Volume occupied by one molecule of water (density = 1 g `cm^(3)`)

Options

A. `3.0xx10^(-23) cm^(3)`
B. `5.5xx10^(-23)cm^(3)`
C. `9.0xx10^(-23)cm^(3)`
D. `6.023xx10^(-23)cm^(3)`

Explanation

To find the volume occupied by one molecule of water, we can follow these steps:

### Step 1: Understand the given data
We know the density of water is 1 g/cm³ and the molecular mass of water (H₂O) is 18 g/mol.

### Step 2: Calculate the mass of one molecule of water
To find the mass of one molecule of water, we use Avogadro's number, which is approximately \(6.022 \times 10^{23}\) molecules/mol.

The mass of one molecule of water can be calculated using the formula:
\[ \text{Mass of one molecule} = \frac{\text{Molar mass}}{\text{Avogadro's number}} = \frac{18 \text{ g/mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \]
Calculating this gives:
\[ \text{Mass of one molecule} \approx 3.0 \times 10^{-23} \text{ g} \]

### Step 3: Use the density to find the volume
We know the formula relating density, mass, and volume:
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies \text{Volume} = \frac{\text{Mass}}{\text{Density}} \]

Substituting the values we have:
\[ \text{Volume} = \frac{3.0 \times 10^{-23} \text{ g}}{1 \text{ g/cm}^3} \]

### Step 4: Calculate the volume
Since the density is 1 g/cm³, the calculation simplifies to:
\[ \text{Volume} = 3.0 \times 10^{-23} \text{ cm}^3 \]

### Conclusion
The volume occupied by one molecule of water is:
\[ \text{Volume} = 3.0 \times 10^{-23} \text{ cm}^3 \]