1. We are given an octahedral complex of formula $MCl_3\cdot 2L$ and on reaction with excess $AgNO_3$, $1$ mol of $AgCl$ is formed per mol of complex.

2. $AgNO_3$ precipitates only the chloride ions present outside the coordination sphere as free counter ions.

   Since $1$ mol of complex gives $1$ mol of $AgCl$, there is exactly one free $Cl^-$ outside the coordination sphere.

3. Total chlorides in the formula = $3$. Therefore:

   • $1$ chloride is outside the coordination sphere.
   • Remaining $2$ chlorides are inside the coordination sphere as ligands.

   So the complex can be written as $[MCl_2L_2]Cl$

4. The complex is stated to be octahedral, so the coordination number of the metal is $6$.

5. Inside the coordination sphere:

   • $2$ chloride ligands are present, each monodentate, contributing $2$ coordination sites.
   • There are $2$ ligands $L$, each of denticity $d$.

   Hence total coordination number is $2 + 2d = 6$

6. Solving: $2d = 4$ $d = 2$

7. Therefore, ligand $L$ is bidentate.

Final integer answer: $2$