A compound AB has a rock salt type structure with A:B=1:1 . The formula weight of AB is 6.023 y amu and the closest A-B distance is `y^(1//3)` nm. Calculate the density of lattice (in `kg//m^(3))`

Explanation

To calculate the density of the compound AB with a rock salt structure, we can follow these steps:

### Step 1: Determine the formula weight in kg
The formula weight of AB is given as 6.023 amu. To convert this into kg, we use the conversion factor:
1 amu = 1.66053906660 × 10^-27 kg.

\[ \text{Formula weight in kg} = 6.023 \, \text{amu} \times 1.66053906660 \times 10^{-27} \, \text{kg/amu} \]
\[ = 1.000 \times 10^{-26} \, \text{kg} \]

### Step 2: Calculate the volume of the unit cell
In a rock salt structure, the unit cell is a cube. The closest A-B distance is given as \( y^{1/3} \) nm. This distance corresponds to the edge length of the unit cell (a) in the rock salt structure.

Convert \( y^{1/3} \) nm to meters:
\[ \text{Edge length (a)} = y^{1/3} \, \text{nm} = y^{1/3} \times 10^{-9} \, \text{m} \]

The volume of the unit cell (V) is given by:
\[ V = a^3 = (y^{1/3} \times 10^{-9})^3 = y \times 10^{-27} \, \text{m}^3 \]

### Step 3: Calculate the number of formula units in the unit cell
In a rock salt structure, there are 4 formula units of AB per unit cell.

### Step 4: Calculate the density
Density (ρ) is given by the formula:
\[ \rho = \frac{\text{mass of the unit cell}}{\text{volume of the unit cell}} \]

The mass of the unit cell can be calculated as:
\[ \text{Mass of the unit cell} = \text{number of formula units} \times \text{formula weight in kg} = 4 \times 1.000 \times 10^{-26} \, \text{kg} = 4.000 \times 10^{-26} \, \text{kg} \]

Now substituting the values into the density formula:
\[ \rho = \frac{4.000 \times 10^{-26} \, \text{kg}}{y \times 10^{-27} \, \text{m}^3} = \frac{4.000}{y} \times 10^{1} \, \text{kg/m}^3 \]
\[ = \frac{40.00}{y} \, \text{kg/m}^3 \]

### Final Answer:
The density of the lattice is \( \frac{40.00}{y} \, \text{kg/m}^3 \).