Chapter 6, Problem 6A

(a)

Interpretation Introduction

Interpretation: The mass of 125 carbon atoms needs to be determined.

Concept Introduction: The mass of any atom can be calculated from its number of moles and molar mass. Also, in 1 mol of any substance there are $6.022\times {10}^{23}$ atoms.

Answer to Problem 6A

  $249.20\times {10}^{-23}\text{ g}$

Explanation of Solution

The number of moles from given number of atoms and number of atoms in 1 mol can be calculated as follows:

  $\text{Moles=}\frac{\text{number of atoms of carbon}}{\text{Avogadro's number}}$

  $=\frac{125}{6.022\times {10}^{23}}$

  $=20.75\times {10}^{-23}$

  $\text{Number of moles of carbon=}\frac{\text{Given mass}}{\text{Atomic mass}}$

  $20.75\times {10}^{-23}=\frac{xgm}{12.01}$

  $x=20.75\times {10}^{-23}\times 12.01$

  $x=249.20\times {10}^{-23}\text{ g}$

(b)

Interpretation Introduction

Interpretation: The mass of 5 million potassium atoms needs to be determined.

Concept Introduction: The mass of any atom can be calculated from its number of moles and molar mass. Also, in 1 mol of any substance there are $6.022\times {10}^{23}$ atoms.

Answer to Problem 6A

  $Massofpotassium=32.45\times {10}^{-17}gm$

Explanation of Solution

  $\text{Moles=}\frac{\text{number of atoms of potassium}}{\text{Avogadro's number}}$

  $=\frac{5\times {10}^6}{6.022\times {10}^{23}}$

  $=0.830\times {10}^{-17}$

  $\text{Number of moles of potassium=}\frac{\text{Given mass}}{\text{Atomic mass}}$

  $0.830\times {10}^{-17}=\frac{xgm}{39.10}$

  $x=0.830\times {10}^{-17}\times 39.10$

  $x=32.45\times {10}^{-17}gm$

(c)

Interpretation Introduction

Interpretation: The mass of given number of moles of Li atoms needs to be determined.

Concept Introduction: The mass of any atom can be calculated from its number of moles and molar mass. Also, in 1 mol of any substance there are $6.022\times {10}^{23}$ atoms.

Answer to Problem 6A

0.119 g

Explanation of Solution

  $\text{Moles=}\frac{\text{number of atoms of Lithium}}{\text{Avogadro's number}}$

  $=\frac{1.04\times {10}^{22}}{6.022\times {10}^{23}}$

  $=0.172\times {10}^{-1}$

  $=0.0172$

  $\text{Number of moles of lithium=}\frac{\text{Given mass}}{\text{Atomic mass}}$

  $0.0172=\frac{xgm}{6.94}$

  $x=0.0172\times 6.94$

  $x=0.119gm$

(d)

Interpretation Introduction

Interpretation: The mass of given number of atoms of magnesium needs to be determined.

Concept Introduction: The mass of any atom can be calculated from its number of moles and molar mass. Also, in 1 mol of any substance there are $6.022\times {10}^{23}$ atoms.

Answer to Problem 6A

  $4.03\times {10}^{-23}gm$

Explanation of Solution

  $\text{Moles=}\frac{\text{number of atoms of magnesium}}{\text{Avogadro's number}}$

  $=\frac{1}{6.022\times {10}^{23}}$

  $=0.166\times {10}^{-23}$

  $\text{Number of moles of magnesium=}\frac{\text{Given mass}}{\text{Atomic mass}}$

  $0.166\times {10}^{-23}=\frac{xgm}{24.31}$

  $x=0.166\times {10}^{-23}\times 24.31$

  $x=4.03\times {10}^{-23}gm$

(e)

Interpretation Introduction

Interpretation: The mass of given number of atoms of iodine needs to be determined.

Concept Introduction: The mass of any atom can be calculated from its number of moles and molar mass. Also, in 1 mol of any substance there are $6.022\times {10}^{23}$ atoms.

Answer to Problem 6A

63.45 g.

Explanation of Solution

  $\text{Moles=}\frac{\text{number of atoms of iodine}}{\text{Avogadro's number}}$

  $=\frac{3.011\times {10}^{23}}{6.022\times {10}^{23}}$

  $=0.5$

  $\text{Number of moles of iodine=}\frac{\text{Given mass}}{\text{Atomic mass}}$

  $0.5=\frac{xgm}{126.9}$

  $x=0.5\times 126.9$

  $x=63.45gm$