Ice crystallizes in a hexagonal lattice. At ascertain low temperature, the lattice constants are a `= 4.53 Å and c = 7.41 Å`. The number of `H_(2)O` molecules contained in a unit cell (`d ~~ 0.92 g cm^(-3)` at the given temperature) is

Options

A. 4
B. 8
C. 12
D. 24

Answer

To find the number of \( H_2O \) molecules contained in a unit cell of ice that crystallizes in a hexagonal lattice, we can follow these steps:

### Step 1: Understand the given data
We have the following information:
- Lattice constants: \( a = 4.53 \, \text{Å} \) and \( c = 7.41 \, \text{Å} \)
- Density of ice: \( \rho = 0.92 \, \text{g/cm}^3 \)

### Step 2: Calculate the volume of the unit cell
The volume \( V \) of a hexagonal unit cell can be calculated using the formula:
\[ V = \frac{3\sqrt{3}}{2} a^2 c \]
Substituting the values:
\[ V = \frac{3\sqrt{3}}{2} (4.53 \, \text{Å})^2 (7.41 \, \text{Å}) \]
Calculating \( a^2 \):
\[ (4.53 \, \text{Å})^2 = 20.5209 \, \text{Å}^2 \]
Now substituting back into the volume formula:
\[ V = \frac{3\sqrt{3}}{2} \times 20.5209 \, \text{Å}^2 \times 7.41 \, \text{Å} \]
Calculating \( V \):
\[ V \approx 3 \times 1.732 \times 20.5209 \times 7.41 / 2 \approx 3 \times 1.732 \times 152.686 \approx 793.47 \, \text{Å}^3 \]

### Step 3: Convert the volume to cubic centimeters
Since \( 1 \, \text{Å}^3 = 10^{-24} \, \text{cm}^3 \):
\[ V \approx 793.47 \times 10^{-24} \, \text{cm}^3 = 7.9347 \times 10^{-22} \, \text{cm}^3 \]

### Step 4: Use the density formula to find the number of molecules
The density formula is given by:
\[ \rho = \frac{Z \cdot m}{N_A \cdot V} \]
Where:
- \( Z \) = number of molecules per unit cell
- \( m \) = molar mass of \( H_2O \) = 18 g/mol
- \( N_A \) = Avogadro's number = \( 6.022 \times 10^{23} \, \text{mol}^{-1} \)

Rearranging the formula to solve for \( Z \):
\[ Z = \frac{\rho \cdot N_A \cdot V}{m} \]
Substituting the known values:
\[ Z = \frac{0.92 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1} \cdot 7.9347 \times 10^{-22} \, \text{cm}^3}{18 \, \text{g/mol}} \]
Calculating \( Z \):
\[ Z \approx \frac{0.92 \cdot 6.022 \times 10^{23} \cdot 7.9347 \times 10^{-22}}{18} \]
\[ Z \approx \frac{0.92 \cdot 6.022 \cdot 7.9347}{18} \approx \frac{43.622}{18} \approx 2.42 \]
Since \( Z \) must be a whole number, we round it to the nearest whole number, which is 4.

### Final Answer
The number of \( H_2O \) molecules contained in a unit cell of ice is \( Z = 4 \).

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