The density of apure substance 'A' whose atoms are in cubic close pack arragement is `1 g//c c`. If the all the tetrahedral voids are occu[pioed by 'B' atom , What is the density of resulting solid in `g//c c`.[ "Atomic mass" =`(A)= 30 g//"mol"` and atomic mass `(B)= 50 g//"mol"`]

Options

A. `33.3`
B. `4.33`
C. `2.33`
D. `5.33`

Answer

To find the density of the resulting solid after all tetrahedral voids in a cubic close-packed (CCP) arrangement of substance A are occupied by substance B, we can follow these steps:

### Step 1: Understand the structure of the cubic close-packed arrangement
In a cubic close-packed (CCP) or face-centered cubic (FCC) arrangement, there are a total of 4 atoms per unit cell. This is derived from:
- 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell: \( 8 \times \frac{1}{8} = 1 \)
- 6 face-centered atoms, each contributing \( \frac{1}{2} \) of an atom to the unit cell: \( 6 \times \frac{1}{2} = 3 \)
Thus, total atoms \( Z = 1 + 3 = 4 \).

### Step 2: Calculate the mass of substance A in the unit cell
Given that the atomic mass of A is \( 30 \, \text{g/mol} \), the mass of A in one unit cell can be calculated using:
\[ \text{Mass of A} = Z \times \text{Molar mass of A} / N_A \]
where \( N_A \) (Avogadro's number) is approximately \( 6.022 \times 10^{23} \, \text{mol}^{-1} \).

### Step 3: Calculate the density of substance A
The density \( \rho \) of substance A is given as \( 1 \, \text{g/cm}^3 \). The formula for density is:
\[ \rho = \frac{Z \times m}{N_A \times a^3} \]
where \( a \) is the edge length of the unit cell.

### Step 4: Calculate the number of tetrahedral voids
In a CCP structure, the number of tetrahedral voids is given by:
\[ \text{Number of tetrahedral voids} = 2 \times Z = 2 \times 4 = 8 \]

### Step 5: Calculate the mass of substance B in the tetrahedral voids
Given that the atomic mass of B is \( 50 \, \text{g/mol} \), the total mass of B occupying the tetrahedral voids is:
\[ \text{Mass of B} = \text{Number of tetrahedral voids} \times \text{Molar mass of B} / N_A = 8 \times \frac{50}{N_A} \]

### Step 6: Calculate the total mass in the unit cell
The total mass in the unit cell after B occupies the tetrahedral voids is:
\[ \text{Total mass} = \text{Mass of A} + \text{Mass of B} \]

### Step 7: Calculate the new density of the resulting solid
The new density \( \rho' \) of the resulting solid can be calculated using the total mass and the volume of the unit cell:
\[ \rho' = \frac{\text{Total mass}}{N_A \times a^3} \]

### Step 8: Substitute and simplify
Using the known values and relationships, we can simplify the calculations to find the new density.

### Final Calculation:
1. From the density of A, we know:
\[ 1 = \frac{4 \times 30}{N_A \times a^3} \]
Rearranging gives:
\[ N_A \times a^3 = 120 \, \text{g} \]

2. The total mass after adding B:
\[ \text{Total mass} = \frac{120}{N_A} + \frac{8 \times 50}{N_A} = \frac{120 + 400}{N_A} = \frac{520}{N_A} \]

3. The new density:
\[ \rho' = \frac{520/N_A}{N_A \times a^3} = \frac{520}{120} = \frac{13}{3} \approx 4.33 \, \text{g/cm}^3 \]

### Conclusion:
The density of the resulting solid after all tetrahedral voids are occupied by B atoms is approximately \( 4.33 \, \text{g/cm}^3 \).

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