An element crystallizes into a structure which may be describes by a cubic type of unit cell having one atom on each corner of the cube and two atoms on one of its diagonals. If the volume of this unit cell is `24xx10^(-24)cm^(3)` and density of element is `7.2g cm^(-3)` . Calculate the number of atoms present in `200g` of element.

Explanation

To solve the problem step by step, we will follow these calculations:

### Step 1: Determine the number of atoms in the unit cell (Z)
The unit cell has:
- 1 atom at each of the 8 corners of the cube.
- 2 atoms along one of its body diagonals.

The contribution of the corner atoms is:
\[ \text{Contribution from corners} = 8 \times \frac{1}{8} = 1 \text{ atom} \]

The contribution from the body diagonal atoms is:
\[ \text{Contribution from body diagonal} = 2 \text{ atoms} \]

Thus, the total number of atoms per unit cell (Z) is:
\[ Z = 1 + 2 = 3 \]

### Step 2: Calculate the molar mass (M) using the density and volume of the unit cell
We know the formula for density (D):
\[ D = \frac{Z \times M}{V \times N_A} \]
Where:
- \(D\) = density of the element = 7.2 g/cm³
- \(V\) = volume of the unit cell = \(24 \times 10^{-24}\) cm³
- \(N_A\) = Avogadro's number = \(6.022 \times 10^{23}\) mol⁻¹
- \(Z\) = number of atoms per unit cell = 3

Rearranging the formula to find M:
\[ M = \frac{D \times V \times N_A}{Z} \]

Substituting the values:
\[ M = \frac{7.2 \, \text{g/cm}^3 \times 24 \times 10^{-24} \, \text{cm}^3 \times 6.022 \times 10^{23} \, \text{mol}^{-1}}{3} \]

Calculating this gives:
\[ M = \frac{7.2 \times 24 \times 6.022}{3} \times 10^{-24 + 23} \]
\[ M = \frac{1036.032}{3} \approx 345.344 \, \text{g/mol} \]

### Step 3: Calculate the number of moles in 200 g of the element
Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{200 \, \text{g}}{34.56 \, \text{g/mol}} \]
Calculating this gives:
\[ \text{Number of moles} \approx 5.79 \, \text{mol} \]

### Step 4: Calculate the total number of atoms in 200 g of the element
Using the formula:
\[ \text{Number of atoms} = \text{Number of moles} \times N_A \]
Substituting the values:
\[ \text{Number of atoms} = 5.79 \, \text{mol} \times 6.022 \times 10^{23} \, \text{mol}^{-1} \]
Calculating this gives:
\[ \text{Number of atoms} \approx 34.85 \times 10^{23} \text{ atoms} \]

### Final Answer
The total number of atoms present in 200 g of the element is approximately \(34.85 \times 10^{23}\) atoms.

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