0.1 molal solution of `Hg(NO_(3))_(2)` freezes at `0.558^(@)` C. The
cryoscopic constant for water is 1.86 KKg `mol^(-1)` then what will
be the percentage ionisation of salt ?

Options

A. `33.33%`
B. `50%`
C. `75%`
D. `100%`

Answer

To solve the problem, we need to determine the percentage ionization of the salt \( \text{Hg(NO}_3\text{)}_2 \) based on the given freezing point depression of a 0.1 molal solution. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the Freezing Point Depression
The freezing point depression (\( \Delta T_f \)) can be calculated using the formula:
\[ \Delta T_f = T_f^0 - T_f \]
where \( T_f^0 \) is the freezing point of pure solvent (water, which is 0°C) and \( T_f \) is the freezing point of the solution.

Given that the solution freezes at \( 0.558^\circ C \):
\[ \Delta T_f = 0 - (-0.558) = 0.558^\circ C \]

### Step 2: Calculate the Theoretical Freezing Point Depression
Using the cryoscopic constant (\( K_f \)) for water, which is \( 1.86 \, \text{K kg mol}^{-1} \), and the molality of the solution (\( m = 0.1 \, \text{mol kg}^{-1} \)), we can calculate the theoretical freezing point depression:
\[ \Delta T_f^{\text{theoretical}} = K_f \times m = 1.86 \times 0.1 = 0.186^\circ C \]

### Step 3: Calculate the Van 't Hoff Factor (i)
The Van 't Hoff factor (\( i \)) can be calculated using the observed and theoretical freezing point depressions:
\[ i = \frac{\Delta T_f^{\text{observed}}}{\Delta T_f^{\text{theoretical}}} \]
Substituting the values:
\[ i = \frac{0.558}{0.186} \approx 3 \]

### Step 4: Determine the Ionization of \( \text{Hg(NO}_3\text{)}_2 \)
The dissociation of \( \text{Hg(NO}_3\text{)}_2 \) in solution can be represented as:
\[ \text{Hg(NO}_3\text{)}_2 \rightarrow \text{Hg}^{2+} + 2 \text{NO}_3^{-} \]
This means that for every 1 mole of \( \text{Hg(NO}_3\text{)}_2 \), it produces 3 moles of ions (1 \( \text{Hg}^{2+} \) and 2 \( \text{NO}_3^{-} \)).

### Step 5: Relate Van 't Hoff Factor to Ionization
The relationship between the Van 't Hoff factor and the degree of ionization (\( \alpha \)) is given by:
\[ i = 1 + n\alpha \]
where \( n \) is the number of particles produced from one formula unit. Here, \( n = 3 \):
\[ 3 = 1 + 2\alpha \]
Solving for \( \alpha \):
\[ 3 - 1 = 2\alpha \implies 2 = 2\alpha \implies \alpha = 1 \]

### Step 6: Calculate Percentage Ionization
The percentage ionization is given by:
\[ \text{Percentage Ionization} = \alpha \times 100\% \]
Substituting the value of \( \alpha \):
\[ \text{Percentage Ionization} = 1 \times 100\% = 100\% \]

### Final Answer
The percentage ionization of \( \text{Hg(NO}_3\text{)}_2 \) is **100%**.

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