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

Options

A. 43.4 J/K-mole
B. 87.2 J/K mole
C. 900 JK/mol
D. Zero

Explanation

To calculate the increase in entropy when 900 J of heat is exchanged at the boiling point of water, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between entropy, heat, and temperature**:
The change in entropy (ΔS) can be calculated using the formula:
\[ \Delta S = \frac{\Delta H}{T} \]
where ΔH is the change in enthalpy (or heat exchanged) and T is the absolute temperature in Kelvin.

2. **Identify the given values**:
- The heat exchanged (ΔH) = 900 J/g
- The boiling point of water = 100°C = 373 K (since we need to convert Celsius to Kelvin, we add 273).

3. **Convert the heat exchanged to per mole**:
We know that 1 mole of water weighs 18 g. Therefore, the total heat exchanged for 1 mole of water is:
\[ \Delta H = 900 \, \text{J/g} \times 18 \, \text{g} = 16200 \, \text{J} \]

4. **Substitute the values into the entropy formula**:
Now we can substitute ΔH and T into the entropy formula:
\[ \Delta S = \frac{16200 \, \text{J}}{373 \, \text{K}} \]

5. **Calculate the increase in entropy**:
Performing the calculation:
\[ \Delta S \approx 43.4 \, \text{J/K} \]

### Final Answer:
The increase in entropy when 900 J of heat is exchanged at the boiling point of water is approximately **43.4 J/K**.

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