# How do we prepare 2*mol*L^-1 HCl from 16*mol*L^-1 HCl?

Sep 2, 2017

We use....approx. $\frac{1}{8} \cdot L$ of the stuff........

#### Explanation:

We use the definition of concentration......

$\text{Concentration"="Moles of solute"/"Volume of solution}$

Now we want $500 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 4 \cdot m o l \cdot {L}^{-} 1$. Agreed?

And thus here the moles of solute ...........$500 \cdot \cancel{m L} \times {10}^{-} 3 \cdot L \cdot \cancel{m {L}^{-} 1} \times 4 \cdot m o l \cdot \cancel{{L}^{-} 1} = 2 \cdot m o l$.

We have $16 \cdot m o l \cdot {L}^{-} 1$ $H C l$ available.....and so from the equation above....

$\text{Volume"="Mole of solute"/"Concentration} = \frac{2 \cdot m o l}{16 \cdot m o l \cdot {L}^{-} 1} = \frac{1}{8} \cdot L$

$= 0.125 \cdot L$

Just to add, that this question is chemically unrealistic. The strongest hydrochloric acid you can buy commercially is approx. $\text{34%(w/w)}$; this has a molar concentration of $12 \cdot m o l \cdot {L}^{-} 1$. And practically, when you mix acid and water, you ALWAYS add acid to water.