Answer :
Answer:
(a) The unit cell edge length 0.421 nm
Explanation:
Lets calculate -:
Given -
density = [tex]3.58g/cm^3[/tex]
Molecular weight of MgO = 24.31 +16
[tex]40.13 g/mole[/tex]
Avagadro number = [tex]6.022\times10^2^3[/tex]
In rock salt structure , there are four anions and four cations , hence Z=4 (Number of formula per unit cell)
Now, using the formula -
[tex]d=\frac{ZM}{N_Aa^3}[/tex]
[tex]a=[\frac{ZM}{N_Ad}]]^\frac{1}{3}[/tex]
Now, putting the given values
[tex]a=\frac{4\times40.13g/mole}{6.022\times10^2^3\times(3.58g/cm^3)}[/tex]
[tex]4.21\times10^-^8 cm[/tex]
= 0.421 nm
Hence , the unit cell edge length = 0.421 nm
The unit cell edge length of magnesium oxide (MgO) is equal to 0.421 nanometer.
Given the following data:
- Density = 3.58 [tex]g/cm^3[/tex]
- Atomic weight of magnesium = 24.31 g/mol.
- Atomic weight of oxygen = 16.00 g/mol.
Scientific data:
- Avogadro's number = [tex]6.02 \times 10^{23}[/tex]
- Z for FCC = 4
- Molar mass of magnesium oxide (MgO) = [tex]24.31 + 16 = 40.31 \;g/mol.[/tex]
To determine the unit cell edge length:
Mathematically, the unit cell edge length for a face-centered cubic (FCC) structure is given by the formula:
[tex]a=\sqrt[3]{\frac{ZM}{\rho N_A} }[/tex]
Where:
- Z is the number of atoms per unit cell.
- [tex]\rho[/tex] is the density.
- M is the molar mass.
- [tex]N_A[/tex] is the Avogadro constant.
Substituting the given parameters into the formula, we have;
[tex]a=\sqrt[3]{\frac{4 \; \times \;40.31}{3.58\; \times \;6.02 \times 10^{23}} }\\\\a=\sqrt[3]{\frac{161.24}{2.16 \times 10^{24}}}\\\\a=\sqrt[3]{7.47 \times 10^{-23}} \\\\a=4.21 \times 10^{-8}\;meter[/tex]
Note: [tex]1 \;nanometer = 1 \times 10^{-9} \;meter[/tex]
a = 0.421 nanometer.
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