A solution of K3P is diluted from its original volume of 1.38 L by adding 0.12 L of solvent. If its new concentration is 0.002 M, what was the original concentration of this solution? M❮❯
Question
A solution of K3P is diluted from its original volume of 1.38 L by adding 0.12 L of solvent. If its new concentration is 0.002 M, what was the original concentration of this solution? M❮❯
Solution
To solve this problem, we can use the formula for dilution: M1V1 = M2V2, where M1 is the initial molarity, V1 is the initial volume, M2 is the final molarity, and V2 is the final volume.
Given in the problem, we have:
M2 = 0.002 M (final molarity) V1 = 1.38 L (initial volume) V2 = V1 + 0.12 L = 1.38 L + 0.12 L = 1.5 L (final volume)
We need to find M1 (initial molarity). We can rearrange the formula to solve for M1:
M1 = M2V2 / V1
Substitute the given values into the formula:
M1 = (0.002 M * 1.5 L) / 1.38 L
M1 = 0.00217 M
So, the original concentration of the solution was 0.00217 M.
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