How do we prepare a $2.0 \cdot m o l \cdot L^{- 1}$ $2.0molL^-1$ of $H C l$ $HCl$ if $12 \cdot m o l \cdot L^{- 1}$ $12molL^-1$ $H C l$ $HCl$ is available...?

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How do we prepare a $2.0 \cdot m o l \cdot L^{- 1}$ $2.0molL^-1$ of $H C l$ $HCl$ if $12 \cdot m o l \cdot L^{- 1}$ $12molL^-1$ $H C l$ $HCl$ is available...?

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Answer 1 · 2017-03-30T18:35:00

Approx. $0.17 \cdot L$ $0.17*L$ of acid......... And note that the strongest concentration of $H C l$ $HCl$ you can normally get is $10.6 \cdot m o l \cdot L^{- 1}$ $10.6*mol*L^-1$ .

Explanation:

We need a two litre volume of $1.0 \cdot m o l \cdot L^{- 1}$ $1.0*mol*L^-1$ $hydrochloric acid$ $"hydrochloric acid"$ .

And this a molar quantity of $2 \cdot L \times 1.0 \cdot m o l \cdot L^{- 1} = 2.0 \cdot m o l$ $2*Lxx1.0*mol*L^-1=2.0*mol$ .

And we thus require a volume of $\frac{2 \cdot m o l}{12 \cdot m o l \cdot L^{- 1}} = \frac{1}{6} \cdot L$ $(2*mol)/(12*mol*L^-1)=1/6*L$

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. $10.6 \cdot m o l \cdot L^{- 1}$ $10.6*mol*L^-1$ .

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

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How do we prepare a $2.0 \cdot m o l \cdot L^{- 1}$ $2.0molL^-1$ of $H C l$ $HCl$ if $12 \cdot m o l \cdot L^{- 1}$ $12molL^-1$ $H C l$ $HCl$ is available...?

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How do we prepare a $2.0 \cdot m o l \cdot L^{- 1}$ $2.0molL^-1$ of $H C l$ $HCl$ if $12 \cdot m o l \cdot L^{- 1}$ $12molL^-1$ $H C l$ $HCl$ is available...?