# How many moles of chloride ions are in 0.50 g of copper(II) chloride?

0.50 g CuCl₂ × $\left(1 \text{mol CuCl₂")/(134.4"g CuCl₂") × (2"mol Cl⁻")/(1"mol CuCl₂}\right)$ = 0.0074 mol Cl⁻