The unit of ebullioscopic 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) or k^(-1) ("molality")^(-1)`
D. K mol` kg^(-1)` or K (molality)

Answer

To determine the unit of the ebullioscopic constant (Kb), we start with the formula that relates the elevation in boiling point (ΔTb) to the molal elevation constant (Kb) and the molality (m) of the solution:

### Step 1: Write the formula
The formula is given by:
\[ \Delta T_b = K_b \times m \]
where:
- \( \Delta T_b \) = elevation in boiling point (measured in °C or K)
- \( K_b \) = ebullioscopic constant (what we want to find)
- \( m \) = molality of the solution (measured in moles of solute per kg of solvent)

### Step 2: Rearrange the formula to isolate Kb
To find the unit of \( K_b \), we can rearrange the formula:
\[ K_b = \frac{\Delta T_b}{m} \]

### Step 3: Identify the units
1. The unit of \( \Delta T_b \) (elevation in boiling point) is degrees Celsius (°C) or Kelvin (K). Since the change in temperature is the same in both scales, we can use either.
2. The unit of molality (m) is moles of solute per kilogram of solvent, which can be expressed as:
\[ \text{molality} = \frac{\text{moles of solute}}{\text{kg of solvent}} \]

### Step 4: Substitute the units into the equation
Now substituting the units into the equation for \( K_b \):
\[ K_b = \frac{\text{°C (or K)}}{\frac{\text{moles}}{\text{kg}}} \]

### Step 5: Simplify the units
This can be simplified to:
\[ K_b = \text{°C} \cdot \frac{\text{kg}}{\text{moles}} \]
or
\[ K_b = \text{K} \cdot \frac{\text{kg}}{\text{moles}} \]

### Conclusion
Thus, the unit of the ebullioscopic constant \( K_b \) is:
\[ \text{°C kg mol}^{-1} \]
or equivalently:
\[ \text{K kg mol}^{-1} \]