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 determine the integer \( x \) in the formula \( \text{HgxIx} \), we can use the freezing point depression equation:

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

where:
- \( \Delta T_f \) is the freezing point depression (1.88 K),
- \( K_f \) is the cryoscopic constant of HgI2,
- \( m \) is the molality of the solution.

First, we need to find \( K_f \) using the enthalpy of fusion and the melting point:

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

Using \( R = 8.314 \, \text{J/(mol·K)} \), \( T_m = 532 \, \text{K} \), and \( \Delta H_f = 41.6 \, \text{J/g} \):

1. Convert \( \Delta H_f \) from J/g to J/mol. The molar mass of \( \text{HgI}_2 \) (approx. \( 454.0 \, \text{g/mol} \)):
   \[
   \Delta H_f \, (\text{J/mol}) = 41.6 \, \text{J/g} \times 454.0 \, \text{g/mol} = 18882.4 \, \text{J/mol}
   \]

2. Now calculate \( K_f \):
   \[
   K_f = \frac{8.314 \cdot (532)^2}{18882.4} = \frac{8.314 \cdot 283024}{18882.4} \approx 123.2 \, \text{K·kg/mol}
   \]

Next, calculate the molality \( m \) using the freezing point depression:

\[
1.88 = 123.2 \cdot m \implies m = \frac{1.88}{123.2} \approx 0.01524 \, \text{mol/kg}
\]

Now, determine the moles of solute in the 100 g of HgI2:

- The mass of the solvent (HgI2) is 100 g = 0.1 kg.
- Therefore, moles of solute \( n \):
\[
n = m \cdot \text{mass of solvent} = 0.01524 \, \text{mol/kg} \cdot 0.1 \, \text{kg} = 0.001524 \, \text{mol}
\]

Now we know that the moles of the solute \( \text{HgxIx} \) are:

1. Calculate the molar mass of the solute:
\[
\text{mass of solute} = 2.177 \, \text{g} \implies \text{molar mass} = \frac{2.177 \, \text{g}}{0.001524 \, \text{mol}} \approx 1430.6 \, \text{g/mol}
\]

2. Assuming \( x \) for \( \text{HgxIx} \):
   - Molar mass of \( \text{Hg} \approx 200.59 \, \text{g/mol} \)
   - Molar mass of \( \text{I} \approx 126.90 \, \text{g/mol} \)
   - Total molar mass: \( 200.59 + 126.90x \)

Setting it equal to the calculated molar mass:
\[
200.59 + 126.90x = 1430.6
\]
\[
126.90x = 1430.6 - 200.59 = 1230.01 \implies x \approx \frac{1230.01}{126.90} \approx 9.69
\]

Since \( x \) must be an integer, we round \( x \) to 10.

Thus, the integer \( x \) is **10**.

Would you like to delve into any specific parts of this calculation or have any questions? Here are some follow-up questions:

1. How does the freezing point depression relate to the properties of solutions?
2. What is the significance of the cryoscopic constant in this context?
3. How would changing the mass of the solute affect the freezing point depression?
4. Can you explain how to determine the molar mass of a compound using freezing point depression?
5. What other properties can be affected by the addition of solutes to solvents? 

Tip: Always check the units in each step to avoid errors in calculations.

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 using freezing point depression. This problem involves a 2.177 g solute in 100 g of HgI2 and a depression of 1.88 K, with the enthalpy of fusion at 41.6 J/g. Ideal for Grades 11-12 students.

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