If the water molecules in `1.0 g` of water were distributed uniformly over the surface of earth, how many such molecules would there be in `1.0 cm^2` of earth's surface ?

Explanation

To solve the problem of how many water molecules would be present in `1.0 cm²` of Earth's surface if `1.0 g` of water were distributed uniformly, we can follow these steps:

### Step 1: Calculate the number of moles of water in `1.0 g`
The molecular mass of water (H₂O) is approximately `18 g/mol`. We can use the formula for the number of moles:

\[ n = \frac{m}{M} \]

Where:
- \( n \) = number of moles
- \( m \) = mass of water = `1.0 g`
- \( M \) = molar mass of water = `18 g/mol`

Substituting the values:

\[ n = \frac{1.0 \, \text{g}}{18 \, \text{g/mol}} \approx 0.05556 \, \text{mol} \]

### Step 2: Calculate the total number of water molecules
Using Avogadro's number, which is approximately \( 6.02 \times 10^{23} \) molecules/mol, we can find the total number of molecules:

\[ N = n \times N_A \]

Where:
- \( N \) = total number of molecules
- \( N_A \) = Avogadro's number

Substituting the values:

\[ N = 0.05556 \, \text{mol} \times 6.02 \times 10^{23} \, \text{molecules/mol} \approx 3.34 \times 10^{22} \, \text{molecules} \]

### Step 3: Calculate the surface area of the Earth
The average radius of the Earth is approximately `6400 km`, which can be converted to centimeters:

\[ r = 6400 \, \text{km} = 6400 \times 10^5 \, \text{cm} \]

The surface area \( A \) of the Earth can be calculated using the formula for the surface area of a sphere:

\[ A = 4 \pi r^2 \]

Substituting the radius:

\[ A = 4 \pi (6400 \times 10^5 \, \text{cm})^2 \]

Calculating this gives:

\[ A \approx 4 \pi (4.096 \times 10^{11} \, \text{cm}^2) \approx 5.15 \times 10^{12} \, \text{cm}^2 \]

### Step 4: Calculate the number of molecules per `1.0 cm²`
To find the number of molecules in `1.0 cm²`, we divide the total number of molecules by the total surface area:

\[ \text{Molecules per } 1.0 \, \text{cm}^2 = \frac{N}{A} \]

Substituting the values:

\[ \text{Molecules per } 1.0 \, \text{cm}^2 = \frac{3.34 \times 10^{22} \, \text{molecules}}{5.15 \times 10^{12} \, \text{cm}^2} \approx 6.48 \times 10^{9} \, \text{molecules/cm}^2 \]

### Final Answer
Approximately \( 6.48 \times 10^{9} \) water molecules would be present in `1.0 cm²` of Earth's surface.

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