Calculating Van't Hoff Factor of [Cr(NH3)6]Cl2
There are a few key points to consider when calculating the Van't Hoff factor of a compound like [Cr(NH3)6]Cl2. Here's a breakdown of the steps involved:

Determining the Number of Particles
- In this compound, [Cr(NH3)6]Cl2, there are two ions that dissociate when it is dissolved in water: [Cr(NH3)6]3+ and 2Cl-.
- The total number of particles formed upon dissociation is 1 (from [Cr(NH3)6]3+) + 2 (from 2Cl-) = 3 particles.

Calculating the Van't Hoff Factor
- The Van't Hoff factor (i) is calculated using the formula: i = 1 + (n-1) * α
- Here, n represents the number of particles formed upon dissociation (in this case, 3), and α is the degree of dissociation.
- Since the compound completely dissociates into its ions in solution, the degree of dissociation (α) is 1.
- Plugging in the values, we get: i = 1 + (3-1) * 1 = 3

Conclusion
- The Van't Hoff factor of [Cr(NH3)6]Cl2 is 3. This means that for every formula unit of [Cr(NH3)6]Cl2 that dissociates in solution, it produces 3 particles.