A solution of CsI is diluted from its original volume of 8000 mL by adding 990 mL of solvent. If its new concentration is 0.2 M, what was the original concentration of this solution?
Question
A solution of CsI is diluted from its original volume of 8000 mL by adding 990 mL of solvent. If its new concentration is 0.2 M, what was the original concentration of this solution?
Solution
To solve this problem, we can use the formula for dilution: M1V1 = M2V2, where M1 is the initial concentration, V1 is the initial volume, M2 is the final concentration, and V2 is the final volume.
Given in the problem, we have:
M2 = 0.2 M (final concentration) V1 = 8000 mL (initial volume) V2 = V1 + 990 mL = 8000 mL + 990 mL = 8990 mL (final volume)
We need to find M1 (initial concentration). We can rearrange the formula to solve for M1:
M1 = M2V2 / V1
Substituting the given values:
M1 = (0.2 M * 8990 mL) / 8000 mL
M1 = 0.224875 M
So, the original concentration of the solution was approximately 0.225 M.
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