0.25 g of an organic compount containing chlorine gave 0.40 g of silver chloride in Carius estimation . The percentage of chlorine present in the compount is ___________ . [in nearest integer]
(Given Molar mass of Ag is 108 `g "mol"^(-1)` and that of Cl is 35.5 g `"mol"^(-1)`)

Explanation

To find the percentage of chlorine present in the organic compound, we can follow these steps:

### Step 1: Calculate the mass of chlorine in silver chloride (AgCl)

Given:
- Mass of AgCl = 0.40 g
- Molar mass of AgCl = Molar mass of Ag (108 g/mol) + Molar mass of Cl (35.5 g/mol) = 143.5 g/mol

Now, we can find the mass of chlorine in 0.40 g of AgCl using the ratio of the mass of chlorine in AgCl to the total mass of AgCl.

\[ \text{Mass of Cl in AgCl} = \left(\frac{\text{Mass of Cl}}{\text{Molar mass of AgCl}}\right) \times \text{Mass of AgCl} \]

\[ \text{Mass of Cl in AgCl} = \left(\frac{35.5 \, \text{g}}{143.5 \, \text{g}}\right) \times 0.40 \, \text{g} \]

Calculating this gives:

\[ \text{Mass of Cl in AgCl} = \left(\frac{35.5}{143.5}\right) \times 0.40 = 0.0986 \, \text{g} \]

### Step 2: Calculate the percentage of chlorine in the organic compound

Now that we have the mass of chlorine, we can calculate the percentage of chlorine in the original organic compound.

\[ \text{Percentage of Cl} = \left(\frac{\text{Mass of Cl}}{\text{Mass of organic compound}}\right) \times 100 \]

Given:
- Mass of organic compound = 0.25 g

Substituting the values:

\[ \text{Percentage of Cl} = \left(\frac{0.0986 \, \text{g}}{0.25 \, \text{g}}\right) \times 100 \]

Calculating this gives:

\[ \text{Percentage of Cl} = 39.44\% \]

Rounding to the nearest integer, we get:

\[ \text{Percentage of Cl} \approx 39\% \]

### Final Answer:
The percentage of chlorine present in the compound is **39%**.

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