In Carius method for estimation of halogens, 180 mg of an organic compound produced 143.5 mg of AgCl . The percentage composition of chlorine in the compound is _______ \%. (Given : molar mass in $$gmol^{-1} \text{ of } Ag: 108, Cl: 35.5$$)

We need to find the percentage of chlorine in an organic compound using the Carius method.

We start by finding the moles of AgCl.

The molar mass of AgCl is 108 + 35.5 = 143.5 g/mol.

The mass of AgCl is 143.5 mg, which is 0.1435 g.

Using these values, we calculate the moles of AgCl:

$$\frac{0.1435}{143.5} = 0.001$$ mol

Next, we determine the mass of chlorine obtained.

Each mole of AgCl contains one mole of Cl, so the moles of Cl are 0.001 mol.

The mass of Cl is then calculated as:

$$0.001 \times 35.5 = 0.0355$$ g, which is 35.5 mg.

Then we calculate the percentage of chlorine in the compound.

$$\% \text{Cl} = \frac{\text{Mass of Cl}}{\text{Mass of compound}} \times 100$$

Substituting the values gives:

$$= \frac{35.5}{180} \times 100 = 19.72\%$$

$$\approx 20\%$$

Therefore, the percentage of chlorine in the compound is 20%.