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A metal has a body-centered cubic lattice with a unit cell edge length of 2.866 Å (1 Å = 10⁻¹⁰ m). The density of the metal is 7.87 g/cm³. What is the mass of an atom of this metal? (1 m = 10¹² pm).

Respuesta :

Answer:

Therefore the mass of an atom of this metal is 55.74 g/mol.

Explanation:

Given, edge length = 2.866 [tex]\AA[/tex]

                                =2.866×10⁻¹⁰ m.

Density of the metal = 7.87 g/ cm³

Body centered cubic has 2 atom per unit cell.

Atomic mass: Atomic mass is mass of the component per atom.

Density: Density is the ratio of mass to volume.

[tex]Density=\frac{mass}{volume}[/tex]

mass = number of atom × atomic mass = Z×M

Volume =( edge length)³× Avogadro Number = [tex]a^3\times N_A[/tex]

z=  number of atom =2

M=atomic mass=?

a=edge length=2.866×10⁻¹⁰ m=[tex]=2.866\times 10^{-10} \times 10^2 cm[/tex]

[tex]N_A[/tex] =Avogadro Number =6.023×10²³

[tex]Desity(d)=\frac{Z\times M}{a^3\times N_A}[/tex]

[tex]\Rightarrow M=\frac{d \times a^3\times N_A}{Z}[/tex]

[tex]\Rightarrow M=\frac{7.87 g/cm^3 \times( 2.866\times 10^{-10} \times 10^2 cm)^3 \times 6.023\times 10^{23}/mol}{2}[/tex]

[tex]\Rightarrow M = 557.94\times 10^{-30}\times 10^{29}[/tex] g/mol

⇒M = 55.74 g/mol

Therefore the mass of an atom of this metal is 55.74 g/mol.

       

           

ajeigbeibraheem

The mass of an atom of this metal is 9.26 × 10⁻²³ grams

Given that:

  • the edge length (a) of the unit cell = 2.366  Å
  • the density of the metal = 7.87 g/cm³

The density [tex]\rho[/tex] of a unit cell can be computed as:

[tex]\mathbf{Density (\rho) = \dfrac{Mass \ of \ the \ unit \ cell }{volume \ of \ the \ unit \ cell }}[/tex]

where;

  • [tex]\mathbf{ mass\ of \ unit\ cell (m) = \dfrac{effective \ no. \ of \ an \ atom (Z) \times \ Molar \ mass (M) }{avogadro's \ No (N_o)}}[/tex]

  • Volume of the unit cell = a³

∴

[tex]\mathbf{ density \ of \ unit\ cell (m) = \dfrac{effective \ no. \ of \ an \ atom (Z) \times \ Molar \ mass (M) }{avogadro's \ No (N_o)\times a^3}}[/tex]

Making the molar mass (M) the subject, we have:

[tex]\mathbf{ Molar \ mass (M) = \dfrac{ avogadro's \ No (N_o)\times a^3 \times \rho}{effective \ no. \ of \ an \ atom (Z) }}[/tex]

where;

  • a = 2.866 Å
  • a = 2.866× 10⁻¹⁰ m
  • a = (2.866 × 10⁻¹⁰) × 10¹² pm  = 286.6 pm
  • a = (286.6 pm × 10⁻¹⁰ cm)/1 pm = 2.866 × 10⁻⁸ cm

Now, the molar mass (M) can be estimated as:

[tex]\mathbf{ Molar \ mass (M) = \dfrac{6.023 \times 10^{23} mol^{-1} \times (2.866 \times 10^{-8} \ cm )^3 \times 7.87 g/cm^3}{2}}[/tex]

[tex]\mathbf{ Molar \ mass (M) = \dfrac{111.5876557}{2}\ g/mol}[/tex]

[tex]\mathbf{ Molar \ mass (M) =55.79 \ g/mol}[/tex]

Finally, the mass of an atom of this metal is:

[tex]\mathbf{mass (m) = \dfrac{55.79 g/mol}{ 6.023 \times 10^{23} \ mol^{-1}}}[/tex]

mass (m) of an atom of this metal = 9.26 × 10⁻²³ grams

Learn more about body-centered cubic structure here:

https://brainly.com/question/12975629?referrer=searchResults

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