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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