A compound X undergoes 100% pentamerisation in a given solvent Y. Correct statements are

Options

A. van't Hoff factor of compound is 0.20
B. Experimental elevation in boiling point
`=("Calculated elevation in boiling point")/(5)`
C. Observed molar mass of solute `=("Normal molar mass")/(5)`
D. Observed freezing pointxx5=Normal freezing point

Answer

To solve the problem regarding the compound X that undergoes 100% pentamerization in a given solvent Y, we need to analyze the implications of this process on the colligative properties of the solution.

### Step-by-Step Solution:

1. **Understanding Pentamerization**:
- Pentamerization means that one molecule of compound X combines with four other molecules to form a pentamer, resulting in a total of five molecules (1 X + 4 X = 5 X).
- This indicates that for every mole of X, we effectively have 5 moles of the pentamer in solution.

2. **Determining the van 't Hoff Factor (i)**:
- The van 't Hoff factor (i) is defined as the ratio of the observed colligative property to the calculated colligative property.
- Since X undergoes complete pentamerization, the effective number of particles in solution is 1/5 of the original moles of X.
- Therefore, \( i = \frac{1}{5} = 0.2 \).

3. **Experimental Elevation in Boiling Point**:
- The elevation in boiling point is a colligative property that depends on the number of solute particles.
- The formula for elevation in boiling point is given by:
\[ \Delta T_b = i \cdot K_b \cdot m \]
- Since \( i = 0.2 \), we can say that the experimental elevation in boiling point is equal to the calculated elevation divided by 5:
\[ \text{Experimental } \Delta T_b = \frac{\text{Calculated } \Delta T_b}{5} \]

4. **Observed Molar Mass**:
- The observed molar mass of the solute can be related to the normal molar mass by the van 't Hoff factor:
\[ i = \frac{\text{Normal Molar Mass}}{\text{Observed Molar Mass}} \]
- Rearranging gives us:
\[ \text{Observed Molar Mass} = \frac{\text{Normal Molar Mass}}{i} = \frac{\text{Normal Molar Mass}}{0.2} = 5 \times \text{Normal Molar Mass} \]

5. **Freezing Point Depression**:
- Similarly, for freezing point depression, we can say:
\[ \Delta T_f = i \cdot K_f \cdot m \]
- Thus, the freezing point depression can also be expressed as:
\[ \text{Experimental } \Delta T_f = 5 \times \text{Normal Freezing Point} \]

### Summary of Correct Statements:
- The correct statements are:
1. The quantum factor is equal to 0.2.
2. The experimental elevation in boiling point is equal to the calculated elevation divided by 5.
3. The observed molar mass is equal to the normal molar mass multiplied by 5.
4. The freezing point depression is equal to the normal freezing point multiplied by 5.