Question

15. What volume of water must be added to 300 mL of 0.75 M HCl to dilute the solution to
0.25 M?

Answer 1

Known :

V1 = 300 mL

M1 = 0.75 M

M2 = 0.25 M

Solution :

M1 • V1 = M2 • V2

(0.75 M) • (300 mL) = (0.25 M) V2

V2 = 900 mL

Water add to this solution is :

∆V = V2 - V1

∆V = 900 - 300

∆V = 600 mL

Answer 2

The answer is 600 ml of water needs to be added to change the concentration to 0.25 M.

What is Dilution ?

A dilution is when you have a solution of a certain concentration and you add more solvent to decrease the concentration.

If you are adding more solvent, the volume of the whole solution is going to increase as the concentration of the solution decreases.

You can solve for the concentration or volume of the concentrated or dilute solution using the equation:

M₁V₁ = M₂V₂,

where M₁ is the concentration in molarity (moles/Liters) of the concentrated solution,

V₁ is the volume of the concentrated solution,

M₂ is the concentration in molarity of the dilute solution (after more solvent has been added), and

V₂ is the volume of the dilute solution.

It is given that

V₁ = 300 ml

V₂ = ?

M₁= 0.75 M

M₂ = 0.25 M

Substituting the values in the equation

0.75 * 300 = 0.25 * V₂

V₂= 900 ml

So 900 - 300 , 600 ml of water needs to be added to change the concentration to 0.25 M.

To know more about dilution

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