During the dilution, the number of the moles of the hydrochloric acid HCl is conserved. This means that the number of the moles of HCl in the needed volume of 12 M solution and the number of the moles of HCl in 250 mL of 3 M solution are equal:
.
The number of the moles of HCl in 250 mL of 3M solution is volume times concentration. The volume must be converted in liters, so L:
mol.
Using the same relation, the volume of 12 M HCl needed is:
L, or 62.5 mL.
Answer: the volume of 12 M HCl of solution needed to prepare 250 mL of 3 M HCl solution is 62.5 mL.
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