C₂V₂

Where:

C₁: Concentration of the given solution (stock)

V₁: volume required to prepare the dilluted solution

C₂; Concentration of the dilluted solution

V₂: Total volume of the dilluted solution.

Now that we know the expression to use and the meaning of each value, let's prepare the solutions:

To prepare 10 mL of 0.1 M using a 0.25 M, we will replace these values in the above expression; from there, we will solve for V₁, that value will tell us the required volume to prepare solution 2, and then, by difference we can calculate the volume of water:

Volume of water (Vw) =V₂ - V₁  

Now replacing the values:

0.25V₁ = 0.1 * 10

V₁ = 1/0.25 = 4 mL

This means that we need 4 mL of the stock to prepare the 0.1 M of dilluted solution, therefore, the volume of water required is:

Vw = 10 - 4

Using these same steps for the other two solutions we will get V1 and V2 for both of them. In this case, I will go straight to the procedure without further explanation because it's the same of this one.

For solution 2:

0.1V₁ = 0.05 * 10

V₁ = 0.5/0.1

Vw = 10 - 5

Finally for solution 3:

V₁ = 0.01 * 10 / 0.05

Vw = 10 - 2 mL

Hope this helps

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