Calculate `K_(b)` of water when 1 mole of the solute is dissolved in 1000g of water. The latent heat of vaporisation of water is `539.9` calories per gram.

Answer

To calculate the ebullioscopic constant \( K_b \) of water when 1 mole of solute is dissolved in 1000 g of water, we can use the formula:

\[ K_b = \frac{R \cdot T_b^2}{1000 \cdot L} \]

where:
- \( R \) is the gas constant,
- \( T_b \) is the boiling point of the solvent in Kelvin,
- \( L \) is the latent heat of vaporization in calories per gram.

### Step 1: Identify the values
- The gas constant \( R = 0.0821 \, \text{L} \cdot \text{atm} / \text{K} \cdot \text{mol} \).
- The boiling point of water \( T_b = 100^\circ C = 373 \, K \).
- The latent heat of vaporization \( L = 539.9 \, \text{cal/g} \).

### Step 2: Convert latent heat to the appropriate units
Since we need to use calories per gram, we keep \( L \) as \( 539.9 \, \text{cal/g} \).

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

\[ K_b = \frac{0.0821 \cdot (373)^2}{1000 \cdot 539.9} \]

### Step 4: Calculate \( T_b^2 \)
Calculate \( T_b^2 \):

\[ T_b^2 = 373^2 = 139129 \]

### Step 5: Substitute \( T_b^2 \) into the equation
Now substitute \( T_b^2 \) back into the equation:

\[ K_b = \frac{0.0821 \cdot 139129}{1000 \cdot 539.9} \]

### Step 6: Calculate the numerator
Calculate the numerator:

\[ 0.0821 \cdot 139129 = 11441.07949 \]

### Step 7: Calculate the denominator
Calculate the denominator:

\[ 1000 \cdot 539.9 = 539900 \]

### Step 8: Final calculation
Now divide the numerator by the denominator:

\[ K_b = \frac{11441.07949}{539900} \approx 0.0212 \]

### Step 9: Convert to degrees Celsius
Since \( K_b \) is typically expressed in degrees Celsius, we can conclude that:

\[ K_b \approx 0.0212 \, ^\circ C \]

### Final Answer
Thus, the value of \( K_b \) for water is approximately \( 0.0212 \, ^\circ C \).

---