At 100°C and 1 atm, if the density of liquid water is `1.0 g cm^(-3)` and that of water vapour is 0.0006 gcm-3 , then the volume occupied by water molecules in 1 litre of steam at that temperature is:

Options

A. `6 cm^(3)`
B. `60 cm^(3)`
C. `0.6 cm^(3)`
D. `0.06 cm^(3)`

Explanation

To solve the problem, we need to find the volume occupied by water molecules in 1 liter of steam at 100°C and 1 atm.

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Density of liquid water = 1.0 g/cm³
- Density of water vapor (steam) = 0.0006 g/cm³
- Volume of steam = 1 liter = 1000 mL

2. **Calculate the Mass of 1 Liter of Steam:**
- We know that density = mass/volume. Therefore, we can rearrange the formula to find mass:
\[ \text{Mass of steam} = \text{Density of steam} \times \text{Volume of steam} \]
- Substituting the values:
\[ \text{Mass of steam} = 0.0006 \, \text{g/cm}^3 \times 1000 \, \text{cm}^3 = 0.6 \, \text{g} \]

3. **Calculate the Volume of Water Corresponding to the Mass of Steam:**
- Now, we need to find out how much volume this mass of steam would occupy if it were in the liquid state (water).
- We use the density of liquid water to find the volume:
\[ \text{Volume of water} = \frac{\text{Mass of steam}}{\text{Density of liquid water}} \]
- Substituting the values:
\[ \text{Volume of water} = \frac{0.6 \, \text{g}}{1.0 \, \text{g/cm}^3} = 0.6 \, \text{cm}^3 \]

4. **Conclusion:**
- The volume occupied by water molecules in 1 liter of steam at 100°C and 1 atm is **0.6 cm³**.

### Final Answer:
The volume occupied by water molecules in 1 liter of steam at 100°C is **0.6 cm³**.