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

### Step 2: Rearrange the formula to solve for Z
We can rearrange the density formula to isolate \( Z \):

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

### Step 3: Substitute the known values into the equation
Now we can substitute the values we have into the rearranged formula:

\[ Z = \frac{0.92 \, \text{g/cm}^3 \cdot (132 \times 10^{-24} \, \text{cm}^3) \cdot (6.022 \times 10^{23} \, \text{molecules/mol})}{18 \, \text{g/mol}} \]

### Step 4: Calculate the numerator
First, calculate the numerator:

\[ 0.92 \cdot 132 \times 10^{-24} \cdot 6.022 \times 10^{23} \]

Calculating step-by-step:
- \( 0.92 \cdot 132 = 121.44 \)
- \( 121.44 \times 10^{-24} \cdot 6.022 \times 10^{23} = 121.44 \cdot 6.022 \times 10^{-1} = 73.07168 \)

### Step 5: Calculate Z
Now, divide by the molar mass:

\[ Z = \frac{73.07168}{18} \approx 4.06 \]

### Step 6: Round to the nearest whole number
Since the number of molecules must be a whole number, we round \( 4.06 \) to \( 4 \).

### Conclusion
Thus, the number of water molecules per unit cell in ice is:

\[ \boxed{4} \]