An element with molecular weight 200 gm/mole crystallizes in a simple cubic cell. If the unit cell length is 3Å and Avogardo’s number is `6xx10^(23)"mole"^(-1)` , the density of the element is:

Options

A. 15 g /cc
B. `13.12` g /cc
C. `12.34` g/cc
D. `15.9`g/cc

Answer

To find the density of the element that crystallizes in a simple cubic cell, we can use the formula for density in terms of the number of atoms per unit cell (Z), the molecular weight (M), Avogadro's number (NA), and the volume of the unit cell (A³).

### Step-by-Step Solution:

1. **Identify the parameters:**
- Molecular weight (M) = 200 g/mol
- Unit cell length (A) = 3 Å = \(3 \times 10^{-8}\) cm (since 1 Å = \(10^{-10}\) m and 1 m = 100 cm)
- Avogadro's number (NA) = \(6 \times 10^{23}\) mol\(^{-1}\)

2. **Determine the number of atoms per unit cell (Z):**
- In a simple cubic cell, there is 1 atom per unit cell. This is because there are 8 corners, and each corner contributes \( \frac{1}{8} \) of an atom to the unit cell:
\[ Z = 8 \times \frac{1}{8} = 1 \]

3. **Calculate the volume of the unit cell (A³):**
- Convert the unit cell length from Ångströms to cm:
\[ A = 3 \text{ Å} = 3 \times 10^{-8} \text{ cm} \]
- Calculate the volume:
\[ V = A^3 = (3 \times 10^{-8} \text{ cm})^3 = 27 \times 10^{-24} \text{ cm}^3 = 2.7 \times 10^{-23} \text{ cm}^3 \]

4. **Use the density formula:**
- The density (D) can be calculated using the formula:
\[ D = \frac{Z \times M}{N_A \times V} \]
- Substitute the known values:
\[ D = \frac{1 \times 200 \text{ g/mol}}{6 \times 10^{23} \text{ mol}^{-1} \times 2.7 \times 10^{-23} \text{ cm}^3} \]

5. **Calculate the density:**
- First, calculate the denominator:
\[ 6 \times 10^{23} \times 2.7 \times 10^{-23} = 16.2 \]
- Now calculate the density:
\[ D = \frac{200}{16.2} \approx 12.34 \text{ g/cm}^3 \]

### Final Answer:
The density of the element is approximately **12.34 g/cm³**.