Question

For an electrolyte consisting of a monovalent cation and a monovalent anion undergoing dissociation in water, the van't Hoff factor (i) is :

• A. less than 1
• B. equal to 1
• C. greater than 1
• D. greater than 2

Hint:

Dissociation increases particle count, so $i \gt 1$. Association decreases it, so $i \lt 1$.

Answer 1

The Correct Option is C

Step 1: Concept
The van't Hoff factor ($i$) relates the actual number of particles in solution after dissociation or association to the initial number of formula units dissolved.

Step 2: Meaning
For dissociation, the formula is $i = 1 + \alpha(n-1)$, where $\alpha$ is the degree of dissociation and $n$ is the number of ions formed from one formula unit.

Step 3: Analysis
An electrolyte with a monovalent cation and anion (like NaCl) yields $n = 2$ particles upon complete dissociation. Since it undergoes dissociation in water, $\alpha \gt 0$. Therefore, $i = 1 + \alpha(2-1) = 1 + \alpha$.

Step 4: Conclusion
Because $\alpha$ is a positive fraction (or 1 for complete dissociation), $i$ will be strictly between 1 and 2, which makes it greater than 1.

Final Answer: (C)