The electrolyte solutions show abnormal colligative porperties.To account for this effect we define a quantity called the Van't Hoff factor given by
`i=("Actual number of particles in solution after dissociation")/("Number of formula units initially dissolved in solution")`
`i=1 ("for non-electrolytes")`
`igt1 ("for electrolytes, undergoing dissociation")`
`ilt1 ("for solutes, undergoing association")`
Answer the following questions:
`0.1 M K_(4)[Fe(CN)_(6)]` is `60%` ionized. What will be its Van't Hoff factor?

Options

A. `1.4`
B. `3.4`
C. `2.4`
D. `4.4`

Explanation

To find the Van't Hoff factor (i) for the given solution of \(0.1 \, M \, K_4[Fe(CN)_6]\) that is 60% ionized, we can follow these steps:

### Step 1: Determine the dissociation of the solute
The formula \(K_4[Fe(CN)_6]\) dissociates in solution as follows:
\[ K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-} \]
This means one formula unit of \(K_4[Fe(CN)_6]\) produces 5 particles in total (4 potassium ions and 1 hexacyanoferrate ion).

### Step 2: Calculate the number of particles produced after ionization
Since the solution is 60% ionized, we can calculate the effective concentration of the ions produced:
- The initial concentration of \(K_4[Fe(CN)_6]\) is \(0.1 \, M\).
- The degree of ionization is 60%, which means:
\[ \text{Ionized concentration} = 0.1 \, M \times 0.6 = 0.06 \, M \]

### Step 3: Calculate the concentration of each ion
From the dissociation, for every 1 mole of \(K_4[Fe(CN)_6]\) that ionizes, we get:
- 4 moles of \(K^+\)
- 1 mole of \([Fe(CN)_6]^{4-}\)

Thus, the total concentration of particles produced from the ionization is:
\[ \text{Total concentration of ions} = 4 \times 0.06 \, M + 0.06 \, M = 0.24 \, M + 0.06 \, M = 0.30 \, M \]

### Step 4: Calculate the Van't Hoff factor (i)
The Van't Hoff factor is given by the formula:
\[ i = \frac{\text{Actual number of particles in solution after dissociation}}{\text{Number of formula units initially dissolved in solution}} \]
Substituting the values we have:
\[ i = \frac{0.30 \, M}{0.1 \, M} = 3 \]

### Step 5: Adjust for the remaining undissociated solute
Since only 60% of the solute is ionized, we need to account for the remaining 40% that does not dissociate:
- The remaining concentration of undissociated \(K_4[Fe(CN)_6]\) is:
\[ 0.1 \, M - 0.06 \, M = 0.04 \, M \]
This means we have an additional 0.04 M of undissociated particles.

### Step 6: Adjust the total particle count
Now, the total number of particles in solution becomes:
\[ \text{Total particles} = 0.30 \, M + 0.04 \, M = 0.34 \, M \]

### Final Calculation of Van't Hoff factor
Now we can recalculate the Van't Hoff factor:
\[ i = \frac{0.34 \, M}{0.1 \, M} = 3.4 \]

Thus, the Van't Hoff factor \(i\) for \(0.1 \, M \, K_4[Fe(CN)_6]\) that is 60% ionized is **3.4**.