Consider the three solvents of identical molar masses. Match their boiling point with their `K_(b)` values
`{:(ul(bar("Solvents" " ""Boiling point" " "K_(b) "values" " "))),(ul(X " " 100^(@)C " " 0.92 " ")),(ul(Y " " 27^(@)C " "0.63" ")),(ul(Z " " 283^(@)C " "0.53 " ")):}`

Answer

To solve the problem of matching the boiling points of the three solvents with their respective \( K_b \) values, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship**: The boiling point elevation (\( \Delta T_b \)) is given by the formula:
\[ \Delta T_b = K_b \cdot m \cdot i \]
where \( K_b \) is the ebullioscopic constant, \( m \) is the molality, and \( i \) is the van 't Hoff factor. Since the molar masses of the solvents are identical, we can assume the molalities are the same for comparison.

2. **Identify the Variables**: We are given three solvents (X, Y, Z) with their boiling points and \( K_b \) values:
- Solvent X: Boiling Point = 100°C, \( K_b = 0.92 \)
- Solvent Y: Boiling Point = 27°C, \( K_b = 0.63 \)
- Solvent Z: Boiling Point = 283°C, \( K_b = 0.53 \)

3. **Analyze the \( K_b \) Values**: Since \( \Delta T_b \) is directly proportional to \( K_b \) (assuming constant molality and \( i \)), a higher \( K_b \) value indicates a greater boiling point elevation. Thus, we can rank the solvents based on their \( K_b \) values:
- Highest \( K_b \): 0.92 (X)
- Moderate \( K_b \): 0.63 (Y)
- Lowest \( K_b \): 0.53 (Z)

4. **Match \( K_b \) Values with Boiling Points**: Since higher \( K_b \) values correspond to higher boiling points, we can match them as follows:
- Solvent with \( K_b = 0.92 \) (X) should have the highest boiling point, which is 283°C (Z).
- Solvent with \( K_b = 0.63 \) (Y) should have a moderate boiling point, which is 100°C (X).
- Solvent with \( K_b = 0.53 \) (Z) should have the lowest boiling point, which is 27°C (Y).

5. **Final Matching**:
- Solvent X (100°C) matches with \( K_b = 0.63 \)
- Solvent Y (27°C) matches with \( K_b = 0.53 \)
- Solvent Z (283°C) matches with \( K_b = 0.92 \)

### Summary of Matches:
- X: 100°C → \( K_b = 0.63 \)
- Y: 27°C → \( K_b = 0.53 \)
- Z: 283°C → \( K_b = 0.92 \)