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In a school's laboratory, students require [tex]50.0 \, \text{mL}[/tex] of [tex]2.50 \, \text{M} \, H_2SO_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_j V_j = 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 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]

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