The unit of ebillioscopic constant is

Options

A. K kg `mol^(-1)` or K `"(molality)"^(-1)`
B. mol kg `K^(-1)` or `K^(-1)` (molality )
C. kg `mol^(-1) K^(-1)` or `K^(-1) "(molality)"^(-1)`
D. K mol `kg^(-1)` or K (molality )

Answer

To determine the unit of the ebullioscopic constant (Kb), we can follow these steps:

### Step 1: Understand the relationship
The elevation of boiling point is given by the formula:
\[ \Delta T_b = K_b \times M \]
where:
- \(\Delta T_b\) is the elevation in boiling point,
- \(K_b\) is the ebullioscopic constant,
- \(M\) is the molality of the solution.

### Step 2: Rearrange the formula
From the equation, we can rearrange it to solve for \(K_b\):
\[ K_b = \frac{\Delta T_b}{M} \]

### Step 3: Identify the units
Now, we need to identify the units of \(\Delta T_b\) and \(M\):
- The unit of \(\Delta T_b\) (change in temperature) is in Kelvin (K).
- The unit of molality (M) is moles of solute per kilogram of solvent, which can be expressed as:
\[ \text{molality} = \text{moles of solute} \, \text{per} \, \text{kg of solvent} = \text{mol} \, \text{kg}^{-1} \]

### Step 4: Substitute the units into the equation
Substituting the units into the rearranged formula for \(K_b\):
\[ K_b = \frac{\text{K}}{\text{mol} \, \text{kg}^{-1}} = \text{K} \, \text{kg}^{-1} \, \text{mol} \]

### Step 5: Final unit of Kb
Thus, the unit of the ebullioscopic constant \(K_b\) is:
\[ \text{K} \, \text{kg}^{-1} \, \text{mol}^{-1} \]

### Conclusion
The unit of the ebullioscopic constant (Kb) is Kelvin per kilogram per mole (K kg⁻¹ mol⁻¹).

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