Chapter 2, Problem 154GQ

Estimating the radius of a lead atom.

1. (a) You are given a cube of lead that is 1.000 cm on each side. The density of lead is 11.35 g/ cm³. How many atoms of lead are in the sample?
2. (b) Atoms are spherical; therefore, the lead atoms in this sample cannot fill all the available space As an approximation, assume that 60% of the space of the cube is filled with spherical lead atoms. Calculate the volume of one lead atom from this information. From the calculated volume (V) and the formula (4/3) πτ³ for the volume of a sphere, estimate the radius (r) of a lead atom.

Interpretation Introduction

Interpretation: The number of atoms for lead in given sample of lead cube, with each side of value $\text{1}\text{cm}$ needed to be calculated.

Concept introduction:

Conversion formula for mass of a molecule and number moles,

$Numberofmoles=\frac{Massingrams}{Molarmass}$

Equation for number of atoms is,

    $\text{Number of moles}\times \text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\text{atoms}\text{=}\text{number of atoms}$

Equation for density from volume and mass is,

    $Density=\frac{Mass}{Volume}$

Equation for finding Volume of sphere is,

    $Volume=\left(4/3\right)\pi r^3$

Answer to Problem 154GQ

The number of atoms of lead is $3.3\text{×}{\text{10}}^{\text{22}}\text{atoms}$

Explanation of Solution

The side of the lead cube is given that $1cm$.

Therefore, the volume of the lead cube is,

    ${(1cm)}^3=1c{m}^3$

The density of the lead cube is given as $11.35g/c{m}^3$.

Equation for mass from volume and density is,

    $Density\times Volume=Mass$

Therefore, the mass of lead cube is,

    $11.35g/c{m}^3\times 1c{m}^3=11.35g$

Conversion formula for mass of a molecule and number moles,

$Numberofmoles=\frac{Massingrams}{Molarmass}$

Therefore, the number of lead atoms is,

    $Numberofmoles=\frac{11.35g}{207.2g/mol}=0.05477mol$

Equation for number of atoms is,

    $\text{Number of moles}\times \text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\text{atoms}\text{=}\text{number of atoms}$

Therefore, the number of lead atoms in the sample is,

    $\text{0}\text{.05477}\times \text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\text{atoms}\text{=}3.3\text{×}{\text{10}}^{\text{22}}\text{atoms}$

Interpretation Introduction

Interpretation: The volume of one lead atom and its radius have to be calculated under given conditions.

Concept introduction:

Conversion formula for mass of a molecule and number moles,

$Numberofmoles=\frac{Massingrams}{Molarmass}$

Equation for number of atoms is,

    $\text{Number of moles}\times \text{6}\text{.022}\text{×}{\text{10}}^{\text{23}}\text{atoms}\text{=}\text{number of atoms}$

Equation for density from volume and mass is,

    $Density=\frac{Mass}{Volume}$

Equation for finding Volume of sphere is,

    $Volume=\left(4/3\right)\pi r^3$

Answer to Problem 154GQ

The volume and radius of one lead atom is $1.8\text{×}{\text{10}}^{\text{-23}}c{m}^3$ and $0.7572\text{×}{\text{10}}^{\text{-23}}cm$ respectively.

Explanation of Solution

The volume of the lead cube is found that $1c{m}^3$.

If 60% of the cube is filled with $3.3\text{×}{\text{10}}^{\text{22}}$ lead atom spheres, then the volume of one lead atom is,

    $\frac{60}{100}\times 1c{m}^3\times \frac{1}{3.3\text{×}{\text{10}}^{\text{22}}}=1.8\text{×}{\text{10}}^{\text{-23}}c{m}^3$

Equation for finding Volume of sphere is,

    $Volume=\left(4/3\right)\pi r^3$

Therefore, the radius of one lead atom is,

    $\begin{array}{l}1.8\text{×}{\text{10}}^{\text{-23}}c{m}^3=\left(4/3\right)\times 3.14\times r^3\\ r^3=\frac{1.8\text{×}{\text{10}}^{\text{-23}}c{m}^3}{\left(4/3\right)\times 3.14}=0.4342\text{×}{\text{10}}^{\text{-23}}cm\\ r=0.7572\text{×}{\text{10}}^{\text{-23}}cm\end{array}$