Chapter 12, Problem 12.97AP

Interpretation Introduction

Interpretation: The given two alloys $\text{AuZn}$ and $\text{AgZn}$ have the same unit cell. The alloy that is denser is to be identified.

Concept introduction: An alloy is a solution that is made up of mixing of atoms of an element with other element which results in change in properties of the host element. Gold and silver both crystallize in a face centered cubic unit cell.

The density of the unit cell is calculated by the formula,

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

To determine: The alloy that is denser.

Answer to Problem 12.97AP

Solution

The alloy $\text{AuZn}$ is denser than the alloy $\text{AgZn}$.

Explanation of Solution

Explanation

Gold and silver both crystallize in a face centered cubic unit cell.

There are 4 atoms in an fcc unit cell.

The molar mass of $\text{AuZn}$ is $262.29\text{g}/\text{mol}$.

The mass of a unit cell is calculated by the formula,

$\begin{array}{c}\text{Mass}\text{of}\text{the}\text{unit}\text{cell}=\text{Mass}\text{of}\text{each}\text{atom}\times \text{Number}\text{of}\text{atoms}\text{in}\text{unit}\text{cell(}n\text{)}\\ =\frac{\text{Atomic}\text{mass}}{\text{Avogadro's}\text{number}}\times \text{number}\text{of}\text{atoms}\text{in}\text{unit}\text{cell(}n)\end{array}$

Substitute the value of the molar mass of $\text{AuZn}$ in the formula of the mass of the unit cell,

$\begin{array}{c}\text{Mass}\text{of}\text{AuZn}\text{unit}\text{cell}=\frac{\text{262}\text{.29}\text{g}\text{AuZn}}{\text{mol}\text{AuZn}}\times \frac{\text{1}\text{mol}}{\text{6}\text{.023}\times {\text{10}}^{\text{23}}\text{atoms}}\times \text{4}\text{atoms}\\ =\text{1}\text{.741}\times {\text{10}}^{\text{-21}}\text{g}\end{array}$

The given edge length is $\left(l\right)$ $\text{319}\text{pm}$.

The volume of an $\text{AuZn}$ unit cell is calculated by the formula,

$v=l^3$

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

$\begin{array}{c}v=l^3\\ ={\left(\text{319}\text{pm}\right)}^{\text{3}}\times \frac{{\left({\text{10}}^{\text{-10}}\text{cm}\right)}^{\text{3}}}{{\text{pm}}^{\text{3}}}\\ =\text{3}\text{.246}\times {\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\end{array}$

Density is calculated by the formula,

$\begin{array}{c}\text{Density}\text{of}\text{the}\text{unit}\text{cell}=\frac{\text{Mass}\text{of}\text{the}\text{unit}\text{cell}}{\text{Volume}\text{of}\text{the}\text{unit}\text{cell}}\\ d=\frac{m}{v}\end{array}$

Substitute the value of the mass and volume of the unit cell in the above expression.

$\begin{array}{c}d=\frac{m}{v}\\ =\frac{\text{1}\text{.741}\times {\text{10}}^{\text{-21}}\text{g}}{3.246\times {10}^{-23}{\text{cm}}^{\text{3}}}\\ =53.63\text{g}/{\text{cm}}^{\text{3}}\end{array}$

Therefore, the density of $\text{AuZn}$ is $53.63\text{g}/{\text{cm}}^{\text{3}}$.

The mass of a unit cell is calculated by the formula,

$\begin{array}{c}\text{Mass}\text{of}\text{the}\text{unit}\text{cell}=\text{Mass}\text{of}\text{each}\text{atom}\times \text{Number}\text{of}\text{atoms}\text{in}\text{unit}\text{cell(}n\text{)}\\ =\frac{\text{Atomic}\text{mass}}{\text{Avogadro's}\text{number}}\times \text{number}\text{of}\text{atoms}\text{in}\text{unit}\text{cell(}n)\end{array}$

The mass of $\text{AgZn}$ is calculated by the formula,

$\begin{array}{c}\text{Mass}\text{of}\text{AgZn}\text{unit}\text{cell}=\frac{173.25\text{g}\text{AgZn}}{\text{mol}\text{AgZn}}\times \frac{\text{1}\text{mol}}{\text{6}\text{.023}\times {\text{10}}^{\text{23}}\text{atoms}}\times \text{4}\text{atoms}\\ =\text{1}\text{.150}\times {\text{10}}^{\text{-21}}\text{g}\end{array}$

The given edge length is $\left(l\right)$ $\text{316}\text{pm}$.

The volume of an $\text{AuZn}$ unit cell is calculated by the formula,

$v=l^3$

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

$\begin{array}{c}v=l^3\\ ={\left(\text{316}\text{pm}\right)}^{\text{3}}\times \frac{{\left({\text{10}}^{\text{-10}}\text{cm}\right)}^{\text{3}}}{{\text{pm}}^{\text{3}}}\\ =\text{3}\text{.155}\times {\text{10}}^{\text{-23}}{\text{cm}}^{\text{3}}\end{array}$

Density is calculated by the formula,

$\begin{array}{c}\text{Density}\text{of}\text{the}\text{unit}\text{cell}=\frac{\text{Mass}\text{of}\text{the}\text{unit}\text{cell}}{\text{Volume}\text{of}\text{the}\text{unit}\text{cell}}\\ d=\frac{m}{v}\end{array}$

Substitute the value of the mass and volume of the unit cell in the above expression.

$\begin{array}{c}d=\frac{m}{v}\\ =\frac{\text{1}\text{.150}\times {\text{10}}^{\text{-21}}\text{g}}{3.155\times {10}^{-23}{\text{cm}}^{\text{3}}}\\ =36.45\text{g}/{\text{cm}}^{\text{3}}\end{array}$

Thus, the density of $\text{AgZn}$ is $53.63\text{g}/{\text{cm}}^{\text{3}}$.

Therefore, the alloy $\text{AuZn}$ is denser than the alloy $\text{AgZn}$.

Conclusion

The alloy $\text{AuZn}$ is denser than the alloy $\text{AgZn}$.