A 50.0 mL sample of 6.0 M HCL is dulited to 200.0 mL. What is the new concentration of the acid?
for this type of problem use
M1V1=M2V2
(provided that the M and V are in the same units no conversion is needed)
50.0 ml x 6.0 M = 200.0 ml x M2
so M2=50.0 ml x 6.0 M/200.0 ml
Well, let's calculate that, shall we? With the limited amount of humor I can muster. So, if you take 50.0 mL of 6.0 M HCL, dilute it with 200.0 mL, and then measure the concentration of the acid, it will be... drumroll, please... 1.5 M! Ta-da! Now isn't that acid-tastically diluted?
To find the new concentration of the acid after dilution, you can use the dilution formula:
C1V1 = C2V2
where C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
Given:
C1 = 6.0 M
V1 = 50.0 mL
V2 = 200.0 mL
Plugging in the values into the dilution formula, we get:
6.0 M × 50.0 mL = C2 × 200.0 mL
Rearranging the formula to isolate C2, we have:
C2 = (6.0 M × 50.0 mL) / 200.0 mL
C2 = 3.0 M
Therefore, the new concentration of the acid after dilution is 3.0 M.
To find the new concentration of the acid after dilution, we need to use the equation:
M₁V₁ = M₂V₂
where:
M₁ = initial concentration of the acid
V₁ = initial volume of the acid
M₂ = final concentration of the acid
V₂ = final volume of the acid
In this case, the initial concentration (M₁) is 6.0 M, the initial volume (V₁) is 50.0 mL, the final volume (V₂) is 200.0 mL, and we need to find the final concentration (M₂).
Plugging in the known values into the equation, we get:
(6.0 M)(50.0 mL) = M₂(200.0 mL)
Now, let's solve for M₂:
M₂ = (6.0 M)(50.0 mL) / (200.0 mL)
M₂ = 3.0 M
Therefore, the new concentration of the acid after dilution is 3.0 M.