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 the process of calculating the freezing point depression and then determine the fraction of the solute that dissociates.

### Step 1: Calculate the Freezing Point Depression (ΔTf)

The freezing point depression can be calculated using the formula:
\[ \Delta T_f = T_f^{\text{pure solvent}} - T_f^{\text{solution}} \]
Given:
- \( T_f^{\text{pure solvent}} = 273 \, K \) (freezing point of pure water)
- \( T_f^{\text{solution}} = 271 \, K \)

Calculating ΔTf:
\[ \Delta T_f = 273 \, K - 271 \, K = 2 \, K \]

### Step 2: Calculate the Molality of the Solution

The molality (m) can be calculated using the formula:
\[ \Delta T_f = K_f \cdot m \]
Where:
- \( K_f \) (freezing point depression constant for water) = 1.86 K kg/mol

Rearranging the formula to find molality:
\[ m = \frac{\Delta T_f}{K_f} = \frac{2 \, K}{1.86 \, K \cdot \text{kg/mol}} \approx 1.075 \, \text{mol/kg} \]

### Step 3: Calculate the Moles of Solute (HXO2)

To find the number of moles of HXO2, we use the formula:
\[ \text{moles of HXO2} = \frac{\text{mass of HXO2}}{\text{molar mass of HXO2}} \]
Given:
- Mass of HXO2 = 6.44 g
- Molar mass of HXO2 = 68.5 g/mol

Calculating moles of HXO2:
\[ \text{moles of HXO2} = \frac{6.44 \, g}{68.5 \, g/mol} \approx 0.0942 \, \text{mol} \]

### Step 4: Calculate the Mass of Solvent in kg

The mass of the solvent (water) is given as 94.0 g, which we convert to kg:
\[ \text{mass of water} = 94.0 \, g = 0.094 \, kg \]

### Step 5: Calculate the Total Moles of Solute Using Molality

Using the molality we calculated:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Rearranging gives:
\[ \text{moles of solute} = m \cdot \text{mass of solvent in kg} = 1.075 \, \text{mol/kg} \cdot 0.094 \, kg \approx 0.101 \, \text{mol} \]

### Step 6: Calculate the Van 't Hoff Factor (i)

The van 't Hoff factor (i) is related to the degree of dissociation (α) and the number of particles produced from the dissociation. For HXO2:
\[ i = 1 + (n - 1) \cdot \alpha \]
Where n is the number of particles produced upon dissociation. For HXO2:
- n = 2 (it dissociates into H⁺ and XO₂⁻)

From the previous steps, we have:
\[ i = \frac{\text{moles of solute (calculated from freezing point)}}{\text{moles of solute (calculated from mass)}} = \frac{0.101 \, \text{mol}}{0.0942 \, \text{mol}} \approx 1.075 \]

### Step 7: Calculate the Degree of Dissociation (α)

Now we can rearrange the equation for i to find α:
\[ 1.075 = 1 + (2 - 1) \cdot \alpha \]
\[ 1.075 - 1 = \alpha \]
\[ \alpha = 0.075 \]

### Step 8: Calculate the Fraction of HXO2 that Undergoes Dissociation

The fraction of HXO2 that dissociates is given by α:
\[ \text{Fraction dissociated} = \alpha = 0.075 \]

### Final Answer

Thus, the fraction of HXO2 that undergoes dissociation is approximately **0.075** or **7.5%**.

---