I think I managed it, would be great if someone could check and see If this makes logic.
Side note before I write down what I calculated: The Ammonia Volume and Water I want to keep constant, because the purpose is to do a calibration curve with a UV-VIS. For Ammonia I calculated I would need 8,33mL of NH 30% to have 2,5mL of Ammonia. For water I chose 20 mL.
I started with fixating the concentration of Cu(NO3)2, since I want to measure the Cu in further solutions. Here I chose 0,005M.
With this I calculated the needed mass of Cu(NO3)2 to achieve this concentration.
With the obtained mass of Cu(NO3)2, I calculated the amount of mols of Cu(NO3)2.
With the amount of mols I can calculate the amount of Molecules of Water in my solute. Since my Copper(II) nitrate is trihydrated I will have 3 mols of H2O for every mol of Cu(NO3)2. And with the number of mols I can obtain the mass and Volume of Water.
VCu,H2O= mH2O / ρH2O I assumed the density (ρ) as 0,997 g/mL
Adding the mCu(NO3)2 and mH2O I'll have the amount of mass Cu(NO3)2 · 3H2O (mCu(NO3)2 · 3H2O) I have to weight.
mCu(NO3)2 · 3H2O = mCu(NO3)2 + mH2O
Now to the Volume of Ammonia and Water. To obtain the total of Water I have to add to my system, I first calculated the amount of Water that is add when I want 2,5 mL of Ammonia.
VNH3,H2O: Volume of Water contained in the Volume of NH3 30%
VNH3,H2O = (2,5×100)/30 - 2,5
The total amount of Water I have to add to have constant 20 mL is:
VT = VH2O - VCu,H2O - VNH3,H2O The value of VH2O is 20mL
I got to a Value of:
mCu(NO3)2 · 3H2O = 0,00302 g
VT = 14,16598 mL
Thanks again for any help.