The units of ebullioscopic constant is.

Options

A. ` K kg mol^(-1)`
B. ` mol kg k^(-1)`
C. `k mol kg^(-1)`
D. none of the above

Answer

To determine the units of the ebullioscopic constant (K_b), we can start by analyzing the relationship given in the formula for boiling point elevation:

### Step-by-Step Solution:

1. **Understand the Formula**: The formula for boiling point elevation is given by:
\[ \Delta T_b = i \cdot K_b \cdot m \]
where:
- \(\Delta T_b\) = change in boiling point (in Kelvin)
- \(i\) = van 't Hoff factor (unitless)
- \(K_b\) = ebullioscopic constant (what we want to find)
- \(m\) = molality (in mol/kg)

2. **Identify the Units**:
- The unit of \(\Delta T_b\) is Kelvin (K).
- The unit of \(i\) is unitless (it is a ratio).
- The unit of \(m\) (molality) is moles of solute per kilogram of solvent, which can be expressed as:
\[ m = \text{mol/kg} \quad \text{or} \quad \text{mol kg}^{-1} \]

3. **Rearranging the Formula**: To find the unit of \(K_b\), we can rearrange the formula:
\[ K_b = \frac{\Delta T_b}{i \cdot m} \]

4. **Substituting the Units**: Since \(i\) is unitless, we can simplify the equation:
\[ K_b = \frac{\Delta T_b}{m} = \frac{\text{K}}{\text{mol kg}^{-1}} \]

5. **Final Unit Calculation**: This simplifies to:
\[ K_b = \text{K} \cdot \text{kg} \cdot \text{mol}^{-1} \]
Therefore, the units of the ebullioscopic constant \(K_b\) are:
\[ \text{K kg mol}^{-1} \]

### Final Answer:
The units of the ebullioscopic constant (K_b) are \( \text{K kg mol}^{-1} \).