To determine the volume of a [tex]$1.00 \, \text{M} \, \text{Fe}\left(\text{NO}_3\right)_3$[/tex] solution that can be diluted to prepare [tex]$1.00 \, \text{L}$[/tex] of a solution with a concentration of [tex]$0.250 \, \text{M}$[/tex], we can use the dilution formula:
[tex]\[ C_1 V_1 = C_2 V_2 \][/tex]
where:
- \( C_1 \) is the initial concentration ([tex]$1.00 \, \text{M}$[/tex]),
- \( V_1 \) is the initial volume (which we need to find),
- \( C_2 \) is the final concentration ([tex]$0.250 \, \text{M}$[/tex]),
- \( V_2 \) is the final volume ([tex]$1.00 \, \text{L}$[/tex]).
To find \( V_1 \), we rearrange the dilution formula to solve for \( V_1 \):
[tex]\[ V_1 = \frac{C_2 V_2}{C_1} \][/tex]
Now, substitute the given values into the equation:
[tex]\[ V_1 = \frac{(0.250 \, \text{M}) \times (1.00 \, \text{L})}{1.00 \, \text{M}} \][/tex]
[tex]\[ V_1 = \frac{0.250 \, \text{M}}{1.00 \, \text{M}} \times 1.00 \, \text{L} \][/tex]
[tex]\[ V_1 = 0.250 \, \text{L} \][/tex]
Thus, you need \(0.25 \, \text{L}\) (or 250 mL) of the [tex]$1.00 \, \text{M} \, \text{Fe}\left(\text{NO}_3\right)_3$[/tex] solution to prepare [tex]$1.00 \, \text{L}$[/tex] of a solution with a concentration of [tex]$0.250 \, \text{M}$[/tex].
To summarize, the initial volume [tex]\( V_1 \)[/tex] is [tex]\(0.25 \, \text{L}\)[/tex], so the required volume of [tex]$1.00 \, \text{M} \, \text{Fe}\left(\text{NO}_3\right)_3$[/tex] solution is [tex]\(0.25 \, \text{L}\)[/tex] to get a final volume [tex]\( V_2 \)[/tex] of [tex]\(1.00 \, \text{L}\)[/tex] with a concentration [tex]\( C_2 \)[/tex] of [tex]\(0.250 \, \text{M}\)[/tex].
