How do we prepare a 2.0*mol*L^-1 of HCl if 12*mol*L^-1 HCl is available...?

1 Answer
anor277
Mar 30, 2017

Approx. 0.17*L of acid......... And note that the strongest concentration of HCl you can normally get is 10.6*mol*L^-1.

Explanation:

We need a two litre volume of 1.0*mol*L^-1 "hydrochloric acid".

And this a molar quantity of 2*Lxx1.0*mol*L^-1=2.0*mol.

And we thus require a volume of (2*mol)/(12*mol*L^-1)=1/6*L

Just as an aside, this question has no chemical reality. The conc. laboratory reagent is approx. 33% (w/w); and this translates to a molar concentration of approx. 10.6*mol*L^-1.

As always with these dilutions, we ADD ACID TO WATER, AND NEVER THE REVERSE.

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