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Before start calculating, we have to know density and molar mass of iron: 

d(Fe)=7.874 g/cm3

M(Fe)=55.9 g/mole

From iron density, we can see that there is 7.874 g of iron in 1 cm3, hence we can calculate the number of moles in 1 cm3 of iron:

n(Fe) = m(Fe) / M(Fe) 

n(Fe) = 7.874 / 55.9 = 0.14 moles 

Then we can determine the number of particles:

N(Fe) = n(Fe) x NA 

N(Fe) = 0.14 x 6,02 x 10^23

N(Fe) = 8.45 x 10^22 atoms of Fe 

If we know that body-centered cubic system has 2 lattice points per unit cell, we can calculate the number of cells in 1 cubic centimeter of Fe:

N(cell) = N(Fe) / 2 = (8.45 x 10^22) / 2 = 4.225 x 10^22 cells in 1 cm3

Answer:

4.245 × 10²¹ cubic centimetre of Fe

Explanation:

First step is to find the number of atoms that is present in Fe(Iron)

The atomic mass of Fe(iron) is  55.845

We  have to convert  the atomic weight into grams

1 atomic mass = 1.66 x 10⁻²⁴grams

55.845(atomic mass of Fe) =

55.845 x 1.66 x 10 ⁻²⁴ grams = 9.27 x 10²² grams.

Therefore, the number of atoms in a cubic centimetre of Fe (Iron) =

Density of Fe(solid) ÷ number of grams of Fe

Density of Fe (solid) is known as = 7.874g/cm³

The number of atoms in a cubic centimetre of Fe (Iron =

7.874g/cm³ ÷9.27 x 10²²grams

= 8.494 × 10²¹atoms.

In the question we are told Fe adopted a body centered cubic unit cell

Hence , in Body centered cubic unit cell, we have:

We have one atom at the 8 corners of a cube

We also have one body atom the cube's center

8 corners × 1/8 per corner atom = 8 × 1/8 = 1 atom

(8 corners × 1/8 per corner atom) + (1× 1) = 2 atoms

Therefore, the total number of featoms present per unit cell = 2 atoms.

The number of unit cells are present per cubic centimeter of Fe =

Number of Fe atoms per cubic centimeter ÷ Number of Fe per unit cell

= 8.494 × 10²¹ atoms ÷ 2 atoms.

= 4.247 × 10²¹ cubic centimetre of Fe

Hence the number of unit cells that are present per cubic centimeter of Fe is

4.247 × 10²¹ cubic centimetre of Fe.

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