Chapter 4.8, Problem 37AAP

(a)

To determine

The at%of each element in overall crystal.

Answer to Problem 37AAP

The at% of nickel and copper in overall crystal is $\text{19}\text{.4\%}\text{and}\text{77}\text{.7\%}$ respectively.

Explanation of Solution

Express the at% of nickel.

 $\text{Ni\%}=\frac{N_{\text{Ni}}}{N}\times 100\%$                                                                    (I)

Here, number of atoms of nickel is $N_{\text{Ni}}$ and overall number of atoms is $N$.

Express the at% of copper.

 $\text{Cu\%}=\frac{N_{\text{Cu}}}{N}\times 100\%$                                                                    (II)

Here, number of atoms of copperis $N_{\text{Cu}}$.

Conclusion:

Substitute $\text{7}\text{atoms}$ for $N_{\text{Ni}}$ and $\text{36}\text{atoms}$ for $N$ in Equation (I).

 $\begin{array}{c}\text{Ni\%}=\frac{\text{7}\text{atoms}}{\text{36}\text{atoms}}\times 100\%\\ =0.194\times 100\%\\ =19.4\%\end{array}$

Substitute $\text{28}\text{atoms}$ for $N_{\text{Cu}}$ and $\text{36}\text{atoms}$ for $N$ in Equation (II).

 $\begin{array}{c}\text{Cu\%}=\frac{28\text{atoms}}{\text{36}\text{atoms}}\times 100\%\\ =0.777\times 100\%\\ =77.7\%\end{array}$

Hence, the at% of nickel and copper in overall crystal is $\text{19}\text{.4\%}\text{and}\text{77}\text{.7\%}$ respectively.

(b)

To determine

The defect density in at%.

Answer to Problem 37AAP

The defect density in at%is $2.78\%$.

Explanation of Solution

Express the defect density in at%.

 $\text{Defect\%}=\frac{1}{N}\times 100\%$                                                                           (III)

Conclusion:

Substitute $\text{36}\text{atoms}$ for $N$ in Equation (III).

 $\begin{array}{c}\text{Defect\%}=\frac{1}{36}\times 100\%\\ =2.78\%\end{array}$

Hence, the defect density in at% is $2.78\%$.

(c)

To determine

The wt% of each metal.

Answer to Problem 37AAP

The wt% of nickel is $18.7\%$ and copper is $81.3\%$.

Explanation of Solution

Express the mass of each nickel atom.

 $m_{\text{Ni}}=\frac{M_{\text{Ni}}}{N_{\text{Av}}}\times \text{Ni\%}$    (IV)

Here, molar mass of nickel is $M_{\text{Ni}}$ and number of atom per mole or Avogadro number is $N_{\text{Av}}$.

Express the mass of each copper atom.

 $m_{\text{cu}}=\frac{M_{\text{Cu}}}{N_{\text{Av}}}\times \text{Cu\%}$                                                                               (V)

Here, molar mass of copperis $M_{\text{Cu}}$.

Express the total mass of nickel and copper atoms.

 $m=m_{\text{Ni}}+m_{\text{Cu}}$                                                                               (VI)

Express the wt% of nickel.

 $\text{wt\%Ni}=\frac{m_{\text{Ni}}}{m}\times 100\%$                                                                              (VII)

Express the wt% of copper.

 $\text{wt\%Cu}=\frac{m_{\text{Cu}}}{m}\times 100\%$                                                                             (VIII)

Conclusion:

Write the molar mass of copper, nickel and number of atom per mole or Avogadro number.

 $\begin{array}{l}{M}_{\text{Cu}}=63.55\text{g}\\ M_{\text{Ni}}=58.69\text{g}\\ N_{\text{Av}}=6.02\times {10}^{23}\text{atoms}\end{array}$

Substitute $58.69\text{g}$ for $M_{\text{Ni}}$, $0.194$ for $\text{Ni\%}$ and $6.02\times {10}^{23}\text{atoms}$ for $N_{\text{Av}}$ in Equation (IV).

 $\begin{array}{c}{m}_{\text{Ni}}=\frac{58.69\text{g}}{6.02\times {10}^{23}\text{atoms}}\times 0.194\\ =1.891\times {10}^{-23}\text{g}\end{array}$

Substitute $63.55\text{g}$ for $M_{\text{Cu}}$, $0.777$ for $\text{Cu\%}$ and $6.02\times {10}^{23}\text{atoms}$ for $N_{\text{Av}}$ in Equation (V).

 $\begin{array}{c}{m}_{\text{Cu}}=\frac{63.55\text{g}}{6.02\times {10}^{23}\text{atoms}}\times 0.777\\ =8.202\times {10}^{-23}\text{g}\end{array}$

Substitute $1.891\times {10}^{-23}\text{g}$ for $m_{\text{Ni}}$ and $8.202\times {10}^{-23}\text{g}$ for $m_{\text{Cu}}$ in Equation (VI).

 $\begin{array}{c}m=1.891\times {10}^{-23}\text{g}+8.202\times {10}^{-23}\text{g}\\ =1.009\times {10}^{-22}\text{g}\end{array}$

Substitute $1.891\times {10}^{-23}\text{g}$ for $m_{\text{Ni}}$ and $1.009\times {10}^{-22}\text{g}$ for $m$ in Equation (VII).

 $\begin{array}{c}\text{wt\%Ni}=\frac{1.891\times {10}^{-23}\text{g}}{1.009\times {10}^{-22}\text{g}}\times 100\%\\ =18.7\%\end{array}$

Substitute $8.2021\times {10}^{-23}\text{g}$ for $m_{\text{Cu}}$ and $1.009\times {10}^{-22}\text{g}$ for $m$ in Equation (VIII).

 $\begin{array}{c}\text{wt\%Cu}=\frac{8.2021\times {10}^{-23}\text{g}}{1.009\times {10}^{-22}\text{g}}\times 100\%\\ =81.3\%\end{array}$

Hence, the wt% of nickel is $18.7\%$ and copper is $81.3\%$.