A solution of TiCl3 is diluted from its original volume of 4450 mL by adding 5520 mL of solvent. If its new concentration is 1 M, what was the original concentration of this solution?
Question
A solution of TiCl3 is diluted from its original volume of 4450 mL by adding 5520 mL of solvent. If its new concentration is 1 M, what was the original concentration of this solution?
Solution
To solve this problem, we need to 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 = 4450 mL + 5520 mL = 9970 mL (final volume) V1 = 4450 mL (initial volume)
We need to find C1 (initial concentration).
So, we rearrange the formula to solve for C1:
C1 = C2V2 / V1
Substitute the known values into the equation:
C1 = (1 M * 9970 mL) / 4450 mL
C1 = 2.24 M
So, the original concentration of the solution was 2.24 M.
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