If 40 mL of a 6 {\rm M} \;{\rm NaOH} solution is diluted to a final volume of 200 mL, what is the resulting concentration of the solution?
Use proportions:
6M * (40 mL/200 mL)
= 1.2M
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To determine the resulting concentration of the solution, we can use the formula for dilutions:
C1V1 = C2V2
Where:
C1 = initial concentration of the solution
V1 = initial volume of the solution
C2 = resulting concentration of the solution
V2 = final volume of the solution
Given:
C1 = 6 M (concentration of the NaOH solution)
V1 = 40 mL (initial volume of the NaOH solution)
V2 = 200 mL (final volume of the NaOH solution)
Plugging the values into the formula, we have:
(6 M)(40 mL) = C2(200 mL)
Simplifying the equation:
240 M mL = C2(200 mL)
Now we can solve for C2:
C2 = (240 M mL) / (200 mL)
Cancelling out the unit "mL," we have:
C2 = 1.2 M
Therefore, the resulting concentration of the solution after dilution is 1.2 M.
To find the resulting concentration of the solution after dilution, we can use the formula:
C1V1 = C2V2
Where:
C1 = initial concentration
V1 = initial volume
C2 = resulting concentration
V2 = resulting volume
In this case, we have:
C1 = 6 M (molar concentration) (given)
V1 = 40 mL (given)
V2 = 200 mL (given)
Now we can substitute the given values into the formula and solve for C2:
6 M * 40 mL = C2 * 200 mL
Simplifying the equation:
240 mL • M = C2 • 200 mL
Dividing both sides of the equation by 200 mL:
C2 = (240 mL • M) / 200 mL
Cancelling out the mL unit:
C2 = 1.2 M
Therefore, the resulting concentration of the solution after dilution is 1.2 M (molar concentration).