An element (density 6.8 g ` cm^(-3) ` and the length of the side of the unit cell is 316 pm. The unit cell in the most important crystalline form of tungsten is the body centred unit cell. How many atoms of the element does 50 g of the element contain ?

Explanation

To solve the problem of how many atoms of the element are contained in 50 g of tungsten, we will follow these steps:

### Step 1: Understand the given data
- Density of the element (tungsten) = 6.8 g/cm³
- Length of the side of the unit cell (A) = 316 pm = 316 x 10^(-12) m
- Type of unit cell = Body-Centered Cubic (BCC)
- Molar mass of tungsten (W) = 184 g/mol

### Step 2: Use the formula for density
The formula for density (d) in terms of the unit cell is given by:
\[ d = \frac{Z \times M}{N_A \times A^3} \]
Where:
- \( Z \) = number of atoms per unit cell (for BCC, \( Z = 2 \))
- \( M \) = molar mass of the element (tungsten, \( M = 184 \) g/mol)
- \( N_A \) = Avogadro's number \( (6.022 \times 10^{23} \text{ atoms/mol}) \)
- \( A \) = length of the side of the unit cell in cm (convert pm to cm)

### Step 3: Convert the length of the unit cell to cm
\[ A = 316 \text{ pm} = 316 \times 10^{-12} \text{ m} = 316 \times 10^{-10} \text{ cm} = 3.16 \times 10^{-8} \text{ cm} \]

### Step 4: Calculate the volume of the unit cell
\[ A^3 = (3.16 \times 10^{-8})^3 \text{ cm}^3 \]
\[ A^3 = 3.16^3 \times 10^{-24} \text{ cm}^3 \]
\[ A^3 \approx 31.62 \times 10^{-24} \text{ cm}^3 = 3.162 \times 10^{-23} \text{ cm}^3 \]

### Step 5: Substitute values into the density formula
Now, substituting the known values into the density formula:
\[ 6.8 = \frac{2 \times 184}{6.022 \times 10^{23} \times 3.162 \times 10^{-23}} \]

### Step 6: Solve for the number of atoms in 50 g
First, calculate the number of moles in 50 g of tungsten:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{50 \text{ g}}{184 \text{ g/mol}} \approx 0.2717 \text{ moles} \]

### Step 7: Calculate the total number of atoms
Now, multiply the number of moles by Avogadro's number to find the total number of atoms:
\[ \text{Number of atoms} = \text{Number of moles} \times N_A \]
\[ \text{Number of atoms} = 0.2717 \times 6.022 \times 10^{23} \approx 1.63 \times 10^{23} \text{ atoms} \]

### Final Answer
The number of atoms in 50 g of tungsten is approximately \( 1.63 \times 10^{23} \) atoms.

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