A molecule `A_(2)B (Mw = 166.4)` occupies triclinic lattice with `a = 5 Å, b = 8 Å`, and `c = 4 Å`,. If the density of `A_(2)B` is `5.2 g cm^(-3)`, the number of molecules present in one unit cell is

Options

A. `2`
B. `3`
C. `4`
D. `5`

Explanation

To find the number of molecules present in one unit cell (denoted as \( Z \)), we can use the relationship between density, molecular weight, and the volume of the unit cell. Here’s a step-by-step solution:

### Step 1: Write down the given data
- Molecular weight (\( Mw \)) of \( A_2B = 166.4 \) g/mol
- Density (\( \rho \)) of \( A_2B = 5.2 \) g/cm³
- Lattice parameters:
- \( a = 5 \) Å = \( 5 \times 10^{-8} \) cm
- \( b = 8 \) Å = \( 8 \times 10^{-8} \) cm
- \( c = 4 \) Å = \( 4 \times 10^{-8} \) cm

### Step 2: Calculate the volume of the unit cell
The volume \( V \) of the unit cell in a triclinic lattice is given by:
\[ V = a \times b \times c \]
Substituting the values:
\[ V = (5 \times 10^{-8} \, \text{cm}) \times (8 \times 10^{-8} \, \text{cm}) \times (4 \times 10^{-8} \, \text{cm}) = 160 \times 10^{-24} \, \text{cm}^3 = 1.6 \times 10^{-22} \, \text{cm}^3 \]

### Step 3: Use the density formula
The density formula is given by:
\[ \rho = \frac{Z \times Mw}{V \times N_A} \]
Where:
- \( N_A \) is Avogadro's number, \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)

Rearranging the formula to solve for \( Z \):
\[ Z = \frac{\rho \times V \times N_A}{Mw} \]

### Step 4: Substitute the known values into the equation
Substituting the values we have:
\[ Z = \frac{5.2 \, \text{g/cm}^3 \times 1.6 \times 10^{-22} \, \text{cm}^3 \times 6.022 \times 10^{23} \, \text{mol}^{-1}}{166.4 \, \text{g/mol}} \]

### Step 5: Calculate \( Z \)
Calculating the numerator:
\[ 5.2 \times 1.6 \times 6.022 \approx 50.0864 \]
Now, substituting:
\[ Z = \frac{50.0864 \times 10^{-22} \times 10^{23}}{166.4} \]
\[ Z = \frac{50.0864}{166.4} \approx 0.301 \]
Now multiplying by \( 10^{-1} \) (since we have \( 10^{-22} \times 10^{23} = 10^{-1} \)):
\[ Z \approx 3 \]

### Conclusion
Thus, the number of molecules present in one unit cell is approximately \( Z = 3 \).