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

Mar 30, 2017

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

#### Explanation:

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

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

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

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

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