To solve this problem, we need to determine the volume of the 18.0 M stock solution required to make 50.0 mL of a 2.50 M solution of [tex]\( H_2SO_4 \)[/tex].
We can use the dilution formula, which is:
[tex]\[ M_i \times V_i = M_f \times V_f \][/tex]
Here:
- [tex]\( M_i \)[/tex] is the initial concentration of the stock solution, which is 18.0 M.
- [tex]\( V_i \)[/tex] is the volume of the stock solution we need to find.
- [tex]\( M_f \)[/tex] is the final concentration required, which is 2.50 M.
- [tex]\( V_f \)[/tex] is the final volume required, which is 50.0 mL.
Let's plug these values into the formula and solve for [tex]\( V_i \)[/tex]:
[tex]\[ 18.0 \times V_i = 2.50 \times 50.0 \][/tex]
First, calculate the right-hand side:
[tex]\[ 2.50 \times 50.0 = 125.0 \][/tex]
So the equation becomes:
[tex]\[ 18.0 \times V_i = 125.0 \][/tex]
Now, solve for [tex]\( V_i \)[/tex]:
[tex]\[ V_i = \frac{125.0}{18.0} \][/tex]
[tex]\[ V_i \approx 6.9444 \, \text{mL} \][/tex]
Therefore, the volume of the stock solution required is approximately 6.9444 mL.
Among the given choices:
- 0.900 mL
- 1.11 mL
- 6.94 mL
- 7.20 mL
The closest match to 6.9444 mL is 6.94 mL.
Thus, the volume of the 18.0 M stock solution the students should use to prepare 50.0 mL of 2.50 M [tex]\( H_2SO_4 \)[/tex] is:
6.94 mL.
