In Carius method 0.2425 g of an organic compound gave 0.5253 g silver chloride.
The percentage of chlorine in the organic compound is

The Carius method involves converting the chlorine in an organic compound to silver chloride (AgCl). The mass of chlorine in the AgCl precipitate comes from the organic compound.

Given: - Mass of organic compound = 0.2425 g - Mass of AgCl obtained = 0.5253 g

The molar mass of AgCl is calculated using atomic masses: Ag = 108 g/mol, Cl = 35.5 g/mol. Thus, molar mass of AgCl = 108 + 35.5 = 143.5 g/mol.

In 143.5 g of AgCl, the mass of chlorine is 35.5 g. Therefore, the mass of chlorine in 0.5253 g of AgCl is found using proportion:

Mass of chlorine = $$\left( \frac{35.5}{143.5} \right) \times 0.5253$$

First, compute the ratio: $$\frac{35.5}{143.5} = \frac{355}{1435} = \frac{71}{287} \approx 0.2473867595818815$$

Then, multiply by the mass of AgCl: $$0.2473867595818815 \times 0.5253 = 0.1299522648083624 \text{ g}$$

This mass of chlorine comes from 0.2425 g of the organic compound. The percentage of chlorine is:

Percentage of chlorine = $$\left( \frac{\text{mass of chlorine}}{\text{mass of compound}} \right) \times 100 = \left( \frac{0.1299522648083624}{0.2425} \right) \times 100$$

Compute the division: $$\frac{0.1299522648083624}{0.2425} = 0.5358762886597938$$

Multiply by 100: $$0.5358762886597938 \times 100 = 53.58762886597938\%$$

Rounding to two decimal places gives 53.59%. However, comparing with the options, 53.58% is the closest match.

The options are: - A. 37.57% - B. 53.58% - C. 87.65% - D. 34.79%

Thus, the correct answer is option B, 53.58%.