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!




Jaroslav already gave you a good hint: Assume full dissociation, which leads to the following equation: $$\ce{CuCl2 <=> Cu^2+ + 2Cl^-}$$


Since you know the initial concentration of copper chloride, $c(\ce{CuCl2})=0.74~\mathrm{mol/L}$, what can you say about the ratio of $n(\ce{CuCl2})/n(\ce{Cu^2+})$ and $n(\ce{CuCl2})/n(\ce{Cl^-})$?


Hint: In one mole of of a salt of the composition $\ce{AB}$, you will find one mole of $\ce{A+}$ and one mole of $\ce{B-}$, according to $\ce{AB <=> A+ + B-}$.


Therefore:



> 
>  $n(\ce{CuCl2})/n(\ce{Cu^2+}) = 1$, thus $c(\ce{CuCl2})=n(\ce{Cu^2+})=0.74~\mathrm{mol/L}$  
> 
>  and  
> 
> $n(\ce{CuCl2})/n(\ce{Cl^-}) =\frac12$, thus $c(\ce{CuCl2})=2\cdot c(\ce{Cl^-})=1.48~\mathrm{mol/L}$
> 
> 
> 


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When a salt is fully dissociated, than the atoms in a solution exist separately:
$$\ce{CuCl2<=>Cu^{2+}}+2\ce{Cl-}$$
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. $$ 10~\mathrm{g}- 4.7~\mathrm{g} = 5.3~\mathrm{g}$$ 


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