Chapter 2, Problem 2.8P

Interpretation Introduction

(a)

Interpretation:

The number of nickel atoms for the given plate should be determined.

Concept Introduction:

The mass of nickel is determined using the formula:

i.e. Mass of nickel

  $\begin{array}{l}=volume\times density\\ =area\times thickness\times density\\ =NumberofNickelAtoms=moles\times Avogadrosnumber\end{array}$

Answer to Problem 2.8P

The number of nickel atoms required are around $6\times {10}^{23}$ atoms.

Explanation of Solution

Given:

  $\begin{array}{l}area=200i{n}^2=1290.32c{m}^2\\ thikness=0.002in=0.00508cm\end{array}$ ---------------[1 in =2.54 cm]

Volume of nickel

  $\begin{array}{l}=area\times thikness\\ =1290.32\times 0.00508\\ =6.555c{m}^3\end{array}$

Density of nickel $=8.9191c{m}^2$

Mass if nickel

  $\begin{array}{l}=volume\times density\\ =6.555\times 8.91\\ =58.41gram\end{array}$

Number of nickel atoms

  $\begin{array}{l}=moles\times Avogadrosnumber\\ =0.955\times 6.022\times {10}^{23}\end{array}$

  $\begin{array}{l}=5.99\times {10}^{23}\\ =6\times {10}^{23}\end{array}$ atoms.

Interpretation Introduction

(b)

Interpretation:

Number of moles of nickel required for the given plate should be determined.

Concept Introduction:

Moles of nickel is given as per the following formula:

  $molesofnikel=\frac{mass}{molarmass}$

Answer to Problem 2.8P

Number of moles required to plate a steel part is equal to 1 mol.

Explanation of Solution

Given:

  $\begin{array}{l}area=200i{n}^2=1290.32c{m}^2\\ thikness=0.002in=0.00508cm\end{array}$ ---------------[1 in =2.54 cm]

Volume of nickel

  $\begin{array}{l}=area\times thikness\\ =1290.32\times 0.00508\\ ~6.555c{m}^3\end{array}$

Density of nickel $=8.9191c{m}^2$

Mass if nickel

  $\begin{array}{l}=volume\times density\\ =6.555\times 8.91\\ =58.41gram\end{array}$

  $\begin{array}{l}massofnickel=\frac{mass}{molarmass}\\ =\frac{58.40}{58.69}\\ =0.955mol\\ ~1mol\end{array}$