Match the boiling point with `K_(b)` for `x,y` and `z`, if molecular weight of `x,y` and `z` are same.
`{:(,b.pt,,k_(b),),(x,100,,0.68,),(y,27,,0.53,),(z,253,,0.98,):}`

Explanation

To solve the problem of matching the boiling points with their respective \( K_b \) values for substances \( x \), \( y \), and \( z \) given that their molecular weights are the same, we can follow these steps:

### Step 1: Understand the relationship between boiling point and \( K_b \)
The relationship we need to consider is that \( K_b \) is directly proportional to the square of the boiling point when the molecular weight and other constants are the same. This means that if the boiling point increases, \( K_b \) also increases.

### Step 2: List the boiling points and \( K_b \) values
From the question, we have:
- Boiling points:
- \( x: 100 \, °C \)
- \( y: 27 \, °C \)
- \( z: 253 \, °C \)

- \( K_b \) values:
- \( x: 0.68 \)
- \( y: 0.53 \)
- \( z: 0.98 \)

### Step 3: Compare boiling points
Now we will compare the boiling points to determine their order:
- Highest boiling point: \( z (253 \, °C) \)
- Middle boiling point: \( x (100 \, °C) \)
- Lowest boiling point: \( y (27 \, °C) \)

### Step 4: Match boiling points with \( K_b \) values
Since \( K_b \) is directly proportional to the square of the boiling point, we can match them as follows:
- The highest boiling point \( z (253 \, °C) \) will correspond to the highest \( K_b \) value \( 0.98 \).
- The middle boiling point \( x (100 \, °C) \) will correspond to the middle \( K_b \) value \( 0.68 \).
- The lowest boiling point \( y (27 \, °C) \) will correspond to the lowest \( K_b \) value \( 0.53 \).

### Final Matching:
- \( z \) matches with \( 0.98 \)
- \( x \) matches with \( 0.68 \)
- \( y \) matches with \( 0.53 \)

### Summary of Matches:
- \( z (253 \, °C) \) → \( K_b = 0.98 \)
- \( x (100 \, °C) \) → \( K_b = 0.68 \)
- \( y (27 \, °C) \) → \( K_b = 0.53 \)