Answer to Question #141276 in General Chemistry for yara

Question #141276
A chemist need 250mL of 3 mol/L hydrochloric acid. What volume of 12mol/L solution would be needed to be diluted to get the needed solution?
1
Expert's answer
2020-11-02T08:49:45-0500

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 n12Mn_{12M} in the needed volume of 12 M solution and the number of the moles of HCl n3Mn_{3M} in 250 mL of 3 M solution are equal:

n12M=n3Mn_{12M} = n_{3M} .

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 V=250103V = 250·10^{-3} L:

n=cV=3250103=0.75n=cV = 3·250·10^{-3} = 0.75 mol.

Using the same relation, the volume of 12 M HCl needed is:

V12M=nc=0.7512=0.0625V_{12M} = \frac{n}{c} = \frac{0.75}{12} = 0.0625 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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