An organic compound is subjected to chlorination to get compound A using 5.0 g of chlorine. When 0.5 g of compound A is reacted with AgNO$$_3$$ [Carius Method], the percentage of chlorine in compound A is when it forms 0.3849 g of AgCl. (Round off to the Nearest Integer)
(Atomic masses of Ag and Cl are 107.87 and 35.5 respectively)

We are given that $$0.5\ \text{g}$$ of compound A produces $$0.3849\ \text{g}$$ of silver chloride ($$\text{AgCl}$$) when treated with $$\text{AgNO}_3$$ in the Carius method. Our task is to calculate the percentage of chlorine present in compound A.

First, we note the atomic masses supplied in the question:

$$\text{Atomic mass of Ag}=107.87,\qquad \text{Atomic mass of Cl}=35.5$$

So, the molar mass of silver chloride is obtained by simple addition:

$$M_{\text{AgCl}} = M_{\text{Ag}} + M_{\text{Cl}} = 107.87 + 35.5 = 143.37\ \text{g mol}^{-1}$$

Next, we determine the number of moles of $$\text{AgCl}$$ actually formed. The basic formula for moles is

$$\text{moles} = \dfrac{\text{given mass}}{\text{molar mass}}$$

Substituting the values, we get

$$n_{\text{AgCl}} = \dfrac{0.3849}{143.37} = 0.002685\ \text{mol}\ (\text{approximately}).$$

In every mole of $$\text{AgCl}$$ there is exactly one mole of chlorine. Hence, the moles of chlorine present are the same as the moles of $$\text{AgCl}$$ formed:

$$n_{\text{Cl}} = n_{\text{AgCl}} = 0.002685\ \text{mol}.$$

We now convert these moles of chlorine to mass using the relation

$$\text{mass} = \text{moles} \times \text{atomic mass}.$$

Thus,

$$m_{\text{Cl}} = 0.002685 \times 35.5 = 0.0953\ \text{g}\ (\text{to four significant figures}).$$

Finally, the percentage of chlorine in compound A is calculated by the standard definition

$$\%\text{Cl} = \dfrac{\text{mass of Cl in the sample}}{\text{mass of the sample}} \times 100.$$

Substituting our values,

$$\%\text{Cl} = \dfrac{0.0953}{0.5} \times 100 = 19.06\%.$$

Rounding off to the nearest integer as instructed, we obtain $$19\%.$

Hence, the correct answer is Option 19.