Calculate the ebullioscopic constant for water. The heat of vaporisation is 40.685 kJ ` mol^(-1)`

Options

A. 0.512 K kg `mol^(-1)`
B. 1.86 K kg `mol^(-1)`
C. 5.12 K kg `mol^(-1)`
D. 3.56 K kg `mol^(-1)`

Answer

To calculate the ebullioscopic constant (K_b) for water using the provided heat of vaporization, we can follow these steps:

### Step 1: Write down the formula for K_b
The formula for the ebullioscopic constant (K_b) is given by:

\[ K_b = \frac{R \cdot T_b^2 \cdot M}{\Delta H_{vap}} \]

Where:
- \( R \) = universal gas constant
- \( T_b \) = boiling point of the solvent (in Kelvin)
- \( M \) = molar mass of the solvent (in kg/mol)
- \( \Delta H_{vap} \) = heat of vaporization (in Joules/mol)

### Step 2: Identify the values needed for the calculation
- **R**: The universal gas constant \( R = 8.314 \, \text{J/mol·K} \)
- **T_b**: The boiling point of water is \( 100^\circ C \) which is \( 373 \, \text{K} \)
- **M**: The molar mass of water is \( 18 \, \text{g/mol} \) which can be converted to kg:
\[ M = 18 \times 10^{-3} \, \text{kg/mol} \]
- **ΔH_vap**: The heat of vaporization given is \( 40.685 \, \text{kJ/mol} \), which needs to be converted to Joules:
\[ \Delta H_{vap} = 40.685 \times 10^3 \, \text{J/mol} \]

### Step 3: Substitute the values into the formula
Now we can substitute the values into the formula:

\[ K_b = \frac{8.314 \, \text{J/mol·K} \cdot (373 \, \text{K})^2 \cdot (18 \times 10^{-3} \, \text{kg/mol})}{40.685 \times 10^3 \, \text{J/mol}} \]

### Step 4: Calculate \( K_b \)
1. Calculate \( (373 \, \text{K})^2 \):
\[ 373^2 = 139129 \, \text{K}^2 \]

2. Substitute this back into the equation:
\[ K_b = \frac{8.314 \cdot 139129 \cdot (18 \times 10^{-3})}{40.685 \times 10^3} \]

3. Calculate the numerator:
\[ 8.314 \cdot 139129 \cdot 18 \times 10^{-3} = 8.314 \cdot 139129 \cdot 0.018 = 22.095 \, \text{J·K·kg/mol} \]

4. Calculate the denominator:
\[ 40.685 \times 10^3 = 40685 \, \text{J/mol} \]

5. Now calculate \( K_b \):
\[ K_b = \frac{22.095}{40.685} \approx 0.543 \, \text{K·kg/mol} \]

### Step 5: Final result
After performing the calculations, we find that the ebullioscopic constant \( K_b \) for water is approximately \( 0.543 \, \text{K·kg/mol} \).