Answer:
1) 0.4045nm
2) 0.0027096117g/cm³
Step-by-step explanation:
1) Using the value of ionic radius of aluminum ion r=0.143 nm, the length of each edge of the unit cell can be calculated as what nm?
In the question above, using a face centered cubic structure, we have:
For a face centered site of the cube,
The diagonal = 4r
Where r =Atomic or ionic radius of Aluminum
Let the edge length of the cube be represented by X
Therefore, we have based on Pythagoras theorem,
X² + X² = (4r)²
2X² = 16r²
Divide both sides by 2
X² = 8r²
Find the square root of both sides
X = √8 × r
Since r = 0.143nm
The length of each edge of the unit cell can be calculated as
X = √8 × 0.143nm
X = 0.4044650788nm
Approximately = 0.4045nm
2) Using atomic weight value of aluminum 27.0, the density of aluminum can be calculated as what g/cm^3
Density of an object = Mass of the Object / Volume of the Object.
The object in this question = Cube
Step 1
Volume of a cube = (Length of the cube)³
In the question above, side length of the cube = 0.4045nm
When would convert 0.4045nm to centimeters
= 1 nm = 1 × 10^-7 cm
0.4045nm =
Cross multiply
= 4.045 × 10^-7 cm
Volume of the Aluminum cube =(4.045 × 10^-7cm)³
= 6.618439112 ×10^-20cm³
Step 2
The atomic weight value of aluminum is given as 27.0 in the question
A face centered cubic structure has 4 atoms per unit cell.
1 Atomic mass or weight value = 1.6605 × 10^-24 grams
Hence, the mass of Aluminum is calculated as:
(4 atoms /1 cell )× 1 unit × (27/ 1 atom of Aluminium) × (1.6605 × 10^-24g/ 1 Atomic mass value)
= 1.79334 ×10^-22g
Density = Mass/Volume
= 1.79334 ×10^ -22g /6.618439112 ×10^-20cm³
= 0.0027096117g/cm³