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

`28.75\ \text{mL}`

Step-by-step 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=(C_2V_2)/(C_1)\\\Rightarrow V_1=(0.23* 250)/(2)\\\Rightarrow V_1=28.75\ \text{mL}`

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