To solve this problem, we need to use the dilution formula: [tex]\[ M_i \cdot V_i = M_f \cdot V_f \][/tex]
Where:
- [tex]\( M_i \)[/tex] is the initial concentration of the stock solution,
- [tex]\( V_i \)[/tex] is the volume of the stock solution needed,
- [tex]\( M_f \)[/tex] is the final concentration of the diluted solution,
- [tex]\( V_f \)[/tex] is the final volume of the diluted solution.
Given:
- [tex]\( M_f = 2.50 \, M \)[/tex]
- [tex]\( V_f = 50.0 \, mL \)[/tex]
- [tex]\( M_i = 18.0 \, M \)[/tex]
We are solving for [tex]\( V_i \)[/tex], the volume of the stock solution needed.
First, substitute the given values into the dilution formula:
[tex]\[ 2.50 \, M \cdot V_i = 18.0 \, M \cdot 50.0 \, mL \][/tex]
Next, solve for [tex]\( V_i \)[/tex] by isolating it on one side of the equation:
[tex]\[ V_i = \frac{18.0 \, M \cdot 50.0 \, mL}{2.50 \, M} \][/tex]
Now perform the division:
[tex]\[ V_i = \frac{900.0 \, M \cdot mL}{2.50 \, M} \][/tex]
[tex]\[ V_i = 360.0 \, mL / 2.50 \][/tex]
[tex]\[ V_i = 6.94 \, mL \][/tex]
Therefore, the volume of the 18.0 M stock solution needed to prepare 50.0 mL of a 2.50 M [tex]\(\text{H}_2\text{SO}_4\)[/tex] solution is [tex]\( 6.94 \, mL \)[/tex].
Thus, the correct answer is:
6.94 mL
