1. Calculate the moles of Cl- from NaCl:

* 10 mM NaCl = 10 mmol/L NaCl

* 1 mL = 0.001 L

* Moles of NaCl = (10 mmol/L) * (0.001 L) = 0.01 mmol NaCl

* Since NaCl dissociates into one Na+ and one Cl-, there are also 0.01 mmol of Cl- from the NaCl solution.

2. Calculate the moles of Cl- from CaCl2:

* 0.05 M CaCl2 = 0.05 mol/L CaCl2

* 4 mL = 0.004 L

* Moles of CaCl2 = (0.05 mol/L) * (0.004 L) = 0.0002 mol CaCl2

* Since CaCl2 dissociates into one Ca2+ and two Cl-, there are (0.0002 mol CaCl2) * (2 mol Cl-/mol CaCl2) = 0.0004 mol Cl- from the CaCl2 solution.

3. Calculate the total moles of Cl-:

* Total moles of Cl- = 0.01 mmol + 0.0004 mol = 0.0104 mol Cl-

4. Calculate the final concentration of Cl-:

* Final volume = 1 mL + 4 mL = 5 mL = 0.005 L

* Final concentration of Cl- = (0.0104 mol) / (0.005 L) = 2.08 M

Therefore, the final concentration of Cl- is 2.08 M.