# What volume of a solution that is 6.0*mol*L^-1 HCl is required to make 1.00*L of a 0.10*mol*L^-1 solution?

Jun 1, 2017

We need under $20 \cdot m L$ of stock solution.

#### Explanation:

We use the old relationship, ${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$. The product in each side gives units of $\text{moles}$.

And so $6.0 \cdot m o l \cdot {L}^{-} 1 \times {V}_{1} = 1.000 \cdot L \times 0.1 \cdot m o l \cdot {L}^{-} 1$

and thus..............................

${V}_{1} = \frac{1.000 \cdot L \times 0.1 \cdot m o l \cdot {L}^{-} 1}{6.0 \cdot m o l \cdot {L}^{-} 1} = 0.0167 \cdot L = 16.7 \cdot m L .$

Do you add the water to the acid, or the acid to water? (I am not taking the p, the order of addition is important!)