How many milliliters of 11.5 M HCl(aq) are needed to prepare 310.0 mL of 1.00 M HCl(aq)?
Use the dilution formula.
c1v1 = c2v2
c = concn
v = volume.
To determine the number of milliliters (mL) of 11.5 M HCl(aq) needed to prepare 310.0 mL of 1.00 M HCl(aq), we can use a dilution equation.
The dilution equation is given as:
M1V1 = M2V2
Where:
M1 = initial concentration of the solution (in this case, 11.5 M)
V1 = initial volume of the solution (unknown in this case)
M2 = final concentration of the solution (in this case, 1.00 M)
V2 = final volume of the solution (310.0 mL)
Rearranging the equation to solve for V1, we get:
V1 = (M2 * V2) / M1
Now, substituting the given values:
V1 = (1.00 M * 310.0 mL) / 11.5 M
V1 = 26.9565 mL
Therefore, approximately 26.96 mL of 11.5 M HCl(aq) are needed to prepare 310.0 mL of 1.00 M HCl(aq).