Ask Question
26 October, 03:58

Consider the unit cell of aluminum with aluminum ion on every corner and every face-centered site of the cube.

Question

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?

2. Using atomic weight value of aluminum 27.0, the density of aluminum can be calculated as what g/cm^3

+2
Answers (1)
  1. 26 October, 04:20
    0
    1) 0.4045nm

    2) 0.0027096117g/cm³

    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “Consider the unit cell of aluminum with aluminum ion on every corner and every face-centered site of the cube. Question 1. Using the value ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers