[tex]4.4\; {\rm g}[/tex].
Explanation
Refer to a modern periodic table for relative atomic mass:
Formula of copper (II) chloride: CuCl₂.
Formula mass of copper (II) chloride:
63.546 + 2 × 35.45 = [tex]134.446\; {\rm g \cdot mol^{-1}}[/tex].
How many [tex]{\rm mol}[/tex] of formula units in 8.3 grams of CuCl₂?
[tex](8.3\; {\rm g}) / (134.446\; {\rm g \cdot mol^{-1}}) \approx 0.0617\; {\rm mol}[/tex]
Note the subscript "2" in the formula CuCl₂. There are [tex]2\; {\rm mol}[/tex] of chloride ions in every [tex]1\; {\rm mol}[/tex] formula unit of CuCl₂.
There would be [tex]2 \times 0.0617\; {\rm mol} \approx 0.123\; {\rm mol}[/tex] of [tex]{\rm Cl^{-}}[/tex] ions in [tex]0.0617\; {\rm mol}[/tex] formula units of CuCl₂.
The mass of [tex]1\; {\rm mol}[/tex] of chloride ion is 35.45 grams.
The mass of [tex]0.123\; {\rm mol}[/tex] of [tex]{\rm Cl^{-}}[/tex] ions will be [tex]35.45 \times 0.123 \approx 4.4\; {\rm g}[/tex].
Round the final value to two sig. fig. since the mass in the question is also 2 sig. fig.
