to what volume should 25.0 mL of 15 M nitric acid be diluted to prepare a 3.0 M solution
mL1 x M1 = mL2 x M
25.0 x 15 = mL2 x 3.0
To solve this problem, we can use the dilution equation:
(C1)(V1) = (C2)(V2)
Where:
C1 = initial concentration
V1 = initial volume
C2 = final concentration
V2 = final volume
Given:
C1 = 15 M (concentration of the stock solution)
V1 = 25.0 mL (initial volume of the stock solution)
C2 = 3.0 M (desired concentration of the final solution)
To find V2 (final volume of the diluted solution), we rearrange the equation as follows:
(V2) = (C1)(V1) / (C2)
Substituting the given values into the equation:
V2 = (15 M)(25.0 mL) / (3.0 M)
Now we can calculate V2:
V2 = 375 mL
Therefore, you need to dilute the 25.0 mL of 15 M nitric acid to a final volume of 375 mL to prepare a 3.0 M solution.
To find the volume to which the 25.0 mL of 15 M nitric acid should be diluted to prepare a 3.0 M solution, we can use the formula for dilution:
M1 * V1 = M2 * V2
Where:
M1 = initial concentration
V1 = initial volume
M2 = final concentration
V2 = final volume
In this case, we know:
M1 = 15 M (concentration of the nitric acid)
V1 = 25.0 mL (initial volume of the nitric acid)
M2 = 3.0 M (desired concentration of the final solution)
We need to solve for V2 (final volume). Rearranging the formula, we have:
V2 = (M1 * V1) / M2
Substituting the known values, we have:
V2 = (15 M * 25.0 mL) / 3.0 M
Now we simplify the equation:
V2 = (375 mL * M) / 3.0 M
V2 = 125 mL
Therefore, to prepare a 3.0 M solution, the 25.0 mL of 15 M nitric acid should be diluted to a final volume of 125 mL.