How much water must be added to 482. mL of 0.201 M HCl to produce a 0.147 M solution? (Assume that the volumes are additive.)
Use the dilution formula of
c1v1 = c2v2
c = concn
v = volume
after using the dilution formula M1V1=M2V2, i got .6L of water
To solve this question, we need to use the equation for dilution:
M1V1 = M2V2
where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.
First, let's assign the values:
M1 = 0.201 M (concentration of the initial HCl solution)
V1 = 482. mL (initial volume of the HCl solution)
M2 = 0.147 M (desired final concentration of the solution)
We need to find V2, the final volume of the diluted solution.
Rearranging the equation, we have:
V2 = (M1 * V1) / M2
Now, let's substitute the values into the equation and solve for V2:
V2 = (0.201 M * 482. mL) / 0.147 M
Multiplying the concentration and volume values:
V2 = (0.097482 mL*M) / 0.147 M
Dividing to get the final volume in mL:
V2 ≈ 662.47 mL
Therefore, approximately 662.47 mL of water must be added to 482 mL of 0.201 M HCl to produce a 0.147 M solution.