Assuming that the volumes are additive , how much water should be added to75.0 mL of 6.00 M HCl to prepare 0.500 M HCl?
825 mL
To determine how much water should be added to a given volume of a concentrated solution to prepare a desired dilute solution, we can use the formula for dilution:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = final concentration of the solution
V2 = final volume of the solution
In this case, we have the following information:
C1 = 6.00 M (concentration of the concentrated solution)
V1 = 75.0 mL (initial volume of the concentrated solution)
C2 = 0.500 M (desired concentration of the dilute solution)
V2 = ?
To find V2, we rearrange the formula:
V2 = (C1V1) / C2
Substituting the given values:
V2 = (6.00 M * 75.0 mL) / 0.500 M
To simplify the calculation, we need to ensure that the units cancel out properly. We convert mL to L:
V2 = (6.00 M * 0.075 L) / 0.500 M
Now we can calculate:
V2 = 0.900 L
So, to prepare 0.500 M HCl, you need to add 0.900 L (or 900 mL) of water to the 75.0 mL of 6.00 M HCl solution.