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

Options

A. `3.0 xx 10^(-23) cm^3`
B. `5.5 xx 10^(-23) cm^3 `
C. `9.0 xx 10^(-23) cm^3`
D. `6.023 xx 10^(-23)cm^3`

Answer

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

### Step 1: Understand the given data
We are given:
- Density of water = 1 g/cm³
- Molar mass of water (H₂O) = 18 g/mol

### Step 2: Calculate the molar volume of water
The molar volume is the volume occupied by one mole of a substance and can be calculated using the formula:
\[ \text{Molar Volume} = \frac{\text{Molar Mass}}{\text{Density}} \]

Substituting the values:
\[ \text{Molar Volume} = \frac{18 \text{ g/mol}}{1 \text{ g/cm}^3} = 18 \text{ cm}^3/\text{mol} \]

### Step 3: Use Avogadro's number to find the volume of one molecule
Avogadro's number (Nₐ) is approximately \( 6.022 \times 10^{23} \) molecules/mol. To find the volume occupied by one molecule, we divide the molar volume by Avogadro's number:
\[ \text{Volume of one molecule} = \frac{\text{Molar Volume}}{N_a} \]

Substituting the values:
\[ \text{Volume of one molecule} = \frac{18 \text{ cm}^3/\text{mol}}{6.022 \times 10^{23} \text{ molecules/mol}} \]

### Step 4: Perform the calculation
Calculating the above expression:
\[ \text{Volume of one molecule} = \frac{18}{6.022 \times 10^{23}} \]
\[ \text{Volume of one molecule} \approx 2.989 \times 10^{-23} \text{ cm}^3 \]

### Step 5: Round the result
We can round this to:
\[ \text{Volume of one molecule} \approx 3 \times 10^{-23} \text{ cm}^3 \]

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

---