To what volume must 1.0 l of a 6.0 m solution of hcl be diluted in order to prepare a 0.2 m solution?
I just need to find wht steps i need to take to get the answer
To calculate the volume of the 6.0 M HCl solution that needs to be diluted to prepare a 0.2 M solution, you can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration (6.0 M)
V1 = initial volume (unknown)
C2 = final concentration (0.2 M)
V2 = final volume (1.0 L)
Now, let's break down the steps:
Step 1: Write down the known values:
C1 = 6.0 M
C2 = 0.2 M
V2 = 1.0 L
Step 2: Solve for V1 using the formula mentioned above:
C1V1 = C2V2
6.0 M * V1 = 0.2 M * 1.0 L
Step 3: Rearrange the equation to solve for V1:
V1 = (C2V2) / C1
V1 = (0.2 M * 1.0 L) / 6.0 M
Step 4: Calculate the volume (V1):
V1 = (0.2 L) / (6.0 M)
Step 5: Simplify the equation:
V1 = 0.033 L
Therefore, 1.0 L of the 6.0 M HCl solution must be diluted to a volume of 0.033 L (or approximately 33 mL) to prepare a 0.2 M solution.
To find the volume to which the 1.0 L of a 6.0 M solution of HCl must be diluted in order to prepare a 0.2 M solution, follow these steps:
Step 1: Use the formula for dilution:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution (6.0 M)
V1 = initial volume of the solution (1.0 L)
C2 = final concentration of the solution (0.2 M)
V2 = final volume of the solution (unknown)
Step 2: Rearrange the formula to solve for V2:
V2 = (C1V1) / C2
Step 3: Plug in the values:
V2 = (6.0 M * 1.0 L) / 0.2 M
Step 4: Calculate:
V2 = 30 L
Therefore, you need to dilute the 1.0 L of a 6.0 M solution of HCl to a final volume of 30 L to prepare a 0.2 M solution.
you are reducing the concentration by a factor of .2/6
So, you need to increase the volume by a factor of 6/.2