Respuesta :
Answer:
d = 7.89 g/cm³
experimental value = 7.87 g/cm³
% differrence = 0.28
Explanation:
The density of a material is the relation betwEen the mass and the volume it occupies, thus the density of the unit cell compared to its experimental value should agree reasonably well.
to solve this question lets calculate the density of the unit cell and then compare it with the experimental value.
For a body centered cubic unit cell the volume is the edge length, a, cubed.
In the bcc crystal this edge is related to the radius of the atom as
a= 4r/√3
r = (0.124 nm)( 1 x10⁻⁷ cm/nm) = 1.24 x 10⁻⁸ cm
⇒ a = 4(1.24 x 10⁻⁸ cm)/√3 = 2.86 x 10⁻⁸ cm
V = a³ = (2.86 x 10⁻⁸ cm)³ = 2.35 x 10⁻²³ cm³
To calculate the mass of the cell we have to count the number of atoms per cell:
corners: 8 atoms x 1/8 = 1
center of the cube = 1
So we have 2 atoms of iron per unit cell , and the mass m is:
m = (2 atoms/unit cell )(1mol/6.022 x 10²³ atoms)(55.845 g/mol)
m = 1.85 x 10⁻²² g/unit cell
Finally the density will be:
d= m/V = 1.85 x 10⁻²² g/ 2.35 x 10⁻²³ cm³
= 7.89 g /cm³
The experimental value from the literature is 7.87 g /cm³
The percent difference is:
(7.89 - 7.87) / 7.87 = 0.28 %
Following are the calculation for the front cover of the textbook.
In the given scenario, this problem calls for the computation of the density of iron.
As per the Equation 3.5
[tex]\rho = \frac{nA_{Fe}}{V_CN_A}[/tex]
For [tex]BCC,\ n = 2 \frac{atoms}{unit cell}\ ,[/tex] and
[tex]V_C=(\frac{4R}{\sqrt{3}})^3[/tex]
Thus,
[tex]\rho =\frac{n A_{Fe} }{ (\frac{4R}{\sqrt{3}})^3 N_A}[/tex]
[tex]= \frac{(2 \frac{a t o m s}{u n i t c e l l} )(55.85 \frac{g}{mol} )}{\frac{[\frac{(4)(0.124 \times 10^{-7} cm)}{\sqrt{3}}]}{ \text{(unit cell)} (6.022 \times 10^{23} \ \frac{atoms}{mol})}}[/tex]
[tex]= 7.90\ \frac{g}{cm^3}\\\\[/tex]
The value given inside the front cover is [tex]7.87\ \frac{g}{cm^3}\\\\[/tex].
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