How can you find the concentration of copper(II) ions in a solution of 0.5 mole copper(II) chloride and 0.5 mole of copper(II) sulfate? The question I'm trying to answer is below and the answer is C. 


I tried finding the total moles of copper(II) ions first by using mole ratios for copper(II) sulfate and copper(II) chloride, obtaining an answer of 0.42 moles. Then, I used $n=cv$ and obtained the concentration, using the volume in the question in litres, of 0.84, which is incorrect. Now, I'm sure my method incorrect but I can't think of another way to this.
1 mole of copper(II) sulphate contains 1 mole of copper ($\ce{Cu^{2+}}$) and one mole of sulphate ($\ce{SO_4^{2-}}$). Or, if you want, 1 mole of copper, 1 mole sulphur, and 4 moles of oxygen.

This is by the definition of the mole as the unit of measurement for the amount of substance.

From this it follows that 0.5 moles of copper(II) sulphate contain 0.5 moles of copper. Similarly 0.5 moles of copper(II) chloride also contain 0.5 moles of copper. Together these are 1 mole of copper.

1 mole of copper ($\ce{Cu^{2+}}$) in $\mathrm{0.5\,l}$ water gives us a concentration of $\mathrm{2\, mol\,l^{-1}}$