Answered

In a school's laboratory, students require [tex]50.0 \, \text{mL}[/tex] of [tex]2.50 \, \text{M} \, \text{H}_2\text{SO}_4[/tex] for an experiment, but the only available stock solution of the acid has a concentration of [tex]18.0 \, \text{M}[/tex]. What volume of the stock solution would they use to make the required solution?

Use [tex]M_i V_i = M_f V_f[/tex].

A. [tex]0.900 \, \text{mL}[/tex]
B. [tex]1.11 \, \text{mL}[/tex]
C. [tex]6.94 \, \text{mL}[/tex]
D. [tex]7.20 \, \text{mL}[/tex]



Answer :

GinnyAnswer
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.