Question

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

Answer 1

Approx. `100*mL`

Explanation:

The product, `"concentration"xx"volume"=mol*L^-1xxL`, clearly has units of `mol`, i.e. `mol*cancel(L^-1)xxcancelL=mol`.

And thus `C_1V_1=C_2V_2`, and given the equality we can use units of `mL` for volume.

`V_1=(C_2V_2)/C_1` `=` `(500*mLxx2.00*mol*L^-1)/(10.0*mol*L^-1)`

`=??mL`

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