A solution contains `6.44g` of `HXO_(2)` (mole weight 68.5) in 94.0 g of water. The freezing poit of the solution is `271.K`. Calculate the fraction of `HXO_(2)` that undergoes dissociation to `H^(+) and XO_(2) :` (Given `K_(f)` (water) =`1.86k` kg/mol).

Answer

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

### Step 1: Calculate the depression in freezing point (ΔTf)
The freezing point of pure water is 273 K. The freezing point of the solution is given as 271 K. Therefore, the depression in freezing point (ΔTf) is:
\[ \Delta T_f = T_f^{\text{pure}} - T_f^{\text{solution}} = 273 \, \text{K} - 271 \, \text{K} = 2 \, \text{K} \]

### Step 2: Calculate the molality (m) of the solution
To find the molality, we need to calculate the number of moles of HXO₂ and the mass of the solvent (water) in kg.

1. **Calculate moles of HXO₂**:
\[ \text{Moles of HXO}_2 = \frac{\text{mass}}{\text{molar mass}} = \frac{6.44 \, \text{g}}{68.5 \, \text{g/mol}} \approx 0.0942 \, \text{mol} \]

2. **Convert mass of water to kg**:
\[ \text{Mass of water} = 94.0 \, \text{g} = 0.094 \, \text{kg} \]

3. **Calculate molality**:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0942 \, \text{mol}}{0.094 \, \text{kg}} \approx 1.002 \, \text{mol/kg} \]

### Step 3: Use the freezing point depression formula
The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \( \Delta T_f \) is the depression in freezing point.
- \( i \) is the van 't Hoff factor (number of particles the solute dissociates into).
- \( K_f \) is the cryoscopic constant (given as 1.86 k kg/mol).
- \( m \) is the molality.

Substituting the known values:
\[ 2 = i \cdot 1.86 \cdot 1.002 \]

### Step 4: Solve for \( i \)
Rearranging the equation to solve for \( i \):
\[ i = \frac{2}{1.86 \cdot 1.002} \approx 1.075 \]

### Step 5: Calculate the degree of dissociation (α)
The degree of dissociation \( \alpha \) can be calculated using the formula:
\[ \alpha = \frac{i - 1}{n - 1} \]
Where \( n \) is the number of particles the solute can dissociate into. For HXO₂, it dissociates into \( H^+ \) and \( XO_2^- \), so \( n = 2 \).

Substituting the values:
\[ \alpha = \frac{1.075 - 1}{2 - 1} = 0.075 \]

### Step 6: Calculate the fraction of dissociation
The fraction of HXO₂ that undergoes dissociation is equal to \( \alpha \):
\[ \text{Fraction of dissociation} = \alpha = 0.075 \]

### Final Answer
The fraction of HXO₂ that undergoes dissociation is **0.075**.

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