Question

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.

Answer 1

Answer:

$28.75\ \text{mL}$

Explanation:

$C_1$ = Initial concentration = 2 M

$C_2$ = Final concentration = 0.23 M

$V_1$ = Initial volume

$V_2$ = Final volume = 250 mL

We have the relation

$C_1V_1=C_2V_2\\\Rightarrow V_1=\dfrac{C_2V_2}{C_1}\\\Rightarrow V_1=\dfrac{0.23\times 250}{2}\\\Rightarrow V_1=28.75\ \text{mL}$

So, the required volume is $28.75\ \text{mL}$.