To determine the integer $x$ in the formula $\text{HgxIx}$, we will use the freezing point depression formula and the enthalpy of fusion of $\text{HgI}_2$.

Step 1: Calculate the Freezing Point Depression

The freezing point depression ($\Delta T_f$) is given as 1.88 K. The formula for freezing point depression is:

$\Delta T_f = i \cdot K_f \cdot m$

Where:

• $i$ = van 't Hoff factor (number of particles the solute breaks into),
• $K_f$ = cryoscopic constant (not given, but we'll find $m$),
• $m$ = molality of the solution.

Step 2: Calculate the Molar Mass of $\text{HgI}_2$

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

• Molar mass of $\text{Hg}$ = 200.59 g/mol,
• Molar mass of $\text{I}$ = 126.90 g/mol.

Thus, the molar mass of $\text{HgI}_2$ is:

$\text{Molar mass of } \text{HgI}_2 = 200.59 + 2 \times 126.90 = 454.39 \, \text{g/mol}$

Step 3: Calculate the Mass of $\text{HgI}_2$ in the Solution

The total mass of the solution is $100.0 \, \text{g} + 2.177 \, \text{g} = 102.177 \, \text{g}$.

Step 4: Calculate the Molality of the Solution

Using the molar mass of $\text{HgI}_2$:

$\text{Moles of } \text{HgI}_2 = \frac{100.0 \, \text{g}}{454.39 \, \text{g/mol}} \approx 0.220 \, \text{mol}$

The molality $m$ can be found using:

$m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.220 \, \text{mol}}{0.100 \, \text{kg}} = 2.20 \, \text{mol/kg}$

Step 5: Calculate the Van 't Hoff Factor $i$

Assuming $\text{HgxIx}$ dissociates into $x + 1$ ions:

$\Delta T_f = i \cdot K_f \cdot m$

From the freezing point depression:

$1.88 = i \cdot K_f \cdot 2.20$

Step 6: Determine $K_f$

The $K_f$ value for $\text{HgI}_2$ can be derived from the enthalpy of fusion:

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

Using $R = 8.314 \, \text{J/(mol K)}$, $T_m = 532 \, \text{K}$, and $\Delta H_f = 41.6 \, \text{J/g}$ (converted to J/mol for calculation):

$\Delta H_f = 41.6 \times 454.39 \approx 18832.62 \, \text{J/mol}$

Now substitute to find $K_f$:

$K_f = \frac{(8.314)(532^2)}{18832.62} \approx 0.005 \, \text{K kg/mol}$

Step 7: Substitute $K_f$ to find $i$

Now we can find $i$:

$1.88 = i \cdot 0.005 \cdot 2.20 \implies i = \frac{1.88}{0.011} \approx 171.82$

This gives $i = x + 1$:

$x + 1 \approx 172 \implies x \approx 171$

Conclusion

The integer $x$ in the formula $\text{HgxIx}$ is:

$\boxed{171}$

Questions

1. Would you like a detailed explanation of the freezing point depression concept?
2. Do you need help with another chemistry problem?
3. Would you like to learn more about colligative properties?
4. Are you interested in how to calculate molar masses in general?
5. Would you like tips on understanding the van 't Hoff factor?

Tip

Always double-check your units when performing calculations involving molality and other concentration measures to avoid errors.