Chapter 4.8, Problem 56SEP

(a)

To determine

Estimate the density of a $\text{75 wt\% copper}-\text{25 wt\% Ni}$ alloy.

Answer to Problem 56SEP

The density of copper and nickel alloy is $\text{8}\text{.94}\text{g}/{\text{cm}}^{\text{3}}$.

Explanation of Solution

Express the density of alloy.

 ${\rho }_{\text{alloy}}=\left(\text{wt\%Cu}\right)\left({\rho }_{\text{Cu}}\right)+\left(\text{wt\%Ni}\right)\left({\rho }_{\text{Ni}}\right)$                                              (I)

Here, density of copper and nickel is ${\rho }_{\text{Cu}}\text{and}{\rho }_{\text{Ni}}$ respectively.

Conclusion:

Write the density of copper and nickel and also write the number of atom per mole or Avogadro number.

 $\begin{array}{l}{\rho }_{\text{Cu}}=8.96\text{g}/{\text{cm}}^{\text{3}}\\ {\rho }_{\text{Ni}}=8.9\text{g}/{\text{cm}}^{\text{3}}\\ N=6.02\times {10}^{23}\text{atoms}\end{array}$

Write the molar or atomic mass of copper and nickel.

 $\begin{array}{l}{M}_{\text{Cu}}=63.55\text{g}\\ M_{\text{Ni}}=58.69\text{g}\end{array}$

Substitute $\text{25}\text{g}$ for $\text{wt\%Ni}$, $\text{75}\text{g}$ for $\text{wt\%Cu}$, $8.96\text{g}/{\text{cm}}^{\text{3}}$ for ${\rho }_{\text{Cu}}$ and $8.9\text{g}/{\text{cm}}^{\text{3}}$ for ${\rho }_{\text{Ni}}$ in Equation (I).

 $\begin{array}{c}{\rho }_{\text{alloy}}=\left(\text{0}\text{.75}\right)\left(8.96\text{g}/{\text{cm}}^{\text{3}}\right)+\left(0.25\right)\left(8.9\text{g}/{\text{cm}}^{\text{3}}\right)\\ =8.94\text{g}/{\text{cm}}^{\text{3}}\end{array}$

Hence, the density of copper and nickel alloy is $\text{8}\text{.94}\text{g}/{\text{cm}}^{\text{3}}$.

(b)

To determine

The most probable crystal structure for this copper and nickel alloy.

Answer to Problem 56SEP

The most probable crystal structure for this copper and nickel alloy is FCC.

Explanation of Solution

Write the information data for copper and nickel as in Table (1).

Element	Crystal structure
Copper	FCC
Nickel	FCC

As both of the copper and nickel have similar FCC crystal structure, then the formed alloy will also have the FCC structure.

Hence, the most probable crystal structure for this copper and nickel alloy is FCC.

(c)

To determine

The mass in grams of the atoms inside a unit cell of the copper nickel alloy.

Answer to Problem 56SEP

The mass in grams of the atoms inside a unit cell of the copper nickel alloy is $\text{4}\text{.13}\times {\text{10}}^{-22}\text{g}$.

Explanation of Solution

Consider the mass of alloy as $\text{100}\text{g}$.

Express the total atoms of copper.

 $n_{\text{Cu}}=\frac{\left(\text{wt\%Cu}\right)N}{M_{\text{Cu}}}$                                                                           (II)

Here, atomic mass of copper is $M_{\text{Cu}}$ and number of atom per mole or Avogadro number is $N$.

Express the total atoms of nickel.

 $n_{\text{Ni}}=\frac{\left(\text{wt\%Ni}\right)N}{M_{\text{Ni}}}$                                                                          (III)

Here, atomic mass of nickel is $M_{\text{Ni}}$.

Express the mass in grams of the atoms inside a unit cell of the copper nickel alloy.

 $m_{\text{cell}}=n_{\text{Cu}}\frac{M_{\text{cu}}}{N}+n_{\text{Ni}}\frac{M_{\text{Ni}}}{N}$                                                                     (IV)

Conclusion:

Consider the mass of the alloy be $\text{100}\text{g}$.

Substitute $\text{75}\text{g}$ for $\text{wt\%Cu}$, $6.02\times {10}^{23}\text{atoms}$ for $N$ and $63.55\text{g}$ for $M_{\text{Cu}}$ in Equation (II).

 $\begin{array}{c}{n}_{\text{Cu}}=\frac{\left(\text{75}\text{g}\right)\left(6.02\times {10}^{23}\text{atoms}\right)}{63.55\text{g}}\\ =7.10\times {10}^{23}\text{atoms}\end{array}$                                                (V)

Substitute $\text{25}\text{g}$ for $\text{wt\%Ni}$, $6.02\times {10}^{23}\text{atoms}$ for $N$ and $58.69\text{g}$ for $M_{\text{Cu}}$ in Equation (III).

 $\begin{array}{c}{n}_{\text{Ni}}=\frac{\left(\text{25}\text{g}\right)\left(6.02\times {10}^{23}\text{atoms}\right)}{58.69\text{g}}\\ =2.56\times {10}^{23}\text{atoms}\end{array}$                                            (VI)

Thus from Equations (V) and (VI), it is observed that there are $\simeq 73\%\text{Cu}\text{atoms}\text{to}\simeq \text{27\%Ni}\text{atoms}$. Hence, there will be $\text{4}\text{atoms}$ per unit cell based on the calculations that is 3 copper atoms to 1 atom in each cell.

 $\begin{array}{l}{n}_{\text{Cu}}=3\text{atoms}\\ n_{\text{Ni}}=1\text{atoms}\end{array}$

Substitute $\text{3}\text{atoms}$ for $n_{\text{Cu}}$, $1\text{atoms}$ for $n_{\text{Ni}}$, $6.02\times {10}^{23}\text{atoms}$ for $N$, $63.55\text{g}$ for $M_{\text{Cu}}$, and $58.69\text{g}$ for $M_{\text{Cu}}$ in Equation (IV).

 $\begin{array}{c}{m}_{\text{cell}}=\left(\text{3}\text{atoms}\right)\left[\frac{63.55\text{g}}{6.02\times {10}^{23}\text{atoms}}\right]+\left(1\text{atoms}\right)\left[\frac{58.69\text{g}}{6.02\times {10}^{23}\text{atoms}}\right]\\ =\text{4}\text{.13}\times {\text{10}}^{-22}\text{g}\end{array}$

Hence, the mass in grams of the atoms inside a unit cell of the copper nickel alloy is $\text{4}\text{.13}\times {\text{10}}^{-22}\text{g}$.

(d)

To determine

The lattice constant for the copper nickel alloy.

Answer to Problem 56SEP

The lattice constant for the copper nickel alloy is $3.58\times {10}^{-8}\text{cm}$.

Explanation of Solution

Express the lattice constant for the copper nickel alloy.

 $a=\frac{m_{\text{cell}}}{{\rho }_{\text{alloy}}}$                                                                               (VII)

Conclusion:

Substitute $\text{4}\text{.13}\times {\text{10}}^{-22}\text{g}$ for $m_{\text{cell}}$ and $8.94\text{g}/{\text{cm}}^{\text{3}}$ for ${\rho }_{\text{alloy}}$ in Equation (VII).

 $\begin{array}{c}a=\frac{\text{4}\text{.13}\times {\text{10}}^{-22}\text{g}}{8.94\text{g}/{\text{cm}}^{\text{3}}}\\ =3.58\times {10}^{-8}\text{cm}\end{array}$

Hence, the lattice constant for the copper nickel alloy is $3.58\times {10}^{-8}\text{cm}$.