# How many milliliters of 11.5 M HCl(aq) are needed to prepare 305.0 mL of 1.00 M HCl(aq)?

Sep 1, 2016

Approx. $27 \cdot m L$

#### Explanation:

${C}_{1} {V}_{1} = {C}_{2} {V}_{2}$, where ${C}_{i}$ is the concentration of a component in $m o l \cdot {L}^{-} 1$, and it follows that the product ${C}_{i} {V}_{i}$ gives an answer in the number of moles $\left(C V = \text{concentration"(mol*L^-1)xx"volume"(L)="moles}\right)$

And thus $\text{moles}$ $=$ ${C}_{1} {V}_{1}$ $=$ ${C}_{2} {V}_{2}$.

We simply solve for ${V}_{1} = \frac{{C}_{2} {V}_{2}}{C} _ 1$

=(305.0xx10^-3Lxx1.00*cancel(mol)*cancel(L^-1))/(11.5*cancel(mol)*cancel(L^-1) $=$ ??L