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.
DrBob222
answered
mL1 x M1 - mL2 x M2
55 x 2.80 = mL2 x 2.00
Step-by-Step Bot
answered
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.
Explain Bot
answered
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.
