To determine the integer $x$ in the compound $\text{Hg}_x\text{I}_x$, we can use the freezing point depression formula:

$\Delta T_f = K_f \cdot m$

where:

• $\Delta T_f$ is the freezing point depression (1.88 K),
• $K_f$ is the cryoscopic constant (not given directly, but can be derived from the enthalpy of fusion),
• $m$ is the molality of the solution.

Step 1: Calculate the molar mass of $\text{HgI}_2$

The molar mass of $\text{HgI}_2$ can be calculated as follows:

$\text{Molar mass of } \text{HgI}_2 = \text{Molar mass of Hg} + 2 \times \text{Molar mass of I}$

$= 200.59 \, \text{g/mol} + 2 \times 126.90 \, \text{g/mol} = 454.39 \, \text{g/mol}$

Step 2: Calculate the molality $m$

The freezing point depression constant $K_f$ can be found from the enthalpy of fusion:

$K_f = \frac{R \cdot T_m^2}{\Delta H_f}$

where:

• $R = 8.314 \, \text{J/(mol K)}$
• $T_m = 532 \, \text{K}$
• $\Delta H_f = 41.6 \, \text{J/g}$ (convert to J/mol)

To convert $\Delta H_f$ to J/mol:

$\Delta H_f = 41.6 \, \text{J/g} \times 454.39 \, \text{g/mol} = 18815.73 \, \text{J/mol}$

Now substituting into the equation for $K_f$:

$K_f = \frac{8.314 \cdot (532)^2}{18815.73} = \frac{8.314 \cdot 283024}{18815.73} \approx 122.27 \, \text{K kg/mol}$

Step 3: Calculate molality $m$

Using the formula for freezing point depression:

$1.88 = 122.27 \cdot m$

Solving for $m$:

$m = \frac{1.88}{122.27} \approx 0.0154 \, \text{mol/kg}$

Step 4: Calculate the moles of solute

The mass of the solvent $\text{HgI}_2$ is $100.0 \, \text{g} = 0.1 \, \text{kg}$. Thus, the moles of solute (the compound $\text{Hg}_x\text{I}_x$) can be calculated as follows:

$\text{moles of solute} = m \cdot \text{mass of solvent} = 0.0154 \, \text{mol/kg} \cdot 0.1 \, \text{kg} = 0.00154 \, \text{mol}$

Step 5: Determine the number of moles of $\text{Hg}_x\text{I}_x$

The total mass of the compound $\text{Hg}_x\text{I}_x$ is $2.177 \, \text{g}$, and its molar mass is:

$\text{Molar mass of } \text{Hg}_x\text{I}_x = x \cdot 200.59 + x \cdot 126.90 = x(200.59 + 126.90) = x(327.49)$

Setting up the equation for moles:

$\frac{2.177 \, \text{g}}{x \cdot 327.49 \, \text{g/mol}} = 0.00154 \, \text{mol}$

Solving for $x$:

$2.177 = 0.00154 \cdot x \cdot 327.49$

$x = \frac{2.177}{0.00154 \cdot 327.49} \approx \frac{2.177}{0.50478} \approx 4.31$

Rounding to the nearest integer, we get:

$\boxed{4}$

Conclusion

The integer $x$ in the compound $\text{Hg}_x\text{I}_x$ is 4.

Do you want more details on any specific step or have any other questions? Here are some related questions to consider:

1. What other properties of the compound could be relevant?
2. How does freezing point depression relate to colligative properties?
3. Can you calculate the boiling point elevation for this solution?
4. What assumptions are made in this calculation?
5. How would the results change with a different solvent?

Tip: Always check the units to ensure consistency throughout the calculations.