To solve this we use the equation,
M1V1 = M2V2
where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.
2 M x V1 = 0.1 M x .5 L
V1 = 0.025 L or 25 mL of the 2 M KCl solution is needed
To prepare 500. ml of a 0.100 M KCl solution, 25 ml of a 2.00 M KCl solution is required.
For making the dilutions, the following equation can be used:
[tex]\rm M_1V_1\;=\;M_2V_2[/tex]
Where, M1 and V1 are the molarity and the volume of the solution to be diluted. M2 and V2 are the molarity and volume of the final solution.
Here, M1 = 2.0 M KCl
M2 = 0.1 M KCl
V2 = 500 ml
So, the volume of 2.0 M KCl required will be V2 :
V2 = [tex]\rm \dfrac{M_2V_2}{M_1}[/tex]
[tex]\rm V_2\;=\;\dfrac{0.1\;\times\;500}{2}\;ml[/tex]
[tex]\rm V_2[/tex] = 25 ml.
25 ml of a 2.00 M KCl solution is required to prepare 500. ml of a 0.100 M KCl solution.
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