What volume of 12.1 HCL (aq) is needed to prepare 250.0 mL of 1.00 M HCL (aq).
To determine the volume of 12.1 M HCl (aq) needed to prepare 250.0 mL of 1.00 M HCl (aq), we can use the equation:
(M1)(V1) = (M2)(V2)
Where:
M1 = initial concentration of the HCl solution
V1 = initial volume of the HCl solution
M2 = final concentration of the HCl solution
V2 = final volume of the HCl solution
In this case, we have:
M1 = 12.1 M
V1 = ?
M2 = 1.00 M
V2 = 250.0 mL
Rearranging the equation, we have:
V1 = (M2)(V2) / M1
Substituting the given values, we get:
V1 = (1.00 M)(250.0 mL) / 12.1 M
Calculating this, we find:
V1 ≈ 20.661 mL
Therefore, approximately 20.661 mL of 12.1 M HCl (aq) is needed to prepare 250.0 mL of 1.00 M HCl (aq).
mL1 x M1 = mL2 x M2
mL1 x 12.1M = etc