To find the percentage composition of chlorine in an organic compound using the Carius method, we follow these steps:

Calculate Millimoles of AgCl:

Since 143.5 mg of AgCl is produced, and the molar mass of AgCl is 143.5 g/mol, the millimoles of AgCl is:

$ \text{Millimoles of AgCl} = \frac{143.5 \, \text{mg}}{143.5 \, \text{mg/mmol}} = 1 \, \text{mmol} $

Determine Millimoles of Cl:

In AgCl, there is a 1:1 ratio of Ag to Cl, so the millimoles of Cl are equal to that of AgCl:

$ \text{Millimoles of Cl} = 1 \, \text{mmol} $

Calculate Mass of Cl:

Using the molar mass of Cl (35.5 g/mol), the mass of Cl is:

$ \text{Mass of Cl} = 35.5 \times 10^{-3} \, \text{g} = 35.5 \, \text{mg} $

Compute Percentage by Mass of Cl:

The total mass of the organic compound is 180 mg, so the percentage of chlorine is:

$ \% \text{ Cl by mass} = \left( \frac{35.5 \, \text{mg}}{180 \, \text{mg}} \right) \times 100 = 19.72\% $

Round to the Nearest Integer:

Rounding 19.72% to the nearest integer gives 20%.

Thus, the percentage composition of chlorine in the compound is approximately 20%.