1. Principle of Carius estimation

   In Carius estimation, chlorine present in the organic compound is converted into silver chloride, $\mathrm{AgCl}$.

   So, from the mass of $\mathrm{AgCl}$ formed, we first find the mass of chlorine.

2. Molar mass of $\mathrm{AgCl}$

   Given:

   • Atomic mass of $\mathrm{Ag} = 108$
   • Atomic mass of $\mathrm{Cl} = 35.5$

   Therefore, $M(\mathrm{AgCl}) = 108 + 35.5 = 143.5 \text{ g mol}^{-1}$

3. Mass of chlorine in $0.40$ g of $\mathrm{AgCl}$

   In $143.5$ g of $\mathrm{AgCl}$, mass of chlorine $= 35.5$ g.

   Hence, in $0.40$ g of $\mathrm{AgCl}$, mass of chlorine is $\text{Mass of Cl} = 0.40 \times \frac{35.5}{143.5}$

   $= 0.40 \times 0.2474 \approx 0.09895 \text{ g}$

4. Percentage of chlorine in the organic compound

   Mass of organic compound taken $= 0.25$ g.

   Therefore, $\%\text{Cl} = \frac{0.09895}{0.25} \times 100$

   $= 39.58\%$

5. Nearest integer

   $39.58 \approx 40$

Final Answer

The percentage of chlorine present in the compound is: $\boxed{40}$