Density of a unit cell is same as the density of the substance. If the density of the substance is known, number of atoms or dimensions of the unit cell can be calculated . The density of the unit cell is related to its mass(M), no. of atoms per unit cell (Z), edge length (a in cm) and Avogadro number `N_A` as :
`rho = (Z xx M)/(a^3 xx N_A)`
The number of atoms present in 100 g of a bcc crystal (density = `12.5 g cm^(-3))` having cell edge 200 pm is

Options

A. `1 xx 10^(25)`
B. `1 xx 10^(24)`
C. `2 xx 10^(24)`
D. `2 xx 10^(26)`

Explanation

To solve the problem of finding the number of atoms present in 100 g of a bcc crystal with a density of 12.5 g/cm³ and an edge length of 200 pm, we can follow these steps:

### Step 1: Understand the given values
- Density (ρ) = 12.5 g/cm³
- Edge length (a) = 200 pm = 200 x 10^(-12) m = 200 x 10^(-10) cm = 2 x 10^(-8) cm
- For a bcc (body-centered cubic) crystal, the number of atoms per unit cell (Z) = 2
- Mass of the substance (W) = 100 g

### Step 2: Use the density formula
The formula relating density, mass, number of atoms per unit cell, and edge length is:
\[ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \]
Where:
- ρ = density
- Z = number of atoms per unit cell
- M = molar mass (which we need to find)
- a = edge length of the unit cell
- N_A = Avogadro's number (approximately \(6.022 \times 10^{23} \text{ mol}^{-1}\))

### Step 3: Rearranging the formula
We can rearrange the formula to find the mass (M):
\[ M = \frac{\rho \cdot a^3 \cdot N_A}{Z} \]

### Step 4: Calculate the volume of the unit cell
Calculate \(a^3\):
\[ a^3 = (2 \times 10^{-8} \text{ cm})^3 = 8 \times 10^{-24} \text{ cm}^3 \]

### Step 5: Substitute the values into the formula
Now substitute the values into the rearranged formula:
\[ M = \frac{12.5 \text{ g/cm}^3 \cdot 8 \times 10^{-24} \text{ cm}^3 \cdot 6.022 \times 10^{23} \text{ mol}^{-1}}{2} \]

### Step 6: Calculate the molar mass (M)
Calculating the above expression:
\[ M = \frac{12.5 \cdot 8 \cdot 6.022}{2} \times 10^{-24 + 23} \text{ g/mol} \]
\[ M = \frac{12.5 \cdot 48.176}{2} \text{ g/mol} \]
\[ M = \frac{582.2}{2} \text{ g/mol} = 291.1 \text{ g/mol} \]

### Step 7: Calculate the number of moles in 100 g
Now, we can find the number of moles in 100 g of the substance:
\[ \text{Number of moles} = \frac{W}{M} = \frac{100 \text{ g}}{291.1 \text{ g/mol}} \approx 0.343 \text{ mol} \]

### Step 8: Calculate the number of atoms
Finally, to find the total number of atoms, we multiply the number of moles by Avogadro's number:
\[ \text{Number of atoms} = \text{Number of moles} \times N_A = 0.343 \text{ mol} \times 6.022 \times 10^{23} \text{ atoms/mol} \]
\[ \text{Number of atoms} \approx 2.064 \times 10^{23} \text{ atoms} \]

### Final Answer
The number of atoms present in 100 g of the bcc crystal is approximately \(2.064 \times 10^{23}\) atoms.