A 25.00 mL sample of KNO3 solution was diluted to 1.00X103 mL. The concentration of the final solution is 0.00383 M. Calculate the concentration of the original solution?
A solution 0.00383 M has 0.00383 moles/L solution. What if you had 0.00383 moles/0.025 L (which is what you had before dilution) = ??
To solve this problem, we can use the equation for dilution:
C1V1 = C2V2
Where:
C1 = concentration of the original solution
V1 = volume of the original solution
C2 = concentration of the final solution
V2 = volume of the final solution
In this case, we are given:
C2 = 0.00383 M
V2 = 1.00 x 10^3 mL
We need to determine C1, the concentration of the original solution.
First, we need to convert the volumes to the same unit, either mL or L. Let's convert V1 and V2 to liters:
V1 = 25.00 mL = 25.00/1000 = 0.025 L (converting mL to L)
V2 = 1.00 x 10^3 mL = 1.00 x 10^3/1000 = 1.00 L (converting mL to L)
Now we can substitute the values into the dilution equation:
C1 * 0.025 L = 0.00383 M * 1.00 L
Simplifying the equation gives us:
C1 * 0.025 = 0.00383
Now we can solve for C1 by dividing both sides of the equation by 0.025:
C1 = 0.00383 / 0.025
C1 ≈ 0.1532 M
Therefore, the concentration of the original solution is approximately 0.1532 M.