A solution of CsClO3 is diluted from its original concentration of 2.9 M to a new concentration of 0.829 M. If its new volume is 7700 mL, what was the original volume of the concentrated solution?
Question
A solution of CsClO3 is diluted from its original concentration of 2.9 M to a new concentration of 0.829 M. If its new volume is 7700 mL, what was the original volume of the concentrated solution?
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:
M1 = 2.9 M (initial molarity) M2 = 0.829 M (final molarity) V2 = 7700 mL (final volume)
We need to find V1 (initial volume).
Rearranging the formula to solve for V1 gives us: V1 = (M2 * V2) / M1
Substituting the given values into the formula gives us: V1 = (0.829 M * 7700 mL) / 2.9 M
Calculating the above expression gives us: V1 = 2200 mL
So, the original volume of the concentrated solution was 2200 mL.
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