Ice crystallises in hexagonal lattice having volume of unit cell is `132times10^(-24)cm^(3)`.If density is 0.92g `cm^(3)`at a given temperature, then number of water molecules per unit cell is

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To find the number of water molecules per unit cell (denoted as Z) in ice, we can use the formula for density:

\[ D = \frac{Z \cdot M}{V \cdot N_A} \]

Where:
- \( D \) = density of the substance (in g/cm³)
- \( Z \) = number of molecules per unit cell
- \( M \) = molar mass of the substance (in g/mol)
- \( V \) = volume of the unit cell (in cm³)
- \( N_A \) = Avogadro's number (approximately \( 6.022 \times 10^{23} \) molecules/mol)

### Step-by-Step Solution:

1. **Identify Given Values**:
- Volume of unit cell, \( V = 132 \times 10^{-24} \, \text{cm}^3 \)
- Density, \( D = 0.92 \, \text{g/cm}^3 \)
- Molar mass of water, \( M = 18 \, \text{g/mol} \)
- Avogadro's number, \( N_A = 6.022 \times 10^{23} \, \text{molecules/mol} \)

2. **Rearrange the Density Formula**:
We need to solve for \( Z \):
\[ Z = \frac{D \cdot V \cdot N_A}{M} \]

3. **Substitute the Values**:
\[ Z = \frac{0.92 \, \text{g/cm}^3 \times 132 \times 10^{-24} \, \text{cm}^3 \times 6.022 \times 10^{23} \, \text{molecules/mol}}{18 \, \text{g/mol}} \]

4. **Calculate the Numerator**:
\[ \text{Numerator} = 0.92 \times 132 \times 10^{-24} \times 6.022 \times 10^{23} \]
- First, calculate \( 0.92 \times 132 = 121.44 \)
- Then, \( 121.44 \times 6.022 \approx 7312.57 \)
- Finally, \( 7312.57 \times 10^{-24} \)

5. **Calculate the Denominator**:
\[ \text{Denominator} = 18 \]

6. **Final Calculation**:
\[ Z = \frac{7312.57 \times 10^{-24}}{18} \approx 4.06 \]

Since \( Z \) must be a whole number, we round it to \( 4 \).

### Conclusion:
The number of water molecules per unit cell in ice is \( Z = 4 \).