9.0 g of `H_2 O` is vaporized at `100^@`C and 1 atm pressure. If the latent heat of vaporization of water is x J/g , then `Delta S` is given by ______ .

Options

A. `x/373`
B. `(18x)/100`
C. `(18x)/373 `
D. `1/2 xx(18x)/373`

Answer

To solve the problem, we need to calculate the change in entropy (ΔS) when 9.0 g of water is vaporized at 100°C (373 K) and 1 atm pressure. The latent heat of vaporization of water is given as x J/g.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of water (m) = 9.0 g
- Temperature (T) = 100°C = 373 K
- Latent heat of vaporization (L) = x J/g

2. **Calculate the Heat Absorbed (ΔH):**
The heat absorbed during the vaporization of water can be calculated using the formula:
\[ \Delta H = m \times L \]
Substituting the values we have:
\[ \Delta H = 9.0 \, \text{g} \times x \, \text{J/g} = 9x \, \text{J} \]

3. **Calculate the Change in Entropy (ΔS):**
The change in entropy is given by the formula:
\[ \Delta S = \frac{\Delta H}{T} \]
Substituting ΔH and T into the equation:
\[ \Delta S = \frac{9x \, \text{J}}{373 \, \text{K}} = \frac{9x}{373} \, \text{J/K} \]

4. **Final Expression:**
Thus, the change in entropy (ΔS) when 9.0 g of water is vaporized at 100°C is:
\[ \Delta S = \frac{9x}{373} \, \text{J/K} \]