Number of ions present in `2.0 "litre"` of a solution of `0.8 M K_(4)Fe(CN)_(6)` is:

Options

A. `4.8xx10^(22)`
B. `4.8xx10^(24)`
C. `9.6xx10^(24)`
D. `9.6xx10^(22)`

Explanation

To find the number of ions present in 2.0 litres of a 0.8 M solution of \( K_4[Fe(CN)_6] \), we can follow these steps:

### Step 1: Calculate the number of moles of \( K_4[Fe(CN)_6] \)

The molarity (M) of a solution is defined as the number of moles of solute per litre of solution. Therefore, we can calculate the number of moles of \( K_4[Fe(CN)_6] \) in 2.0 litres of a 0.8 M solution using the formula:

\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (L)} \]

Substituting the values:

\[ \text{Number of moles} = 0.8 \, \text{mol/L} \times 2.0 \, \text{L} = 1.6 \, \text{moles} \]

### Step 2: Determine the number of ions produced per formula unit of \( K_4[Fe(CN)_6] \)

The compound \( K_4[Fe(CN)_6] \) dissociates in solution to produce ions. The dissociation can be represented as:

\[ K_4[Fe(CN)_6] \rightarrow 4K^+ + [Fe(CN)_6]^{4-} \]

From this equation, we can see that one formula unit of \( K_4[Fe(CN)_6] \) produces 4 potassium ions (\( K^+ \)) and 1 hexacyanoferrate ion (\( [Fe(CN)_6]^{4-} \)). Therefore, the total number of ions produced per formula unit is:

\[ 4 \, (K^+) + 1 \, ([Fe(CN)_6]^{4-}) = 5 \, \text{ions} \]

### Step 3: Calculate the total number of ions in the solution

Now, we can calculate the total number of ions produced in the solution by multiplying the number of moles of \( K_4[Fe(CN)_6] \) by the number of ions produced per mole:

\[ \text{Total number of ions} = \text{Number of moles} \times \text{Number of ions per mole} \]

Substituting the values:

\[ \text{Total number of ions} = 1.6 \, \text{moles} \times 5 \, \text{ions/mole} = 8.0 \, \text{ions} \]

### Final Answer

The total number of ions present in 2.0 litres of a 0.8 M solution of \( K_4[Fe(CN)_6] \) is **8.0 ions**.

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