# A 0.500 L solution of 6 M HCl has to be made. How much 12 M HCI is needed?

May 20, 2017

We use the relationship, $\text{concentration"="moles of solute"/"volume of solution}$ TWICE to get a volume of ..............$0.25 \cdot L$

#### Explanation:

$\text{concentration"="moles of solute"/"volume of solution}$

And thus $\text{moles of solute} = 6 \cdot m o l \cdot {L}^{-} 1 \times 0.500 \cdot L = 3 \cdot m o l$ required initially.

Since this comes from a volume of $12 \cdot m o l \cdot {L}^{-} 1$ $H C l$.....we get

$\frac{3 \cdot m o l}{12 \cdot m o l \cdot {L}^{-} 1} = \frac{0.25}{\frac{1}{L}} = 0.25 \cdot L$

The concentrated hydrochloric acid used in the lab ($\text{36%w/w}$) is $10.6 \cdot m o l \cdot {L}^{-} 1$ so this is another poor question.