Question

When 0.50 liter of 12 M solution is diluted to 1.0 liter, what is the molarity of the new solution?

Answer 1

The new concentration is HALF that of the original.

Explanation:

`"Concentration (C)"` `=` `"moles of solute (n)"/"volume of solution (V)"`.

Since `C=n/V`, `n=CV`.

And thus `n_"initial"` `=` `0.50*cancelLxx12*mol*cancel(L^-1)=6*mol`

But `V_"final"` `=` `1.0*L`.

So `"concentration"`

`=` `(0.50*cancelLxx12*mol*cancel(L^-1))/(1*L)=??mol*L^-1`

You use the relationships, `C=n/V`, `V=n/C`, and `n=CV` continually in a laboratory.

Concentrated hydrochloric acid is supplied as a `10*mol*L^-1` solution in water. If I have a `2.5*L` bottle of conc. acid, how many litres of `1.0*mol*L^-1` can I prepare?

IMPORTANT: WE WOULD ALWAYS ADD CONC ACID TO WATER AND NEVER THE REVERSE!!