CuCl2 dissolved in water We dissolve $10\ \mathrm{g}$ of $\ce{CuCl2}$ in $0.1\ \mathrm{L}$ of water.
Given that:

$M(\ce{Cl})=35.5\ \mathrm{g/mol}$
$M(\ce{Cu})=63.5\ \mathrm{g/mol}$

How can one calculate the molar concentration of $\ce{Cu^2+}$ and $\ce{Cl-}$?

First of all I calculated the the quantity of matter of $\ce{CuCl2}$:

$n(\ce{CuCl2})=0.074\ \mathrm{mol}$

The the molar concentration:

$c=\frac{0.074\ \mathrm{mol}}{0.1\ \mathrm{L}}=0.74\ \mathrm{mol/L}$

Then I don’t know how can I calculate the molar concentration of $\ce{Cu^2+}$ and $\ce{Cl-}$ because I don’t know the quantity of matter of them!

But here im blocked!
When a salt is fully dissociated, than the atoms in a solution exist separately:
$$\ce{CuCl2&lt;=&gt;Cu^{2+}}+2\ce{Cl-}$$
Amount of substance is 0.074 mol. The molecule is made of 1 atom of Cu and 2 atoms of Cl. 

$$ n=\frac{m}{M}$$
$$ n \cdot M =m$$
Amount of copper in the mixture then is: $$0.074~\mathrm{mol} \cdot 63.5~\mathrm{g/mol} = 4.7~\mathrm{g} $$ and the rest is formed by chlorine. $$ 10~\mathrm{g}- 4.7~\mathrm{g} = 5.3~\mathrm{g}$$