Face Centered Cubic

Coordination number of 12

8 unit cells

0 unit cells (just with an adjacent atom)

4 atoms per unit cell

Mass of unit cell = mass 4 Al atoms = (195.08)(g/mol)(1mol/6.022x10^{23}atoms)(4 atoms/unit cell) = 12.96 x 10^{-22} g/unit cell

1 amu = 1.66054 x 10^{-24 }g

Thus, 12.96 x 10^{-22} g/unit cell / 1.66054 x 10^{-24 }g = 780.47 amu

Edge length 3.924 A. Thus, the Volume unit cell = A^{3} = (3.924x10^{-8}cm)^{3} = 60.42x10^{-24} cm^{3}/unit cell = 60.42 A^{3 }( since 1 A^{3} = (10^{-8 })^{3 }cm^{3 })

74%

100-74 = 26%

As we know, total volume is 100 % and it is 60.42 A^{3}

Then, 74 % (volume of atoms only) would be: 74 x 60.42 A^{3 }/ 100 = 44.71 A^{3}

V = 4/3 πr^{3 }

Thus, r = (V/ (4/3 π))^{1/3 }= (3x44.71/4 π)^{1/3} = 2.202 A

Density = m/V = 780.47 amu /60.42 A^{3}

Thus, 760 amu = 12.96 x 10^{-22} g, since in order to convert to grams: 760 amu x 1.66054 x 10^{-24 }g / 1 amu = 12.96 x 10^{-22} g

1 A^{3} = (10^{-8 })^{3 }cm^{3 }

Thus, 60.42 A^{3 }in cm^{3 }is = 60.42 A^{3 }x (10^{-8 })^{3 }cm^{3 }/ 1 A^{3 }= 60.42x10^{-24} cm^{3 }

Density = m/V = 12.96 x 10^{-22} g/60.42 A^{3 }= 12.96 x 10^{-22} g / 60.42x10^{-24} cm^{3 }= 21.45 g/ cm^{3}

i think your 2 and 4 is wrong? can you check?

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