Respuesta :

yahyathejeneral
To solve this problem, we shall use the role;
M1V1 = M2V2
M1: is the molarity of the solution before dilution.
V1: is the volume of the solution that we need to dilute.
M2: is the molarity of the solution after dilution.
V2: is the volume of the solution after dilution.

By substitution;
12*V1 = 3*5
V1 = 1.25 litters.

The volume of 12.6 M HCl that must be added to enough water to prepare 5.00 litters of 3.00 M HCl is; 1.19 L

Finding Volume of a Mixture

We are given;

Initial volume of HCl; V₁ = ?

Initial concentration of HCl; M₁ = 12.6 M

Final volume of HCl; V₂ = 5 L

Final concentration; M₂ = 3 M

To solve this, we will use one of the popular formulas in titration which is;

V₁ × M₁ = V₂ × M₂

Making V₁ the subject gives;

V₁ = (V₂ × M₂)/M₁

V₁ = (5 × 3)/12.6

V₁ = 1.19L

Thus, Volume of 12.6 M HCl to be added to water is 1.19 liters

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