Question

If a centimolar aqueous solution of K\(_3\)[Fe(CN)\(_6\)] has degree of dissociation 0.78, what is the value of van't Hoff factor?

• A. 3.34
• B. 1.2
• C. 2.5
• D. 4.0

Hint:

The van't Hoff factor is important in colligative properties and depends on the number of particles formed after dissociation.

Answer 1

The Correct Option is A

Step 1: Understanding the van't Hoff factor.
The van't Hoff factor \(i\) is the number of particles the solute dissociates into. For K\(_3\)[Fe(CN)\(_6\)], the dissociation can be represented as: \[ K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-} \] Hence, the total number of ions formed is 4 (3 K\(^+\) ions and 1 [Fe(CN)\(_6\)]\(^{3-}\)).
Step 2: Applying the degree of dissociation.
The degree of dissociation \(\alpha = 0.78\), so the van't Hoff factor is: \[ i = 1 + \alpha (n-1) = 1 + 0.78(4-1) = 1 + 0.78 \times 3 = 3.34 \]
Step 3: Conclusion.
The van't Hoff factor is 3.34.