If the density of water is 1 g `cm^(-3)` then the volume occupied by one molecule of water is approximately

Options

A. `3xx10^(-23)mL`
B. `6xx10^(-22)mL`
C. `3xx10^(-21)mL`
D. `9xx10^(-23)mL`

Explanation

To find the volume occupied by one molecule of water given that the density of water is 1 g/cm³, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Density of water (ρ) = 1 g/cm³
- Molar mass of water (M) = 18 g/mol

2. **Calculate the Volume of One Mole of Water:**
- The volume (V) of one mole of water can be calculated using the formula:
\[ V = \frac{\text{mass}}{\text{density}} = \frac{M}{ρ} \]
- Substituting the values:
\[ V = \frac{18 \text{ g}}{1 \text{ g/cm}^3} = 18 \text{ cm}^3 \]

3. **Determine the Number of Molecules in One Mole:**
- From Avogadro's number, we know that one mole of any substance contains approximately \(6.022 \times 10^{23}\) molecules.

4. **Calculate the Volume Occupied by One Molecule:**
- The volume occupied by one molecule (v) can be calculated by dividing the total volume of one mole by Avogadro's number:
\[ v = \frac{V}{N_A} \]
- Where \(N_A = 6.022 \times 10^{23} \text{ molecules/mol}\).
- Substituting the values:
\[ v = \frac{18 \text{ cm}^3}{6.022 \times 10^{23}} \approx 2.989 \times 10^{-23} \text{ cm}^3 \]

5. **Round the Result:**
- For simplicity, we can round this value to:
\[ v \approx 3 \times 10^{-23} \text{ cm}^3 \]

### Final Answer:
The volume occupied by one molecule of water is approximately \(3 \times 10^{-23} \text{ cm}^3\).

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