The edge length of the fcc(face centered cubic) unit cell of NaCl is 562.8 pm. a) how many unit cells are present in a single crystal of NaCl that ia a cube with edges 1mm in length? b) what amount in moles of NaCl is present in this crystal?
The Correct Answer and Explanation is :
Problem Breakdown:
You are asked to determine:
a) The number of unit cells present in a cubic NaCl crystal with an edge length of 1 mm.
b) The amount in moles of NaCl present in this crystal.
Given Information:
- Edge length of the NaCl fcc unit cell = 562.8 pm (picometers) = ( 562.8 \times 10^{-12} \, \text{m} )
- Edge length of the NaCl crystal = 1 mm = ( 1 \times 10^{-3} \, \text{m} )
- The NaCl crystal has a face-centered cubic (FCC) structure.
a) Number of Unit Cells:
- Volume of the NaCl crystal:
- Since the crystal is a cube with edge length 1 mm, the volume (V_{\text{crystal}}) is:
[
V_{\text{crystal}} = (\text{edge length})^3 = (1 \times 10^{-3} \, \text{m})^3 = 1 \times 10^{-9} \, \text{m}^3
]
- Volume of one unit cell:
- For NaCl, with an fcc structure, the unit cell has 4 formula units. The edge length of the unit cell is 562.8 pm.
[
V_{\text{unit cell}} = (\text{edge length})^3 = (562.8 \times 10^{-12} \, \text{m})^3 = 1.78 \times 10^{-28} \, \text{m}^3
]
- Number of unit cells in the crystal:
The number of unit cells (N_{\text{cells}}) is the ratio of the crystal’s volume to the unit cell’s volume:
[
N_{\text{cells}} = \frac{V_{\text{crystal}}}{V_{\text{unit cell}}} = \frac{1 \times 10^{-9} \, \text{m}^3}{1.78 \times 10^{-28} \, \text{m}^3} = 5.62 \times 10^{18}
]
Therefore, the number of unit cells is approximately (5.62 \times 10^{18}).
b) Moles of NaCl in the Crystal:
- Avogadro’s number:
- Avogadro’s number ((N_A)) is (6.022 \times 10^{23}) unit cells per mole.
- Moles of NaCl:
Since there are 4 NaCl formula units per unit cell in the fcc structure:
[
\text{Moles of NaCl} = \frac{\text{Number of formula units}}{N_A} = \frac{4 \times 5.62 \times 10^{18}}{6.022 \times 10^{23}} = 3.73 \times 10^{-5} \, \text{moles}
]
Final Answers:
- a) The number of unit cells in the crystal is approximately (5.62 \times 10^{18}).
- b) The amount in moles of NaCl in the crystal is approximately (3.73 \times 10^{-5}) moles.
Explanation:
The cubic NaCl crystal’s volume is calculated from its edge length, and the volume of a single unit cell is calculated using its edge length. The number of unit cells is determined by dividing the total volume by the volume of one unit cell. For moles, we use Avogadro’s number to convert the number of formula units into moles, accounting for the number of NaCl formula units per unit cell in the fcc structure.
Let me generate the image to visualize the cubic arrangement of unit cells in the NaCl crystal.
Here is an illustration showing the cubic crystal structure of NaCl with a face-centered cubic (FCC) arrangement of unit cells. It visualizes how the unit cells fit together to form the large cube, with the sodium (Na) and chloride (Cl) ions within each unit cell. The edge length of the cube is marked as 1mm.