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A 569 mL NaCl solution is diluted to a volume of 1.37 L with a concentration of 3.00 M. What was the initial concentration?

Answer :

A 569 mL NaCl solution that has been diluted to a volume of 1.37 L and a concentration of 3.00 M. The initial concentration of the NaCl solution was approximately 7.21 M.

It can be found using the formula:

initial concentration x initial volume = final concentration x final volume

We are given that the final concentration is 3.00 M and the final volume is 1.37 L.

The initial volume can be found by converting the initial volume from milliliters to liters:

569 mL ÷ 1000 mL/L = 0.569 L

Now we can plug in the values we know and solve for the initial concentration:

(initial concentration)(0.569 L) = (3.00 M)(1.37 L)

initial concentration = (3.00 M)(1.37 L) ÷ (0.569 L)

initial concentration ≈ 7.21 M

Therefore, the initial concentration of the NaCl solution was approximately 7.21 M.

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