Question

The density of solid fe is 7.87 g/cm3. how many atoms are present per cubic centimeter of fe?

Answer 1

  The  average  that reflect  the  typical  ratio  of  natural  abundance  of  iron  isotope  is  the  atomic  weight  of  the iron.  Iron  has  four  stable  isotope  that  is  iron  54 ,56,  57, 58.  its  atomic weight  by  mass  is  9.27 x10^-22 grams
therefore  number  of atoms  in  cm^3 is
(7.87g/cm^3)/(9.27x10^-22)=8.49x10^21atoms/cm^3

Answer 2

Answer: The number of iron atoms in given amount is $8.49\times 10^{22}$

Explanation:

Density of a substance is defined as the ratio of its mass and volume. The chemical equation representing density of a substance is:  

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

We are given:

Density of iron = $7.87g/cm^3$

Volume of iron = $1cm^3$

Putting values in above equation, we get:

$7.87g/cm^3=\frac{\text{Mass of iron}}{1g/cm^3}\\\\\text{Mass of iron}=(7.87g/cm^3\times 1cm^3)=7.87g$

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of iron = 7.87 g

Molar mass of iron = 55.65 g/mol

Putting values in above equation, we get:

$\text{Moles of iron}=\frac{7.87g}{55.65g/mol}=0.141mol$

According to mole concept:

1 mole of an element contains $6.022\times 10^{23}$ number of atoms.

So, 0.141 moles of iron will contain = $0.141\times 6.022\times 10^{23}=8.49\times 10^{22}$ number of atoms.

Hence, the number of iron atoms in given amount is $8.49\times 10^{22}$