A solution of 2.177 g of a compound with the formula HgxIx in 100.0 g pure HgI2 has a freezing point that is 1.88 K lower than that of pure HgI2. Determine the integer x if the enthalpy of fusion of HgI2 is 41.6 J/g at its melting point of 532 K. 

To solve the problem, we can use the concept of freezing point depression, which is described by the formula:

\[
\Delta T_f = K_f \cdot m
\]

where:
- \(\Delta T_f\) is the freezing point depression.
- \(K_f\) is the cryoscopic constant (freezing point depression constant) for the solvent.
- \(m\) is the molality of the solution.

### Given Information
- \(\Delta T_f = 1.88 \, \text{K}\)
- Mass of compound = \(2.177 \, \text{g}\)
- Mass of \(HgI_2\) = \(100.0 \, \text{g}\)
- Enthalpy of fusion of \(HgI_2\) = \(41.6 \, \text{J/g}\)
- Melting point of \(HgI_2\) = \(532 \, \text{K}\)

### Steps to Solve

1. **Calculate the moles of \(HgI_2\):**
   - The molar mass of \(HgI_2\) is calculated as:
     \[
     M_{HgI_2} = M_{Hg} + 2 \cdot M_{I} = 200.59 \, \text{g/mol} + 2 \cdot 126.90 \, \text{g/mol} = 454.39 \, \text{g/mol}
     \]
   - Moles of \(HgI_2\):
     \[
     n_{HgI_2} = \frac{100.0 \, \text{g}}{454.39 \, \text{g/mol}} \approx 0.220 \, \text{mol}
     \]

2. **Calculate the molality of the solution:**
   - Since molality \(m\) is defined as:
     \[
     m = \frac{n_{solute}}{mass_{solvent \, (kg)}}
     \]
   - Here, the solute is \(HgxIx\) and the solvent is \(HgI_2\), which we will treat as the solvent mass in kg:
     \[
     m = \frac{n_{HgxIx}}{0.1 \, \text{kg}} = \frac{n_{HgxIx}}{0.1}
     \]

3. **Calculate the moles of the solute \(HgxIx\):**
   - The molar mass of \(HgxIx\) can be represented as:
     \[
     M_{HgxIx} = M_H + x \cdot M_I = 1.008 \, \text{g/mol} + x \cdot 126.90 \, \text{g/mol}
     \]
   - Moles of \(HgxIx\):
     \[
     n_{HgxIx} = \frac{2.177 \, \text{g}}{M_{HgxIx}} = \frac{2.177}{1.008 + x \cdot 126.90}
     \]

4. **Plug values into the freezing point depression formula:**
   - Set \(\Delta T_f\) equal to \(K_f \cdot m\):
     \[
     1.88 = K_f \cdot \left(\frac{n_{HgxIx}}{0.1}\right)
     \]
   - Assuming \(K_f\) is derived from the enthalpy of fusion:
     \[
     K_f = \frac{R \cdot T^2}{\Delta H_{fus}} \quad (R = 8.314 \, \text{J/(mol K)}, \Delta H_{fus} = 41.6 \, \text{J/g})
     \]

5. **Substituting and solving for \(x\):**
   - Find \(n_{HgxIx}\) in terms of \(x\) and set it equal to the previously established relationship. After substituting and simplifying, you can solve for \(x\).

### Final Calculation
1. **Calculate \(K_f\) using the values provided:**
   - This calculation will provide you with the precise \(K_f\) value to plug into the depression formula and help isolate \(x\).

2. **Isolate and solve for \(x\) in the equation set above.**

By completing these calculations, you will arrive at the integer value of \(x\).

If you have any questions or need further details, feel free to ask! Here are some additional questions you might consider:

1. What is the significance of the cryoscopic constant in this calculation?
2. How would changing the mass of \(HgI_2\) affect the freezing point depression?
3. Can you explain how the enthalpy of fusion relates to the freezing point depression?
4. What would happen if more \(HgxIx\) were added to the solution?
5. How do real solutions differ from ideal solutions in terms of colligative properties?

**Tip:** Always ensure to keep track of units during calculations to avoid confusion.

A solution of 2.177 g of a compound with the formula HgxIx in 100.0 g pure HgI2 has a freezing point that is 1.88 K lower than that of pure HgI2. Determine the integer x if the enthalpy of fusion of HgI2 is 41.6 J/g at its melting point of 532 K.

Explore how to determine the integer x in the compound HgxIx from a freezing point depression of 1.88 K in a HgI2 solution. This problem utilizes key concepts such as molality, freezing point depression, and the enthalpy of fusion of HgI2. Ideal for high school chemistry students.

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