$$0.25$$ g of an organic compound containing chlorine gave $$0.40$$ g of silver chloride in Carius estimation. The percentage of chlorine present in the compound is ______ [in nearest integer]
(Given: Molar mass of Ag is $$108$$ g mol$$^{-1}$$ and that of Cl is $$35.5$$ g mol$$^{-1}$$)

We need to find the percentage of chlorine in an organic compound using Carius estimation. The mass of the organic compound is 0.25 g, the mass of AgCl obtained is 0.40 g, and the molar masses are Ag = 108 g/mol and Cl = 35.5 g/mol.

The molar mass of AgCl is calculated as $$M_{AgCl} = M_{Ag} + M_{Cl} = 108 + 35.5 = 143.5 \text{ g/mol}$$, and the moles of AgCl are given by $$\text{Moles of AgCl} = \dfrac{0.40}{143.5} = 2.787 \times 10^{-3} \text{ mol}$$.

Since 1 mole of AgCl contains 1 mole of Cl, the moles of Cl are $$\text{Moles of Cl} = \text{Moles of AgCl} = 2.787 \times 10^{-3} \text{ mol}$$ and its mass is $$\text{Mass of Cl} = 2.787 \times 10^{-3} \times 35.5 = 0.09894 \text{ g}$$.

The percentage of chlorine is $$\% \text{ of Cl} = \dfrac{\text{Mass of Cl}}{\text{Mass of compound}} \times 100 = \dfrac{0.09894}{0.25} \times 100 = 39.58\%$$.

Alternatively, using the direct formula $$\% \text{ of Cl} = \dfrac{\text{Mass of AgCl} \times M_{Cl}}{\text{Mass of compound} \times M_{AgCl}} \times 100$$ we obtain $$\% \text{ of Cl} = \dfrac{0.40 \times 35.5}{0.25 \times 143.5} \times 100 = \dfrac{14.2}{35.875} \times 100 = 39.58\%$$.

Rounding to the nearest integer gives $$\approx 40\%$$.

The percentage of chlorine present in the compound is 40%.