A chemist must dilute
55.5mL
of
2.80
M aqueous nickel(II) chloride
NiCl2
solution until the concentration falls to
2.00
M . He'll do this by adding distilled water to the solution until it reaches a certain final volume.
Your method of posting makes it difficult to read. Also, you don't ask a question but if you want to know how much is the final volume use the dilution formula.
mL1 x M1 - mL2 x M2
55 x 2.80 = mL2 x 2.00
To dilute the 55.5 mL of 2.80 M aqueous nickel(II) chloride (NiCl2) solution to a concentration of 2.00 M, you can follow these steps:
Step 1: Use the dilution equation to calculate the final volume of the diluted solution:
M1 x V1 = M2 x V2
Here,
M1 = initial concentration = 2.80 M
V1 = initial volume = 55.5 mL
M2 = final concentration = 2.00 M
V2 = final volume (unknown)
Step 2: Substitute the given values into the dilution equation and solve for V2:
(2.80 M) x (55.5 mL) = (2.00 M) x V2
Step 3: Rearrange the equation and solve for V2:
V2 = (2.80 M x 55.5 mL) / 2.00 M
Step 4: Calculate the final volume (V2):
V2 = 77.7 mL
Therefore, in order to dilute 55.5 mL of 2.80 M aqueous nickel(II) chloride solution to a concentration of 2.00 M, you need to add enough distilled water to make the final volume reach 77.7 mL.
To determine the final volume of the solution after dilution, we can use the dilution formula:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution = 2.80 M
V1 = initial volume of the solution = 55.5 mL
C2 = final concentration of the solution = 2.00 M
V2 = final volume of the solution (which we need to find)
Rearranging the formula, we get:
V2 = (C1 * V1) / C2
Substituting in the given values:
V2 = (2.80 M * 55.5 mL) / 2.00 M
V2 = 77.7 mL
Therefore, in order to dilute the 55.5 mL of 2.80 M nickel(II) chloride solution to a concentration of 2.00 M, the chemist needs to add enough distilled water to reach a final volume of 77.7 mL.