A solution of KClO is diluted from its a volume of 7000 mL to its new volume of 7200 mL. If its new concentration is 1 M, what was the original concentration of this solution?
Question
A solution of KClO is diluted from its a volume of 7000 mL to its new volume of 7200 mL. If its new concentration is 1 M, what was the original concentration of this solution?
Solution
To solve this problem, we can use the formula for dilution:
C1V1 = C2V2
where: C1 = initial concentration V1 = initial volume C2 = final concentration V2 = final volume
We know that: C2 = 1 M (final concentration) V2 = 7200 mL (final volume) V1 = 7000 mL (initial volume)
We need to find C1 (initial concentration).
Rearranging the formula to solve for C1 gives us:
C1 = C2V2 / V1
Substituting the known values into the equation gives us:
C1 = (1 M * 7200 mL) / 7000 mL
Solving this equation gives us:
C1 = 1.02857 M
So, the original concentration of the KClO solution was approximately 1.03 M.
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