Question

Van’t Hoff factor for $K_3[Fe(CN)_6]$ is $3.333$. What is its percentage dissociation in water?

• A. $77.7%$
• B. $70%$
• C. $83%$
• D. $58%$

Hint:

For complete dissociation into $n$ ions, van’t Hoff factor approaches $n$. Partial dissociation lowers its value.

Answer 1

The Correct Option is A

Step 1: Write the dissociation of the electrolyte.
\[ K_3[Fe(CN)_6] \rightarrow 3K^+ + [Fe(CN)_6]^{3-} \]
Total ions formed $= 4$
Step 2: Use the van’t Hoff factor relation.
\[ i = 1 + \alpha (n - 1) \]
where $n = 4$
Step 3: Substitute the given values.
\[ 3.333 = 1 + \alpha(4 - 1) \]
\[ 3.333 = 1 + 3\alpha \]
\[ 3\alpha = 2.333 \]
\[ \alpha = 0.7777 \]
Step 4: Convert into percentage dissociation.
\[ %\text{ dissociation} = 0.7777 \times 100 = 77.7% \]
Step 5: Conclusion.
The percentage dissociation of $K_3[Fe(CN)_6]$ in water is $77.7%$.