Answer :
To determine the volume of [tex]\(3.00 \, \text{M} \, \text{HCl}\)[/tex] needed to prepare a [tex]\(50.0 \, \text{mL}\)[/tex] sample of [tex]\(1.80 \, \text{M} \, \text{HCl}\)[/tex], we can use the dilution formula:
[tex]\[ M_i V_i = M_f V_f \][/tex]
where:
- [tex]\(M_i\)[/tex] is the initial concentration of the HCl solution.
- [tex]\(V_i\)[/tex] is the initial volume of the HCl solution that we need to find.
- [tex]\(M_f\)[/tex] is the final concentration of the diluted HCl solution.
- [tex]\(V_f\)[/tex] is the final volume of the diluted HCl solution.
Given values:
- [tex]\(M_i = 3.00 \, \text{M}\)[/tex]
- [tex]\(M_f = 1.80 \, \text{M}\)[/tex]
- [tex]\(V_f = 50.0 \, \text{mL}\)[/tex]
Substituting these values into the formula:
[tex]\[ 3.00 \, \text{M} \times V_i = 1.80 \, \text{M} \times 50.0 \, \text{mL} \][/tex]
Solving for [tex]\(V_i\)[/tex]:
[tex]\[ V_i = \frac{1.80 \, \text{M} \times 50.0 \, \text{mL}}{3.00 \, \text{M}} \][/tex]
Calculating the right-hand side:
[tex]\[ V_i = \frac{90.0 \, \text{M} \cdot \text{mL}}{3.00 \, \text{M}} \][/tex]
[tex]\[ V_i = 30.0 \, \text{mL} \][/tex]
Therefore, the volume of [tex]\(3.00 \, \text{M} \, \text{HCl}\)[/tex] used by the student to make the [tex]\(50.0 \, \text{mL}\)[/tex] sample of [tex]\(1.80 \, \text{M} \, \text{HCl}\)[/tex] is [tex]\(\boxed{30.0 \, \text{mL}}\)[/tex].
[tex]\[ M_i V_i = M_f V_f \][/tex]
where:
- [tex]\(M_i\)[/tex] is the initial concentration of the HCl solution.
- [tex]\(V_i\)[/tex] is the initial volume of the HCl solution that we need to find.
- [tex]\(M_f\)[/tex] is the final concentration of the diluted HCl solution.
- [tex]\(V_f\)[/tex] is the final volume of the diluted HCl solution.
Given values:
- [tex]\(M_i = 3.00 \, \text{M}\)[/tex]
- [tex]\(M_f = 1.80 \, \text{M}\)[/tex]
- [tex]\(V_f = 50.0 \, \text{mL}\)[/tex]
Substituting these values into the formula:
[tex]\[ 3.00 \, \text{M} \times V_i = 1.80 \, \text{M} \times 50.0 \, \text{mL} \][/tex]
Solving for [tex]\(V_i\)[/tex]:
[tex]\[ V_i = \frac{1.80 \, \text{M} \times 50.0 \, \text{mL}}{3.00 \, \text{M}} \][/tex]
Calculating the right-hand side:
[tex]\[ V_i = \frac{90.0 \, \text{M} \cdot \text{mL}}{3.00 \, \text{M}} \][/tex]
[tex]\[ V_i = 30.0 \, \text{mL} \][/tex]
Therefore, the volume of [tex]\(3.00 \, \text{M} \, \text{HCl}\)[/tex] used by the student to make the [tex]\(50.0 \, \text{mL}\)[/tex] sample of [tex]\(1.80 \, \text{M} \, \text{HCl}\)[/tex] is [tex]\(\boxed{30.0 \, \text{mL}}\)[/tex].
