The entropy of vaporization of a liquid is 58 J`K^(-1) mol^(-1)`.If 100 g of its vapour condenses at its boiling point of `123^(@)C`, the value of entropy change for the process is (molar mass of the liquid = `58 mol^(-1)`).

Options

A. `-100 JK^(-1)`
B. `100 JK^(-1)`
C. `123 JK^(-1)`
D. `123 JK^(-1)`

Explanation

To find the entropy change for the condensation of 100 g of vapor at its boiling point, we can follow these steps:

### Step 1: Identify the given values
- Entropy of vaporization (ΔSvap) = 58 J/K·mol
- Molar mass of the liquid = 58 g/mol
- Mass of vapor condensing = 100 g

### Step 2: Calculate the number of moles of vapor
To find the number of moles of the vapor that condenses, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{100 \, \text{g}}{58 \, \text{g/mol}} \approx 1.724 \, \text{mol} \]

### Step 3: Determine the entropy change for condensation
The entropy change for condensation (ΔScond) is the negative of the entropy of vaporization:
\[ \Delta S_{\text{cond}} = -\Delta S_{\text{vap}} \]
Substituting the value:
\[ \Delta S_{\text{cond}} = -58 \, \text{J/K·mol} \]

### Step 4: Calculate the total entropy change for 100 g of vapor
Now, we can calculate the total entropy change for the 1.724 moles of vapor that condenses:
\[ \Delta S_{\text{total}} = \Delta S_{\text{cond}} \times \text{Number of moles} \]
Substituting the values:
\[ \Delta S_{\text{total}} = -58 \, \text{J/K·mol} \times 1.724 \, \text{mol} \approx -100 \, \text{J/K} \]

### Final Answer
The value of the entropy change for the process is approximately:
\[ \Delta S_{\text{total}} \approx -100 \, \text{J/K} \]

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