Calculate the entropy change when `3.6g` of liquid water is completely converted into vapour at `100^(@)C` . The molar heat of vaporization is `40.85KJ mol^(-1)` .

Options

A. `6.08 JK^(-1)`
B. `109.5JK^(-1)`
C. `21.89JK^(-1)`
D. `-21.89JK^(-1)`

Explanation

To calculate the entropy change when 3.6 g of liquid water is completely converted into vapor at 100°C, we will follow these steps:

### Step 1: Determine the molar heat of vaporization in joules
The molar heat of vaporization (ΔH) is given as 40.85 kJ/mol. To convert this to joules, we multiply by 1000 (since 1 kJ = 1000 J).

\[ \Delta H = 40.85 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 40850 \, \text{J/mol} \]

### Step 2: Convert the temperature from Celsius to Kelvin
The temperature is given as 100°C. To convert this to Kelvin, we add 273.15.

\[ T = 100 + 273.15 = 373.15 \, \text{K} \]

### Step 3: Calculate the change in entropy for one mole
The formula for the change in entropy (ΔS) is given by:

\[ \Delta S = \frac{\Delta H}{T} \]

Substituting the values we calculated:

\[ \Delta S = \frac{40850 \, \text{J/mol}}{373.15 \, \text{K}} \approx 109.38 \, \text{J/(mol·K)} \]

### Step 4: Calculate the number of moles of water
To find the number of moles of water in 3.6 g, we use the molar mass of water, which is approximately 18 g/mol.

\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{3.6 \, \text{g}}{18 \, \text{g/mol}} = 0.2 \, \text{mol} \]

### Step 5: Calculate the total change in entropy for 3.6 g of water
Now, we need to multiply the change in entropy per mole by the number of moles:

\[ \Delta S_{\text{total}} = \Delta S \times \text{Number of moles} = 109.38 \, \text{J/(mol·K)} \times 0.2 \, \text{mol} = 21.88 \, \text{J/K} \]

### Final Result
The total change in entropy when 3.6 g of liquid water is converted into vapor at 100°C is approximately:

\[ \Delta S_{\text{total}} \approx 21.88 \, \text{J/K} \]