Question

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

Answer 1

Approx. `27*mL`

Explanation:

`C_1V_1=C_2V_2`, where `C_i` is the concentration of a component in `mol*L^-1`, and it follows that the product `C_iV_i` gives an answer in the number of moles `(CV="concentration"(mol*L^-1)xx"volume"(L)="moles")`

And thus `"moles"` `=` `C_1V_1` `=` `C_2V_2`.

We simply solve for `V_1=(C_2V_2)/C_1`

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