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Barium has a body-centered cubic structure. If the atomic radius of barium is 222 pm, calculate the density of solid barium?

Respuesta :

igeclement43

Answer: density =3.377g/cm³

Explanation:

Density =( molecular weight × effective number of atoms per unit cell) / (volume of unit cell × avogadro constant)

D= (M ×n) /(V×A)

M= 137g/mol

n= 2 (For BCC)

V=a³ , where a= 4r/√3

a= (4×222)/√3

a=512.69pm

a= 512.69×10^-10cm

V= ( 512.69×10^-10)^3

V= 1.3476×10^-22cm³

D= (137×2)/(1.3476×10^-22 × 6.02^23)

D= 3.377g/cm³

Therefore the density of barium is 3.377g/cm³

Cricetus

The density of barium will be "3.377 g/cm³".

Given:

Radius of barium,

  • R = 222 pm

We know,

→ [tex]a = \frac{4r}{\sqrt{3} }[/tex]

     [tex]= \frac{4\times 222}{\sqrt{3} }[/tex]

     [tex]= 512.69 \ pm[/tex]

     [tex]= 512.69\times 10^{-10} \ cm[/tex]

Now,

→ [tex]V = a^3[/tex]

      [tex]= (512.69\times 10^{-10})^3[/tex]

      [tex]= 1.3476\times 10^{-22} \ cm^3[/tex]

hence,

The density will be:

→ [tex]D = \frac{M\times n}{V\times A}[/tex]

      [tex]= \frac{137\times 2}{1.3476\times 10^{-22}\times 6.02\times 10^{23}}[/tex]

      [tex]= 3.377 \ g/cm^3[/tex]

Thus the response above is right.

Learn more about barium here:

https://brainly.com/question/16554508

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