The number of unit cells in the `Ca` atom lies on the surface of a cubic crystal that is `1.0 cm` in length is

Options

A. `9.17 xx 10^(23)`
B. `9.17 xx 10^(22)`
C. `2 xx 9.17 xx 10^(23)`
D. `2 xx 9.17 xx 10^(22)`

Explanation

To solve the problem of finding the number of unit cells in a cubic crystal of calcium (Ca) that has a length of 1.0 cm, we will follow these steps:

### Step 1: Calculate the Volume of the Crystal
The volume \( V \) of a cubic crystal can be calculated using the formula:
\[ V = a^3 \]
where \( a \) is the length of the edge of the cube. Given that \( a = 1.0 \, \text{cm} \), we first convert this to meters:
\[ 1.0 \, \text{cm} = 0.01 \, \text{m} \]
Now, we can calculate the volume:
\[ V = (0.01 \, \text{m})^3 = 0.000001 \, \text{m}^3 = 10^{-6} \, \text{m}^3 \]

### Step 2: Determine the Volume of One Unit Cell
The volume of one unit cell for calcium can be approximated using its molar mass and density. The molar mass of calcium (Ca) is approximately 40.08 g/mol, and the density is about 1.54 g/cm³. First, we convert the density to kg/m³:
\[ 1.54 \, \text{g/cm}^3 = 1540 \, \text{kg/m}^3 \]
Now, we can find the volume of one mole of calcium:
\[ \text{Volume of 1 mole} = \frac{\text{Molar mass}}{\text{Density}} = \frac{0.04008 \, \text{kg}}{1540 \, \text{kg/m}^3} \approx 2.60 \times 10^{-5} \, \text{m}^3 \]
Since one mole contains \( 6.022 \times 10^{23} \) atoms (Avogadro's number), the volume of one unit cell can be calculated as:
\[ \text{Volume of one unit cell} = \frac{2.60 \times 10^{-5} \, \text{m}^3}{6.022 \times 10^{23}} \approx 4.32 \times 10^{-29} \, \text{m}^3 \]

### Step 3: Calculate the Number of Unit Cells in the Crystal
Now, we can find the number of unit cells in the crystal by dividing the total volume of the crystal by the volume of one unit cell:
\[ \text{Number of unit cells} = \frac{V_{\text{crystal}}}{V_{\text{unit cell}}} = \frac{10^{-6} \, \text{m}^3}{4.32 \times 10^{-29} \, \text{m}^3} \]
Calculating this gives:
\[ \text{Number of unit cells} \approx 2.31 \times 10^{22} \]

### Final Answer
The number of unit cells in the calcium atom that lies on the surface of a cubic crystal that is 1.0 cm in length is approximately \( 2.31 \times 10^{22} \).