A stock solution of magnesium chloride has a concentration of 120 mg mL. How many milliliters of the stock solution are required to prepare 1.5 L of 25 mg mL solution

312.5 mL of the stock solution are required to prepare 1.5 L of 25 mg mL solution.

First of all, you have to know that when it is desired to prepare a less concentrated solution from a more concentrated one, it is called dilution, as in this case.

In this problem you get that 312.5 mL of the stock solution are required to prepare 1.5 L of 25 mg mL solution.

In first place, you have to know that dilution is the process of reducing the concentration of solute in solution, which is accomplished by simply adding more solvent to the solution at the same amount of solute.

In a dilution the amount of solute does not change, but as more solvent is added, the concentration of the solute decreases, as the volume (and weight) of the solution increases.

A dilution is mathematically expressed as:

Ci×Vi = Cf×Vf

where

In this case, you know:

Replacing in the definition of dilution:

120 × Vi= 25

Solving:

Vi= 312.5 mL

In summary, 312.5 mL of the stock solution are required to prepare 1.5 L of 25 mg mL solution.

Learn more about dilution:

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