Question

A 0.500 L solution of 6 M HCl has to be made. How much 12 M HCI is needed?

Answer 1

We use the relationship, `"concentration"="moles of solute"/"volume of solution"` TWICE to get a volume of ..............`0.25*L`

Explanation:

`"concentration"="moles of solute"/"volume of solution"`

And thus `"moles of solute"=6*mol*L^-1xx0.500*L=3*mol` required initially.

Since this comes from a volume of `12*mol*L^-1` `HCl`.....we get

`(3*mol)/(12*mol*L^-1)=0.25/(1/L)=0.25*L`

The concentrated hydrochloric acid used in the lab (`"36%w/w"`) is `10.6*mol*L^-1` so this is another poor question.