A binary salt AB (formula weight = 6.023 Y amu, where Y is an arbitray number) has rock salt structure with `1 : 1` ratio of A to B. The shortes A - B distance in the unit cell is `Y^(1//3)nm`.
(a) Calculate the density of the salt in `kg m^(-3)`
(ii) Given that the measured denstiy of the salt is `20 kg m^(-3)`, specify the type of point defect present in the crystal

Answer

To solve the problem, let's break it down step by step.

### Step 1: Understanding the Structure and Given Data
The binary salt AB has a rock salt structure (FCC) with a 1:1 ratio of A to B. The formula weight is given as \(6.023Y\) amu, and the shortest A-B distance in the unit cell is \(Y^{1/3}\) nm.

### Step 2: Calculate the Edge Length of the Unit Cell
In a rock salt structure, the edge length \(a\) of the unit cell can be related to the shortest distance between the cations and anions (A and B). The shortest A-B distance is given as:
\[ \text{Shortest A-B distance} = \frac{a}{\sqrt{2}} = Y^{1/3} \text{ nm} \]
To find the edge length \(a\):
\[ a = Y^{1/3} \cdot \sqrt{2} \text{ nm} \]
Converting this to meters:
\[ a = Y^{1/3} \cdot \sqrt{2} \times 10^{-9} \text{ m} \]

### Step 3: Calculate the Volume of the Unit Cell
The volume \(V\) of the unit cell is given by:
\[ V = a^3 = \left(Y^{1/3} \cdot \sqrt{2} \times 10^{-9}\right)^3 = 2^{3/2} Y \times 10^{-27} \text{ m}^3 \]

### Step 4: Calculate the Mass of the Unit Cell
The molar mass of the salt AB in kg is:
\[ \text{Molar mass} = 6.023Y \times 10^{-3} \text{ kg} \]
Since there are 4 formula units of AB in one unit cell (for FCC):
\[ \text{Mass of the unit cell} = 4 \times \text{Molar mass} = 4 \times 6.023Y \times 10^{-3} \text{ kg} \]

### Step 5: Calculate the Density of the Salt
The density \(\rho\) can be calculated using the formula:
\[ \rho = \frac{\text{Mass of the unit cell}}{\text{Volume of the unit cell}} = \frac{4 \times 6.023Y \times 10^{-3}}{2^{3/2} Y \times 10^{-27}} \]
Simplifying this:
\[ \rho = \frac{4 \times 6.023 \times 10^{-3}}{2^{3/2} \times 10^{-27}} = \frac{4 \times 6.023}{2^{3/2}} \times 10^{24} \text{ kg/m}^3 \]

Calculating \(2^{3/2} = 2.828\):
\[ \rho = \frac{24.092}{2.828} \times 10^{24} \approx 8.5 \times 10^{24} \text{ kg/m}^3 \]

### Step 6: Compare with Measured Density
Given that the measured density of the salt is \(20 \text{ kg/m}^3\), we can see that the calculated density is significantly lower than the measured density.

### Step 7: Identify the Type of Point Defect
Since the measured density (20 kg/m³) is greater than the calculated density (approximately 8.5 kg/m³), this indicates that there is an increase in density due to defects. The types of point defects that can lead to an increase in density are:
1. **Interstitial Defects**: Where extra atoms occupy interstitial sites, increasing the mass without significantly increasing the volume.
2. **Substitutional Defects**: Where atoms of a different element replace some of the original atoms, potentially increasing the overall mass.

### Final Answer
(a) The calculated density of the salt is approximately \(8.5 \times 10^{24} \text{ kg/m}^3\).
(b) The type of point defect present in the crystal is likely an **interstitial defect** or a **substitutional defect**.