If `900J//g `of heat is exchanged at boiling point of water then water is the increase in entropy.

Options

A. `43.4J//mol`
B. `87.2J//mol`
C. `900J//mol`
D. Zero

Explanation

To calculate the increase in entropy when heat is exchanged at the boiling point of water, we can use the formula for entropy change (ΔS):

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

where:
- \( \Delta S \) is the change in entropy,
- \( q \) is the heat exchanged,
- \( T \) is the absolute temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the heat exchanged (q)**:
Given that the heat exchanged is \( 900 \, \text{J/g} \), we need to convert this to a per mole basis since entropy is typically expressed in terms of moles. The molar mass of water (H₂O) is approximately \( 18 \, \text{g/mol} \).

\[ q = 900 \, \text{J/g} \times 18 \, \text{g/mol} = 16200 \, \text{J/mol} \]

2. **Determine the temperature (T)**:
The boiling point of water is \( 100^\circ C \). To convert this to Kelvin:

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

3. **Calculate the change in entropy (ΔS)**:
Now we can substitute the values of \( q \) and \( T \) into the entropy change formula:

\[ \Delta S = \frac{q}{T} = \frac{16200 \, \text{J/mol}}{373.15 \, \text{K}} \approx 43.4 \, \text{J/(mol·K)} \]

### Final Answer:
The increase in entropy of water when \( 900 \, \text{J/g} \) of heat is exchanged at the boiling point is approximately \( 43.4 \, \text{J/(mol·K)} \).

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