A solution containing `0.85 g` of `ZnCl_(2)` in `125.0g` of water freezes at `-0.23^(@)C`. The apparent degree of dissociation of the salt is:
`(k_(f)` for water `= 1.86 K kg mol^(-1)`, atomic mass, `Zn = 65.3` and `Cl = 35.5)`

Options

A. `1.36%`
B. `2.47%`
C. `73.5%`
D. `7.35%`

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate the Molecular Weight of ZnCl₂
The molecular weight of ZnCl₂ can be calculated using the atomic masses provided:
- Atomic mass of Zn = 65.3 g/mol
- Atomic mass of Cl = 35.5 g/mol

Molecular weight of ZnCl₂ = Atomic mass of Zn + 2 × Atomic mass of Cl
= 65.3 + 2 × 35.5
= 65.3 + 71
= 136.3 g/mol

### Step 2: Calculate the Moles of ZnCl₂
Using the weight of ZnCl₂ given (0.85 g), we can find the number of moles:
\[ \text{Moles of ZnCl₂} = \frac{\text{Weight of ZnCl₂}}{\text{Molecular weight of ZnCl₂}} = \frac{0.85 \text{ g}}{136.3 \text{ g/mol}} \approx 0.00623 \text{ mol} \]

### Step 3: Convert the Weight of Solvent to kg
The weight of the solvent (water) is given as 125 g. We need to convert this to kg:
\[ \text{Weight of solvent in kg} = \frac{125 \text{ g}}{1000} = 0.125 \text{ kg} \]

### Step 4: Calculate the Depression in Freezing Point (ΔTf)
The freezing point of pure water is 0°C, and the freezing point of the solution is -0.23°C. Thus, the depression in freezing point is:
\[ \Delta T_f = T_f (\text{pure solvent}) - T_f (\text{solution}) = 0 - (-0.23) = 0.23°C \]

### Step 5: Use the Freezing Point Depression Formula
The formula for freezing point depression is given by:
\[ \Delta T_f = K_f \times \text{molality} \times i \]
Where:
- \( K_f \) = 1.86 K kg/mol (given)
- Molality = \(\frac{\text{moles of solute}}{\text{weight of solvent in kg}}\)

First, we calculate the molality:
\[ \text{Molality} = \frac{0.00623 \text{ mol}}{0.125 \text{ kg}} \approx 0.04984 \text{ mol/kg} \]

Now substituting into the freezing point depression formula:
\[ 0.23 = 1.86 \times 0.04984 \times i \]
Solving for \( i \):
\[ i = \frac{0.23}{1.86 \times 0.04984} \approx 2.47 \]

### Step 6: Determine the Degree of Dissociation (α)
For ZnCl₂, it dissociates into 3 ions:
\[ \text{ZnCl}_2 \rightarrow \text{Zn}^{2+} + 2 \text{Cl}^- \]
Thus, \( n = 3 \) (total ions produced).

Using the relationship:
\[ i = 1 + (n - 1) \alpha \]
Substituting the values:
\[ 2.47 = 1 + (3 - 1) \alpha \]
\[ 2.47 = 1 + 2\alpha \]
\[ 2\alpha = 2.47 - 1 = 1.47 \]
\[ \alpha = \frac{1.47}{2} \approx 0.735 \]

### Step 7: Calculate the Percentage of Degree of Dissociation
To find the percentage of degree of dissociation:
\[ \text{Percentage of } \alpha = \alpha \times 100 = 0.735 \times 100 \approx 73.5\% \]

### Final Answer
The apparent degree of dissociation of the salt is approximately **73.5%**.

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