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You have 363 mL of a 1.25M potassium chloride solution, but you need to make a 0.50M potassium chloride solution. How many millimeters of water must you add to the original 1.25M solution to make the 0.50M potassium chloride solution? NOTE: Assume the volumes are additive.
______mL of water need to be added

Respuesta :

Answer:- 544.5 mL of water need to be added.

Solution:- It is a dilution problem. The equation used for solving this type of problems is:

[tex]M_1V_1=M_2V_2[/tex]

where, [tex]M_1[/tex] is initial molarity and  [tex]M_2[/tex] is the molarity after dilution. Similarly,  [tex]V_1[/tex] is the volume before dilution and  [tex]V_2[/tex] is the volume after dilution.

Let's plug in the values in the equation:

[tex]1.25M(363mL)=0.50M(V_2)[/tex]

[tex]V_2=\frac{1.25M(363mL)}{0.50M}[/tex]

[tex]V_2=907.5mL[/tex]

Volume of water added = 907.5mL - 363mL  = 544.5 mL

So, 544.5 mL of water are need to be added to the original solution for dilution.

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