To solve the problem of diluting a 1.0 mL sample of a 2.0 M solution to obtain a final concentration of 0.0080 M, we use the formula for dilution:
[tex]\[ M_1 \times V_1 = M_2 \times V_2 \][/tex]
Where:
- [tex]\( M_1 \)[/tex] is the initial molarity (2.0 M)
- [tex]\( V_1 \)[/tex] is the initial volume (1.0 mL)
- [tex]\( M_2 \)[/tex] is the final molarity (0.0080 M)
- [tex]\( V_2 \)[/tex] is the final volume (in mL), which we need to find
We can rearrange the formula to solve for [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = \frac{M_1 \times V_1}{M_2} \][/tex]
Now, plug in the given values:
[tex]\[ V_2 = \frac{2.0 \, \text{M} \times 1.0 \, \text{mL}}{0.0080 \, \text{M}} \][/tex]
Performing the calculation:
[tex]\[ V_2 = \frac{2.0 \, \text{M} \cdot 1.0 \, \text{mL}}{0.0080 \, \text{M}} \][/tex]
[tex]\[ V_2 = \frac{2.0 \, \text{mL}}{0.0080} \][/tex]
[tex]\[ V_2 = 250.0 \, \text{mL} \][/tex]
Therefore, the final volume [tex]\( V_2 \)[/tex] required to achieve a concentration of 0.0080 M is 250.0 mL.
So, the correct answer is:
[tex]\[ \boxed{250 \, \text{mL}} \][/tex]
