A chlorine compound with a mass of 0.320 g is fused with sodium…

Step 1. $M(\text{AgCl}) = 143.5\ \text{g/mol}$, so $n = \dfrac{0.410\ \text{g}}{143.5\ \text{g/mol}} = 0.00286\ \text{mol}$.

Step 2. All the chlorine came from the compound: $0.00286\ \text{mol} \times 35.5\ \text{g/mol} = 0.101\ \text{g}$.

Step 3. $\%\,\text{Cl} = \dfrac{0.101\ \text{g}}{0.320\ \text{g}} \times 100 = 31.7\ \%$.