Respuesta :
Answer:
Atoms present per cm³ of Ag = [tex]0.105\ mol/cm^3\times 6.023\times 10^{23}\ atoms/mol=6.32\times 10^{22}\ atoms/cm^3[/tex]
Unit cells which are present per cubic centimeter of W = [tex]3.16\times 10^{22}\ unit\ cells /cm^3[/tex]
Volume = [tex]31.6\times 10^{-24}\ cm^3[/tex]
Edge length = [tex]3.16\times 10^{-8}\ cm[/tex]
Explanation:
Given that:-
The density of the solid W = 19.3 g/cm³
Molar mass of W = 183.84 g/mol
So, Moles present per cm³ of Ag = [tex]\frac{19.3\ g/cm^3}{183.84\ g/mol}[/tex]=0.105 mol/cm³
Also, 1 mole = [tex]6.023\times 10^{23}[/tex] atoms.
So,
Atoms present per cm³ of Ag = [tex]0.105\ mol/cm^3\times 6.023\times 10^{23}\ atoms/mol=6.32\times 10^{22}\ atoms/cm^3[/tex]
Thus, answer = [tex]6.32\times 10^{22}\ atoms/cm^3[/tex]
In BCC, the number of atoms in the unit cell = 2
So,
Unit cells which are present per cubic centimeter of W = [tex]\frac{6.32\times 10^{22}\ atoms/cm^3}{2}=3.16\times 10^{22}\ unit\ cells /cm^3[/tex]
Unit cells which are present per cubic centimeter of W = [tex]3.16\times 10^{22}\ unit\ cells /cm^3[/tex]
The reciprocal of the unit cell/cm³ is the volume of the unit cell.
So, [tex]Volume=\frac{1}{3.16\times 10^{22}\ unit\ cells /cm^3}=31.6\times 10^{-24}\ cm^3[/tex]
Volume = [tex]31.6\times 10^{-24}\ cm^3[/tex]
Also, Volume = [tex]{(Edge\ length)}^3[/tex]
Thus, edge length = [tex]{Volume}^{\frac{1}{3}}[/tex] = [tex]\left(31.6\times \:\:10^{-24}\right)^{\frac{1}{3}}\ cm=4.1\times 10^{-8}\ cm[/tex]
Edge length = [tex]3.16\times 10^{-8}\ cm[/tex]
The unit cells present per cubic centimeter of W are [tex]\bold{ 3.16 \times10^2^2\;unit\;cell/cm^3} [/tex]
The volume of a unit cell of this metal is [tex]\bold{31.6\times 10^-^2^4\;cm^3} [/tex].
The edge length is [tex]\bold{4.1 \times 10^8\;cm}
What is density?
Density can be defined as the mass per unit volume.
Given,
The density of W is [tex]\bold{19.3\; g/cm^3}[/tex]
Molar mass of W = 183.84 g/mol
Step 1: Moles present per cm³ of Ag is [tex]\bold{\dfrac{19.3\;g/cm^3}{183.84\;g/mol} = 0.105\;mol/cm^3}[/tex]
1 mole = [tex]\bold{6.023\times 10^2^3\;atoms}[/tex]
Then, Atoms present per cm³ of Ag is
[tex]6.32\times 10^2^2\;atoms/cm^3[/tex]
In BCC, the number of atoms in the unit cell = 2
Unit cells which are present per cubic centimeter of W =
[tex]\bold{\frac{6.32\times 10^2^2\;atoms}{2}= 3.16 \times10^2^2\;unit\;cell/cm^3} [/tex]
Step2: Now, calculating the volume of a unit cell of the metal
[tex]\bold{volume =\dfrac{1}{3.16 \times10^2^2\;unit\;cell/cm^3} = 31.6\times 10^-^2^4\;cm^3} [/tex]
Thus, the volume is [tex]\bold{31.6\times 10^-^2^4\;cm^3} [/tex]
Step 3: calculating the edge length
[tex]\bold{Volume^\dfrac{1}{3} \times(31.6 \times 10^-2^4)^\dfrac{1}{3} \\\\
cm = 4.1 \times 10^8\;cm}
Thus, The unit cells present per cubic centimeter of W are [tex]\bold{ 3.16 \times10^2^2\;unit\;cell/cm^3} [/tex]
The volume of a unit cell of this metal is [tex]\bold{31.6\times 10^-^2^4\;cm^3} [/tex].
The edge length is [tex]\bold{4.1 \times 10^8\;cm}
Learn more about density, here:
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Also, Volume =
Thus, edge length = =
Edge length =
