To address this question, we need to use the dilution formula, which is given by:
[tex]\[ M_j V_j = M_f V_f \][/tex]
where:
- \( M_j \) is the initial concentration (or molarity) of the stock solution,
- \( V_j \) is the initial volume of the stock solution required,
- \( M_f \) is the final concentration (or molarity) of the desired solution,
- \( V_f \) is the final volume of the desired solution.
Given values:
- \( M_f = 2.50 \, M \) (final concentration)
- \( V_f = 50.0 \, mL \) (final volume)
- \( M_j = 18.0 \, M \) (initial concentration of the stock solution)
We need to find \( V_j \), the volume of the stock solution required.
Starting with the given formula:
[tex]\[ M_j V_j = M_f V_f \][/tex]
Plug in the known values:
[tex]\[ 18.0 \, M \times V_j = 2.50 \, M \times 50.0 \, mL \][/tex]
First, calculate the product on the right-hand side:
[tex]\[ 2.50 \, M \times 50.0 \, mL = 125.0 \, M \cdot mL \][/tex]
Now, solve for \( V_j \):
[tex]\[ V_j = \frac{125.0 \, M \cdot mL}{18.0 \, M} \][/tex]
[tex]\[ V_j \approx 6.944444444444445 \, mL \][/tex]
Thus, the volume of the stock solution required is approximately \( 6.94 \, mL \).
Among the options given, the best match for our calculation is:
[tex]\[ \boxed{6.94 \, mL} \][/tex]
