What is the final concentration of the solution produced when 225.5 mL of 0.09988-M solution of Na2CO3 is allowed to evaporate until the solution volume is reduced to 45.00 mL?

Respuesta :

Explanation:

The given data is as follows.

       [tex]V_{1}[/tex] = 225.5 mL,             [tex]V_{2}[/tex] = 45.00 mL

       [tex]M_{1}[/tex] = 0.09988 M,          [tex]M_{2}[/tex] = ?

Therefore, formula to calculate final concentration will be as follows.

                  [tex]M_{1}V_{1}[/tex] = [tex]M_{2}V_{2}[/tex]

Putting the given values into the above formula as follows.

            [tex]M_{1}V_{1}[/tex] = [tex]M_{2}V_{2}[/tex]

   [tex]0.09988 M \times 225.5 mL[/tex] = [tex]M_{2} \times 45.0 mL[/tex]

                    [tex]M_{2}[/tex] = 0.5 M

Thus, we can conclude that the final concentration of the given solution is 0.5 M.

The final concentration of the solution produced when 225.5 mL of 0.09988-M solution of Na2CO3 is allowed to evaporate until the solution volume is reduced to 45.00 mL is 0.5M.

HOW TO CALCULATE CONCENTRATION?

The final concentration of a solution can be calculated using the following formula:

C1V1 = C2V2

Where;

  • C1 = initial concentration
  • C2 = final concentration
  • V1 = initial volume
  • V2 = final volume

225.5 × 0.09988 = 45 × C2

22.52 = 45C2

C2 = 22.52/45

C2 = 0.5M

Therefore, the final concentration of the solution produced when 225.5 mL of 0.09988-M solution of Na2CO3 is allowed to evaporate until the solution volume is reduced to 45.00 mL is 0.5M.

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