Chapter 12, Problem 12.96AP

Interpretation Introduction

Interpretation: The density of $\text{MgSr}$ when the unit cell of an alloy with a $1:1$ ratio of magnesium and strontium is identical to the unit cell of $\text{CsCl}$ is to be calculated.

Concept introduction: The density of the unit cell is defined as the mass of the unit cell divided by the volume of the unit cell.

$\text{Density}\text{of}\text{the}\text{unit}\text{cell}=\frac{\text{Mass}\text{of}\text{the}\text{unit}\text{cell}}{\text{Volume}\text{of}\text{the}\text{unit}\text{cell}}$

To determine: The density of $\text{MgSr}$ when the unit cell of an alloy with a $1:1$ ratio of magnesium and strontium is identical to the unit cell of $\text{CsCl}$.

Answer to Problem 12.96AP

Solution

The density of $\text{MgSr}$ is $\underset{\_}{3.13g/c{m}^3}$.

Explanation of Solution

Explanation

The density of $\text{MgSr}$ is calculated by the formula,

$\text{Density}\text{of}\text{the}\text{unit}\text{cell}=\frac{\text{Mass}\text{of}\text{the}\text{unit}\text{cell}}{\text{Volume}\text{of}\text{the}\text{unit}\text{cell}}$

$\text{MgSr}$ crystallizes in a body centered cubic unit cell (bcc). The number of atoms in bcc unit cell is 2. So, the unit cell contains one $\text{Mg}$ atom and one $\text{Sr}$ atom.

The atomic mass of $\text{Mg}$ atom is $24.3050\text{g}/\text{mol}$.

The atomic mass of $\text{Sr}$ atom is $\text{87}\text{.62}\text{g}/\text{mol}$.

The number of atoms in one mole is $\text{6}\text{.023}\times {\text{10}}^{\text{23}}\text{atoms}$.

Therefore, the mass of one magnesium atom is calculated by the formula,

$\text{Mass}\text{of}\text{Mg}\text{atom}=\frac{\text{Molar}\text{mass}\text{of}\text{Mg}}{\text{6}\text{.023}\times {\text{10}}^{\text{23}}\text{atoms}}$

Substitute the value of the molar mass of Mg in the above expression.

$\begin{array}{c}\text{Mass}\text{of}\text{Mg}\text{atom}=\left(\frac{\text{24}\text{.3050}}{\text{6}\text{.023}\times {\text{10}}^{\text{23}}}\right)\text{g}\\ =\text{4}\text{.035}\times {\text{10}}^{-\text{23}}\text{g}\end{array}$

Substitute the value of the molar mass of Sr in the above expression.

$\begin{array}{c}\text{Mass}\text{of}\text{Mg}\text{atom}=\left(\frac{\text{24}\text{.3050}}{\text{6}\text{.023}\times {\text{10}}^{\text{23}}}\right)\text{g}\\ =\text{4}\text{.035}\times {\text{10}}^{-\text{23}}\text{g}\end{array}$

The mass of one magnesium atom is calculated by the formula,

$\text{Mass}\text{of}\text{Sr}\text{atom}=\frac{\text{Molar}\text{mass}\text{of}\text{Sr}}{\text{6}\text{.023}\times {\text{10}}^{\text{23}}\text{atoms}}$

Substitute all these values in the formula for the mass of the unit cell,

$\begin{array}{c}\text{Mass}\text{of}\text{Sr}\text{atom}=\left(\frac{\text{87}\text{.62}\text{g}}{\text{6}\text{.023}\times {\text{10}}^{\text{23}}}\right)\text{g}\\ =\text{14}\text{.54}\times {\text{10}}^{-\text{23}}\text{g}\end{array}$

The mass of the $\text{MgSr}$ unit cell is calculated by the formula,

$\text{Mass}\text{of}\text{MgSr}\text{unit}\text{cell}=\text{Mass}\text{of}\text{Mg}\text{atom}+\text{Mass}\text{of}\text{Sr}\text{atom}$

Substitute the mass of the $\text{Mg}$ atom and that of the $\text{Sr}$ atom in the above formula,

$\begin{array}{c}\text{Mass}\text{of}\text{MgSr}\text{unit}\text{cell}=\text{Mass}\text{of}\text{Mg}\text{atom}+\text{Mass}\text{of}\text{Sr}\text{atom}\\ =\left(\text{4}\text{.035}\times {\text{10}}^{-\text{23}}\right)\text{g}+\left(\text{14}\text{.54}\times {\text{10}}^{-\text{23}}\right)\text{g}\\ =\left(\text{18}\text{.58}\times {\text{10}}^{-\text{23}}\right)\text{g}\end{array}$

The volume of $\text{MgSr}$ unit cell with an edge length $(l)$ of $390\text{pm}$ is calculated by the formula,

$v=l^3$

Substitute the value of the edge length in the above expression.

$\begin{array}{c}v=l^3\\ ={\left(390\text{pm}\right)}^3\times \frac{{\left({\text{10}}^{\text{-10}}\text{cm}\right)}^3}{{\text{pm}}^3}\\ =5.93\times {10}^{-23}{\text{cm}}^3\end{array}$

The density of $\text{MgSr}$ is calculated by the formula,

$d=\frac{m}{v}$

Substitute the value of the mass and that of volume in the above expression.

$\begin{array}{c}d=\frac{m}{v}\\ =\frac{18.58\times {10}^{-23}\text{g}}{5.93\times {10}^{-23}{\text{cm}}^3}\\ =\underset{\_}{3.13g/c{m}^3}\end{array}$

Conclusion

The density of $\text{MgSr}$ is $\underset{\_}{3.13g/c{m}^3}$.