Calculate the entropy change for vaporization of water if latent heat of vaporization for water is 2.26 kJ/gram. The `K_(b)` for `H_(2)O` is 0.51 K/molality

Options

A. 26.33cal/mole
B. 23.33cal/mole
C. 1KJ/mole
D. 30cal/mole

Explanation

To calculate the entropy change for the vaporization of water, we can follow these steps:

### Step 1: Understand the relationship between latent heat and entropy change
The latent heat of vaporization (ΔH) is related to the entropy change (ΔS) during the phase transition by the equation:
\[ \Delta S = \frac{\Delta H}{T} \]
where \( T \) is the temperature in Kelvin at which the phase change occurs.

### Step 2: Convert the latent heat of vaporization to appropriate units
The latent heat of vaporization for water is given as 2.26 kJ/gram. We need to convert this to calories per mole.

1 kJ = 239.005736 calories, so:
\[ \Delta H = 2.26 \, \text{kJ/g} \times 1000 \, \text{g} = 2260 \, \text{kJ} = 2260 \times 239.005736 \, \text{calories} = 540,000 \, \text{calories} \]

### Step 3: Determine the boiling point of water in Kelvin
The boiling point of water is 100°C, which can be converted to Kelvin:
\[ T = 100 + 273.15 = 373.15 \, \text{K} \]

### Step 4: Calculate the entropy change
Using the formula for entropy change:
\[ \Delta S = \frac{\Delta H}{T} \]
Substituting the values we have:
\[ \Delta S = \frac{540,000 \, \text{calories}}{373.15 \, \text{K}} \approx 1446.8 \, \text{calories/K} \]

### Step 5: Convert to per mole
Since we need the entropy change per mole, we also need to divide by the molar mass of water (18 g/mol):
\[ \Delta S_{\text{per mole}} = \frac{1446.8 \, \text{calories/K}}{18 \, \text{g/mol}} \approx 80.37 \, \text{calories/mol/K} \]

### Step 6: Final result
The final entropy change for the vaporization of water is approximately:
\[ \Delta S \approx 80.37 \, \text{calories/mol/K} \]