Chapter 18, Problem 18.132QA

Interpretation Introduction

To find:

a) If lithium, gold, and titanium can form substitutional alloy with $Al$.

b) The density of $LiAl$.

Answer to Problem 18.132QA

Solution:

a) Lithium, $LiAl$, gold $\left(AuC{l}_2\right),$ and titanium $\left(A{l}_{3}Ti\right)$ will form substitutional alloys with aluminum.

b) Density of $LiAl$ is $2.85 g/c{m}^3$.

Explanation of Solution

a) Substitutional alloys may form when the difference in atomic radii between the alloying elements is less than 15%.

1) Concept:

Aluminum forms alloys with different elements; hence it acts as a host metal. Other metals, such as lithium, gold, and titanium form alloys with a particular metal, so they are called non-host metals. The atomic radii of all the elements is given, and the ratio between the radii of non-host elements and host elements will determine whether they will form substitutional alloys.

2) Formula:

$\frac{r_{non-host- r_{host}}}{r_{non-host}} <0.15$

3) Given:

i) Atomic radii of $Al=143 pm$

ii) Atomic radii of $Li=152 pm$

iii) Atomic radii of $Au=144 pm$

iv) Atomic radii of $Ti=147 pm$

4) Calculations:

For $LiAl$ alloy, the ratio of the radii of $Li$ to $Al$ is

$\frac{r_{non-host}- r_{host}}{r_{non-host}}= \frac{152 pm-143 pm}{152 pm}=0.063 or 6.3\%$

For $A{l}_{2}Au$ alloy, the ratio of the radii of $Au$ to $Al$ is

$\frac{r_{non-host}- r_{host}}{r_{non-host}}= \frac{144 pm-143 pm}{144 pm}=0.0069 or 0.69\%$

For $A{l}_{3}Ti$ alloy, the ratio of the radii of $Ti$ to $Al$ is

$\frac{r_{non-host}- r_{host}}{r_{non-host}}= \frac{147 pm-143 pm}{147 pm}=0.028 or 2.8\%$

Since all the ratios of the above alloys is less than $15\%$, all the metals will form substitutional alloys with aluminum.

b) Calculate the density of $LiAl$

1) Concept:

To calculate the density of LiAl, we will calculate the mass and volume of the unit cell from the number of atoms present in bcc structure. First we will calculate the edge length from the atomic radii and calculate the volume of the unit cell. We will calculate the mass of the unit cell by using number of atoms present in per unit cell and molar mass.

2) Formulae:

i) $r= \frac{l\sqrt{3}}{4}$

ii) $V_{unit cell}=l^3 c{m}^3$

iii) $density= \raisebox{1ex}{${mass}_{unit cell}$}\!\left/ \!\raisebox{-1ex}{$V_{unit cell}$}\right.$

3) Given:

i) Atomic radius of $Li=152 pm$

ii) Atomic radius of $Al=143 pm$

iii) Atomic mass of $Li=6.941 g$

iv) Atomic mass of $Al=26.98 g$

v) Number of atoms in the unit cell $=2$

vi) Avogadro’s Number $1 mol =6.023 \times {10}^{23} atoms$

4) Calculations:

First calculate the edge length of the $LiAl$ alloy from the atomic radii.

The value of body diagonal has been calculated as

$body diagonal, \left(4r\right)= 2r_{Al}+2r_{Li}$

$body diagonal, \left(4r\right)= \left(2 \times 143\right) pm+\left(2 \times 152\right) pm$

$body diagonal, \left(4r\right)=286 pm+304 pm$

$body diagonal, \left(4r\right)=590 pm$

Edge length of the alloy:

$l= \frac{4r}{\sqrt{3}}= \frac{590 pm}{\sqrt{3}}= 340.64 pm=3.41 \times {10}^{-8} cm$

Volume of unit cell:

$V_{unit cell}=l^3 c{m}^3= {\left(3.41 \times {10}^{-8}\right)}^3 c{m}^3=3.965 \times {10}^{-23} c{m}^3$

Molar mass of the alloy

$molar mass of LiAl=molar mass of Li+Molar mass of Al$

$molar mass of LiAl=6.941 g+26.98 g$

$molar mass of LiAl=33.921 g$

Mass of the unit cell:

$2 atoms \times \frac{1 mol}{6.022 \times {10}^{23}atoms} \times \frac{33.921 g }{1 mol}=1.13 \times {10}^{-22} g$

${mass}_{unit cell}=1.13 \times {10}^{-22} g$

Density of the alloy:

$density, d= \raisebox{1ex}{${mass}_{unit cell}$}\!\left/ \!\raisebox{-1ex}{$V_{unit cell}$}\right.$

$d= \frac{1.13 \times {10}^{-22} g}{3.965 \times {10}^{-23} c{m}^3}$

$\therefore d=2.85 g/c{m}^3$

Hence, the density of the $LiAl$ alloy is $2.85 g/c{m}^3$.

Conclusion:

The substitutional alloy is found from the atomic radii, and the density of LiAl is calculated from the volume of unit cell and mass of unit cell.