Calculating Van't Hoff Factor of [Ag(CN)2]
To calculate the Van't Hoff factor of a compound, we need to consider the number of particles that the compound dissociates into when dissolved in a solution. In this case, we are looking at the compound [Ag(CN)2].

1. Understanding the Compound
- The compound [Ag(CN)2] consists of a single silver ion (Ag+) and two cyanide ions (CN-) bound together.
- When this compound dissociates in a solution, it will break into its respective ions: Ag+ and 2CN-.

2. Determining the Van't Hoff Factor
- The Van't Hoff factor (i) for a compound is the ratio of moles of particles in solution after dissociation to moles of formula units initially dissolved.
- In this case, [Ag(CN)2] dissociates into one Ag+ ion and two CN- ions.
- Therefore, the total number of moles of particles after dissociation is 1 (Ag+) + 2 (CN-) = 3 moles.
- The moles of the formula unit [Ag(CN)2] initially dissolved is 1 mole.
- Thus, the Van't Hoff factor (i) for [Ag(CN)2] is 3.

3. Conclusion
- The Van't Hoff factor of [Ag(CN)2] is 3, indicating that this compound dissociates into three particles when dissolved in a solution.
- Understanding the Van't Hoff factor is crucial in various chemical and physical processes, especially in determining colligative properties such as boiling point elevation and freezing point depression in solutions.