Chapter 10.2, Problem 19PP

a.

Interpretation Introduction

Interpretation : The number of atoms in the given sample of Li needs to be determined.

Concept introduction : One mole of an atom contains Avogadro number $\left(6.022\times {10}^{23}\right)$ of atoms.

Answer to Problem 19PP

$\text{Number of atoms in Li is}=4.7\times {10}^{24}$.

Explanation of Solution

Given information:

$\text{Weight of the Li is}=55.2g.$

Formula used:

$\text{Number of atoms}=\frac{\text{weight }\times 6.022\times {10}^{23}}{\text{gram molecular weight}}$

Calculation:

Given $\text{Weight of the Li is}=55.2g.$$\text{Atomic weight of Li is}=6.939$

Substitute in the above formula.

$\begin{array}{l}\text{Number of atoms}=\text{moles}\times 6.022\times {10}^{23}\\ \text{ = }\frac{\text{weight}}{\text{gram molecular weight}}\times 6.022\times {10}^{23}\\ =\frac{55.2}{6.939}\times 6.022\times {10}^{23}\\ =4.7\times {10}^{24}\end{array}$

$\text{Therefore the number of atoms of }55.2g\text{ of Li is }4.7\times {10}^{24}$.

b.

Interpretation Introduction

Interpretation : The number of atoms in the given sample of Pb needs to be determined.

Concept introduction : One mole of an atom contains Avogadro number $\left(6.022\times {10}^{23}\right)$ of atoms.

Answer to Problem 19PP

$\text{Number of atoms in Pb is}=6.68\times {10}^{20}$.

Explanation of Solution

Given information:

$\text{Weight of the Pb is}=0.230g.$

Formula used:

$\text{Number of atoms}=\frac{\text{weight }\times 6.022\times {10}^{23}}{\text{gram molecular weight}}$

Calculation:

$\begin{array}{l}\text{Given weight of Pb is}=0.230g\\ \text{Atomic weight of Pb is}=207.19\end{array}$

Substitute in the above formula

$\begin{array}{l}\text{Number of atoms}=\text{moles}\times 6.022\times {10}^{23}\\ \text{ = }\frac{\text{weight}}{\text{gram molecular weight}}\times 6.022\times {10}^{23}\\ =\frac{0.230}{207.19}\times 6.022\times {10}^{23}\\ =6.68\times {10}^{20}\end{array}$

$\text{Therefore the number of atoms of 0}\text{.230}g\text{ of Pb is 6}\text{.68}\times {10}^{20}$.

c.

Interpretation Introduction

Interpretation : The number of atoms in the given sample of Hg needs to be determined.

Concept introduction : One mole of an atom contains Avogadro number $\left(6.022\times {10}^{23}\right)$ of atoms.

Answer to Problem 19PP

$\text{Number of atoms in Hg is}=3.45\times {10}^{22}$.

Explanation of Solution

Given information:

$\text{Weight of the Hg is}=11.5g.$

Formula used:

$\text{Number of atoms}=\frac{\text{weight }\times 6.022\times {10}^{23}}{\text{gram molecular weight}}$

$\begin{array}{l}\text{Given weight of Hg is}=11.5g.\\ \text{Atomic weight of Hg is}=\mathrm{200.59.}\end{array}$

Substitute in the above formula

$\begin{array}{l}\text{Number of atoms}=\text{moles}\times 6.022\times {10}^{23}\\ \text{ = }\frac{\text{weight}}{\text{gram molecular weight}}\times 6.022\times {10}^{23}\\ =\frac{11.5}{200.59}\times 6.022\times {10}^{23}\\ =3.452\times {10}^{22}\end{array}$

$\text{Therefore the number of atoms of 11}\text{.5}g\text{ of Hg is 3}\text{.452}\times {10}^{22}$.