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 will use the formula for density in terms of the unit cell parameters:

\[ \text{Density} (\rho) = \frac{Z \times M}{N_A \times a^3} \]

Where:
- \( Z \) = number of atoms per unit cell
- \( M \) = molar mass (molecular weight) in grams per mole
- \( N_A \) = Avogadro's number (in moles\(^{-1}\))
- \( a \) = edge length of the unit cell (in centimeters)

### Step 1: Determine the number of atoms per unit cell (Z)
For a simple cubic cell, there is 1 atom per unit cell because each of the 8 corners contributes \( \frac{1}{8} \) of an atom, totaling to:

\[ Z = 1 \]

### Step 2: Identify the molar mass (M)
The molar mass is given in the problem as:

\[ M = 200 \, \text{g/mol} \]

### Step 3: Use Avogadro's number (NA)
Avogadro's number is provided as:

\[ N_A = 6 \times 10^{23} \, \text{mole}^{-1} \]

### Step 4: Convert the unit cell length (a) from Ångströms to centimeters
The unit cell length is given as:

\[ a = 3 \, \text{Å} = 3 \times 10^{-10} \, \text{m} = 3 \times 10^{-8} \, \text{cm} \]

### Step 5: Calculate \( a^3 \)
Now, we need to calculate \( a^3 \):

\[ a^3 = (3 \times 10^{-8} \, \text{cm})^3 = 27 \times 10^{-24} \, \text{cm}^3 \]

### Step 6: Substitute values into the density formula
Now we can substitute the values into the density formula:

\[ \rho = \frac{Z \times M}{N_A \times a^3} = \frac{1 \times 200 \, \text{g/mol}}{(6 \times 10^{23} \, \text{mole}^{-1}) \times (27 \times 10^{-24} \, \text{cm}^3)} \]

### Step 7: Simplify the expression
Calculating the denominator:

\[ N_A \times a^3 = 6 \times 10^{23} \times 27 \times 10^{-24} = 162 \times 10^{-1} = 16.2 \]

Now substituting back into the density equation:

\[ \rho = \frac{200}{16.2} \approx 12.34 \, \text{g/cm}^3 \]

### Final Answer
Thus, the density of the element is approximately:

\[ \boxed{12.34 \, \text{g/cm}^3} \]