Below ${\mathbf{912}}^{\mathbf{°}}\mathit{C}$,pure iron crystallizes in a body-centered cubic structure (ferrite) with a density ofrole="math" localid="1663336837557" ${\mathbf{\text{7.86g/cm}}}^{\mathbf{\text{3}}}$; from${\mathbf{912}}^{\mathbf{°}}\mathit{C}$to${\mathbf{1394}}^{\mathbf{°}}\mathit{C}$, it adopts a face-centered cubic structure (austenite) with a density of$\mathbf{\text{7.40g/cm}}$. Both types of iron form interstitial alloys with carbon. The maximum amount of carbon is 0.0218mass$\mathbf{\%}$in ferrite and 2.08mass$\mathbf{\%}$in austenite. Calculate the density of each alloy.

As a compound or a solution, an alloy is a metallic material made up of two or more elements. The constituents of alloys are usually metals, while carbon, a nonmetal, is an important component of steel.

Smaller atoms (such as carbon, hydrogen, and water) are placed into spaces (interstices) between atoms of the primary element in interstitial alloys. In this case, carbon atoms occupy the vacancies in the iron crystal structure, resulting in a change in density but not in volume.

a) Pure iron crystallises in a body-centered cubic form called ferrite with a density $7.86g/{\text{cm}}^3$of below${912}^{°}C$ . Ferrite can form an interstitial alloy with carbon, resulting in a increase $0.0218\%$in mass. As previously stated, the volume remains constant, hence the density increases as the amount of carbon in the alloy increases. Let's figure out its density:

role="math" localid="1663337401279" $$\begin{gathered} {\rho }_{ferrit{e}_{aloy}}{\text{= 7.86gcm}}^{\text{- 3}}\text{×}\frac{{\displaystyle \text{100\% + 0.0218\%}}}{{\displaystyle \text{100\%}}} \\ {\rho }_{ferrit{e}_{aloy}}={\text{7.86gcm}}^{\text{- 3}} \end{gathered}$$

Hence, the density is${\text{7.86gcm}}^{\text{- 3}}$

b)

Between ${912}^{°}C$and${1394}^{°}C$ , iron crystallises as austenite, a face-centered cubic structure with a density of $7.40{\text{gcm}}^{-3}$.Austenite can form an interstitial alloy with carbon, resulting in a 2.08% increase in mass. Let's figure out its density:

role="math" localid="1663337474524" $$\begin{gathered} {\rho }_{austenit{e}_{aloy}}={\text{7.40gcm}}^{\text{- 3}}\frac{\text{100\% + 2.08\%}}{\text{100\%}} \\ {\rho }_{austenit{e}_{aloy}}={\text{7.55gcm}}^{\text{- 3}} \end{gathered}$$

Therefore, the density is${\text{7.55gcm}}^{\text{- 3}}$