A compound crystallises as follows :
ions ''A'' are at corners of a cubic unit cell and ''B'' ions at face centres of a cubic unit cell and ''C'' ions in `1//4^(th)` of the total tetrahedral void. Assuming if this is dissolved, only the ions in the tetrahedral voids are dissociated completely in water, which one of the following statements is true. (Assuming all are univalent ions) and also A and C are cations and B is an anions.

Options

A. Boiling point of same concentration of `AlCl_(3)` solution (100% dissociation) will be greater than that of this solution.
B. Boiling point of same concentration of `Ca_(3)(PO_(4))_(2)` (100% dissociation) will be greater than that of this solution.
C. Boiling point of same concentration of sucrose will be greater than this solution
D. Data insufficient to predict.

Answer

To solve the problem, we need to analyze the crystallization of the compound and determine the formula based on the positions of the ions in the unit cell. We will then assess the dissociation of the compound in water and compare the boiling points with other compounds.

### Step-by-Step Solution:

1. **Determine the Contribution of Ion A:**
- Ion A is located at the corners of the cubic unit cell. Each corner atom contributes \( \frac{1}{8} \) of an atom to the unit cell.
- There are 8 corners in a cube, so the total contribution of A is:
\[ \text{Contribution of A} = 8 \times \frac{1}{8} = 1 \]

2. **Determine the Contribution of Ion B:**
- Ion B is located at the face centers of the cubic unit cell. Each face-centered atom contributes \( \frac{1}{2} \) of an atom to the unit cell.
- There are 6 faces in a cube, so the total contribution of B is:
\[ \text{Contribution of B} = 6 \times \frac{1}{2} = 3 \]

3. **Determine the Contribution of Ion C:**
- Ion C occupies \( \frac{1}{4} \) of the total tetrahedral voids. In a face-centered cubic (FCC) unit cell, there are a total of 8 tetrahedral voids.
- Therefore, the contribution of C is:
\[ \text{Contribution of C} = \frac{1}{4} \times 8 = 2 \]

4. **Write the Empirical Formula:**
- Based on the contributions calculated:
- A = 1
- B = 3
- C = 2
- The empirical formula of the compound is:
\[ \text{Formula} = A B_3 C_2 \]

5. **Dissociation in Water:**
- When the compound \( A B_3 C_2 \) is dissolved in water, only the ions in the tetrahedral voids (C) dissociate completely. Therefore, the dissociation can be represented as:
\[ A B_3 C_2 \rightarrow A + B_3 + 2C \]
- Here, C (the cation) dissociates completely, while A and B remain intact.

6. **Calculate the Total Number of Particles (i):**
- The total number of particles after dissociation is:
- 1 (A) + 3 (B) + 2 (C) = 6 particles.
- However, since only C dissociates completely, we consider the effective particles:
- The effective dissociation gives us \( 1 + 3 + 2 = 6 \).

7. **Comparison of Boiling Points:**
- The boiling point elevation is a colligative property that depends on the number of particles in solution (i).
- For comparison:
- \( AlCl_3 \) dissociates into \( Al^{3+} + 3Cl^{-} \) (i = 4).
- \( Ca_3(PO_4)_2 \) dissociates into \( 3Ca^{2+} + 2PO_4^{3-} \) (i = 5).
- Sucrose does not dissociate (i = 1).
- Therefore, the boiling points can be ranked based on the number of particles:
- \( AlCl_3 > Ca_3(PO_4)_2 > A B_3 C_2 > Sucrose \).

### Conclusion:
The correct statements regarding the boiling points are:
- The boiling point of \( AlCl_3 \) is greater than that of \( A B_3 C_2 \).
- The boiling point of \( Ca_3(PO_4)_2 \) is also greater than that of \( A B_3 C_2 \).
- The boiling point of sucrose is less than that of \( A B_3 C_2 \).