Chapter 10.2, Problem 20PP

(a)

Interpretation Introduction

Interpretation:

The mass in grams of a given number of atoms of Bi needs to be determined.

Concept introduction:

One mole of an atom contains Avogadro number $\left(6\cdot 022\times {10}^{23}\right)$ of atoms, and weight of an and mass of an atom is product of number of moles and its atomic mass.

Answer to Problem 20PP

The mass in grams of $6\cdot 02\times {10}^{24}$ atoms Bi is $=2\cdot 0898\times {10}^3\text{ g}$.

Explanation of Solution

Given information:

Number of atoms of Bi is $=6\cdot 02\times {10}^{24}$

Formula used:

$\begin{array}{l}\text{weight}=\text{moles}\times \text{Atomic weight}\\ \text{moles }=\frac{\text{number of atoms}}{6\cdot 022\times {10}^{23}}\end{array}$

Calculation:

Given number of atoms of Bi is $=6\cdot 02\times {10}^{24}$

Atomic weight of Bi is $208\cdot 980$

Substitute in the above formula

$\begin{array}{l}\text{moles }=\frac{\text{number of atoms}}{6\cdot 022\times {10}^{23}}\\ \text{weight}=\frac{\text{Number of atoms}}{6\cdot 023\times {10}^{23}}\times \text{atomic weight}\\ \text{=}\frac{6\cdot 02\times {10}^{24}}{6\cdot 023\times {10}^{23}}\times 208\cdot 980\\ =2089\cdot 8\text{ g}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The mass in grams of a given number of atoms of Mn needs to be determined.

Concept introduction:

One mole of an atom contains Avogadro number $\left(6\cdot 022\times {10}^{23}\right)$ of atoms, and weight of an and mass of an atom is product of number of moles and its atomic mass.

Answer to Problem 20PP

The mass in grams of $1\times {10}^{24}$ atoms of Mn is $=9\cdot 124\times {10}^1\text{ g}$.

Explanation of Solution

Given information:

Number of atoms of Mn is $=1\times {10}^{24}$

Formula used:

$\begin{array}{l}\text{weight}=\text{moles}\times \text{Atomic weight}\\ \text{moles }=\frac{\text{number of atoms}}{6\cdot 022\times {10}^{23}}\end{array}$

Calculation:

Given number of atoms of Mn is $=1\times {10}^{24}$

Atomic weight of Mn is $=54\cdot 93$

Substitute in the above formula

$\begin{array}{l}\text{moles }=\frac{\text{number of atoms}}{6\cdot 022\times {10}^{23}}\\ \text{weight}=\frac{\text{Number of atoms}}{6\cdot 023\times {10}^{23}}\times \text{atomic weight}\\ \text{=}\frac{1\times {10}^{24}}{6\cdot 023\times {10}^{23}}\times 54\cdot 93\\ =9\cdot 124\times {10}^1\text{ g}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The mass in grams of a given number of atoms of He needs to be determined.

Concept introduction:

One mole of an atom contains Avogadro number $\left(6\cdot 022\times {10}^{23}\right)$ of atoms, and weight of an and mass of an atom is product of number of moles and its atomic mass.

Answer to Problem 20PP

The mass in grams of $3\cdot 4\times {10}^{22}$ atoms He is $=2\cdot 259\times {10}^{-1}\text{ g}$.

Explanation of Solution

Given information:

Number of atoms of He is $=3\cdot 4\times {10}^{22}$

Formula used:

$\text{weight}=\frac{\text{number of atoms}}{6\cdot 022\times {10}^{23}}\times \text{Atomic weight}$

Calculation:

Given number of atoms of He is $=3\cdot 4\times {10}^{22}$

Atomic weight of He is $=4$

Substitute in the above formula

$\begin{array}{l}\text{weight}=\frac{\text{Number of atoms}}{6\cdot 023\times {10}^{23}}\times \text{atomic weight}\\ \text{=}\frac{3\cdot 4\times {10}^{22}}{6\cdot 023\times {10}^{23}}\times 4\\ =2\cdot 259\times {10}^{-1\text{ g}}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The mass in grams of a given number of atoms of N needs to be determined.

Concept introduction:

One mole of an atom contains Avogadro number $\left(6\cdot 022\times {10}^{23}\right)$ of atoms, and weight of an and mass of an atom is product of number of moles and its atomic mass.

Answer to Problem 20PP

The mass in grams of $1\cdot 5\times {10}^{15}$ atoms of N is $=3\cdot 48\times {10}^{-8}\text{ g}$.

Explanation of Solution

Given information:

Number of atoms of N is $=1\cdot 5\times {10}^{15}$

Formula used:

$\text{weight}=\frac{\text{number of atoms}}{6\cdot 022\times {10}^{23}}\times \text{Atomic weight}$

Calculation:

Given number of atoms of N is $=1\cdot 5\times {10}^{15}$

Atomic weight of N is $=14$

Substitute in the above formula

$\begin{array}{l}\text{weight}=\frac{\text{Number of atoms}}{6\cdot 023\times {10}^{23}}\times \text{atomic weight}\\ \text{=}\frac{1\cdot 5\times {10}^{15}}{6\cdot 023\times {10}^{23}}\times 14\\ =3\cdot 48\times {10}^{-8}\text{ g}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The mass in grams of a given number of atoms of U needs to be determined.

Concept introduction:

One mole of an atom contains Avogadro number $\left(6\cdot 022\times {10}^{23}\right)$ of atoms, and weight of an and mass of an atom is product of number of moles and its atomic mass.

Answer to Problem 20PP

The mass in grams of $1\cdot 5\times {10}^{15}$ atoms U is $=5\cdot 93\times {10}^{-7}\text{ g}$.

Explanation of Solution

Given information:

Number of atoms of U is $=5\cdot 93\times {10}^{-7}\text{ g}$

Formula used:

$\text{weight}=\frac{\text{number of atoms}}{6\cdot 022\times {10}^{23}}\times \text{Atomic weight}$

Calculation:

Given number of atoms of U is $=5\cdot 93\times {10}^{-7}\text{ g}$

Atomic weight of U is $238\cdot 03$

Substitute in the above formula

$\begin{array}{l}\text{weight}=\frac{\text{Number of atoms}}{6\cdot 023\times {10}^{23}}\times \text{atomic weight}\\ \text{=}\frac{1\cdot 5\times {10}^{15}}{6\cdot 023\times {10}^{23}}\times 238\cdot 03\\ =5\cdot 93\times {10}^{-7}\text{ g}\end{array}$