You have 353 mL of a 1.25 M potassium Chloride solution, but you need to make a 0.50 M potassium Chloride solution. How many milliliters of water must you add to the original 1.25 M solution to make the 0.50 M potassium Chloride solution
529.5
mL x M = mL x M
Substitute and solve for mL.
353 x 1.25 = mL x 0.5
Solve for mL and this will be the FINAL VOLUME. To find how much water must be added requires
a. That we assume the volumes are additive.
b. Then final volume in mL - 353 mL = mL to be added.
Well, if you're looking to dilute the 1.25 M solution to 0.50 M, you'll need to add some water. Let's call the volume of water you need to add "X" mL.
Now, let's think about the number of moles of potassium chloride in both solutions. The moles of potassium chloride in the original solution (1.25 M) is equal to the concentration (1.25 mol/L) multiplied by the volume (353 mL).
So, moles(1.25 M solution) = 1.25 mol/L * 353 mL = 441.25 millimoles
In the final 0.50 M solution, we want the same number of moles of potassium chloride. Since the concentration is given (0.50 M) and we're adding X mL of water, we can calculate the moles as:
moles(0.50 M solution) = 0.50 mol/L * (353 mL + X)
Since we want the same moles in both solutions, we have:
441.25 millimoles = 0.50 mol/L * (353 mL + X)
By rearranging the equation and solving for X, we find:
X = (441.25 millimoles / 0.50 mol/L) - 353 mL
Now, let's calculate that out... Hold on a second...
*starts juggling bottles and test tubes*
Okay, my calculations are complete! You need to add approximately 323 mL of water to the original 1.25 M solution to make the 0.50 M potassium chloride solution.
To solve this problem, we need to calculate the amount of water needed to dilute a 353 mL, 1.25 M potassium chloride solution to a 0.50 M potassium chloride solution.
To find the amount of water needed, we first need to calculate the amount of potassium chloride in the original solution. This can be determined by using the formula:
initial amount of solute = initial concentration × initial volume
initial amount of solute = 1.25 M × 353 mL
initial amount of solute = 441.25 millimoles (mmol)
Next, we need to calculate the amount of potassium chloride we want in the final solution. We can determine this by using the formula:
final amount of solute = final concentration × final volume
Since we want the final concentration to be 0.50 M and the final volume to be the sum of the initial volume and the volume of water added, we can write:
final amount of solute = 0.50 M × (353 mL + volume of water)
Now we can set up an equation using the initial and final amounts of solute:
441.25 mmol = 0.50 M × (353 mL + volume of water)
To solve for the volume of water, we can rearrange the equation:
353 mL + volume of water = 441.25 mmol / 0.50 M
volume of water = (441.25 mmol / 0.50 M) - 353 mL
Evaluating this expression will give us the milliliters of water needed to dilute the solution to 0.50 M.
To find out how many milliliters of water you need to add to the original 1.25 M solution, we can use the concept of dilution. The formula for dilution is as follows:
C₁V₁ = C₂V₂
Where:
C₁ = Initial concentration
V₁ = Initial volume
C₂ = Final concentration
V₂ = Final volume
We need to find V₂, the final volume, which is the sum of the original volume (V₁) and the volume of water we add. The final concentration (C₂) is given as 0.50 M, and the initial volume (V₁) is 353 mL.
Plugging in the values into the formula, we have:
(1.25 M)(353 mL) = (0.50 M)(V₂)
(1.25)(353) = (0.50)(V₂)
441.25 = 0.50V₂
Now, solve for V₂:
V₂ = 441.25 / 0.50
V₂ ≈ 882.5 mL
Therefore, you need to add approximately 882.5 mL of water to the original 1.25 M potassium chloride solution in order to obtain a 0.50 M potassium chloride solution.