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A stock solution of Al(CH3COO)3 is available to prepare solutions that are more dilute. Calculate the volume, in mL, of a 2.0-M solution of Al(CH3COO)3 required to prepare exactly 250 mL of a 0.230-M solution of Al(CH3COO)3.

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Answer:

[tex]28.75\ \text{mL}[/tex]

Explanation:

[tex]C_1[/tex] = Initial concentration = 2 M

[tex]C_2[/tex] = Final concentration = 0.23 M

[tex]V_1[/tex] = Initial volume

[tex]V_2[/tex] = Final volume = 250 mL

We have the relation

[tex]C_1V_1=C_2V_2\\\Rightarrow V_1=\dfrac{C_2V_2}{C_1}\\\Rightarrow V_1=\dfrac{0.23\times 250}{2}\\\Rightarrow V_1=28.75\ \text{mL}[/tex]

So, the required volume is [tex]28.75\ \text{mL}[/tex].