Answer :
Answer:
5 mL
Explanation:
As this is a problem regarding dilutions, we can solve it using the following formula:
- C₁V₁=C₂V₂
Where subscript 1 refers to the initial concentration and volume, while 2 refers to the final C and V. Meaning that in this case:
- C₁ = 5.0 M
- V₁ = ?
- C₂ = 0.25 M
- V₂ = 100 mL
We input the data:
- 5.0 M * V₁ = 0.25 M * 100 mL
And solve for V₁:
- V₁ = 5 mL
Answer:
5 mL of 5.0 M H₂SO₄ (aq) are needed to prepare 100 mL of 0.25 M H₂SO₄ (aq).
Explanation:
In chemistry, dilution is the reduction of the concentration of a chemical in a solution.
Then, dilution consists of preparing a less concentrated solution from a more concentrated one, and it consists simply by adding more solvent to the same amount of solute. That is, the amount or mass of the solute is not changed, but the volume of the solvent varies: as more solvent is added, the concentration of the solute decreases, since the volume (and weight) of the solution increases.
A dilution is calculated by the expression:
Ci*Vi = Cf*Vf
where:
- Ci: initial concentration
- Vi: initial volume
- Cf: final concentration
- Vf: final volume
In this case, you know:
- Ci=5 M
- Vi= ?
- Cf= 0.25 M
- Vf= 100 mL
Replacing:
5 M*Vi = 0.25 M* 100 mL
Solving:
[tex]Vi= \frac{0.25 M*100 mL}{5 M}[/tex]
Vi= 5 mL
5 mL of 5.0 M H₂SO₄ (aq) are needed to prepare 100 mL of 0.25 M H₂SO₄ (aq).