# How many mL of 10.0 M HCI are needed to prepare 500. mL of 2.00 M HCl?

Jan 30, 2017

Approx. $100 \cdot m L$

#### Explanation:

The product, $\text{concentration"xx"volume} = m o l \cdot {L}^{-} 1 \times L$, clearly has units of $m o l$, i.e. $m o l \cdot \cancel{{L}^{-} 1} \times \cancel{L} = m o l$.

And thus ${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, and given the equality we can use units of $m L$ for volume.

${V}_{1} = \frac{{C}_{2} {V}_{2}}{C} _ 1$ $=$ $\frac{500 \cdot m L \times 2.00 \cdot m o l \cdot {L}^{-} 1}{10.0 \cdot m o l \cdot {L}^{-} 1}$

=??mL

AND YOU ADD THE CONCENTRATED ACID TO THE WATER NEVER THE REVERSE. WHY NOT?