Chapter 3, Problem 3.17QP

(1-a)

Interpretation Introduction

Interpretation:

The number of candy pieces in $\text{0}\text{.2 kg}$ of candy should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.17QP

Number of candy pieces in $\text{0}\text{.2 kg}$ of candy is $\text{100}\text{candy}\text{pieces}$.

Explanation of Solution

To calculate the number of candy pieces in $\text{0}\text{.2 kg}$ of candy.

Given,

$\text{1}\text{.0 kg}$ of candy has $\text{500}$ pieces.

From the above mole concept, 1 mole of elemental sample contains $6.022136\times {10}^{23}$ number of atoms, same as $\text{1}\text{.0 kg}$ of candy has $\text{500}$ pieces.

Number of candy pieces in $\text{0}\text{.2 kg}$ of candy is,

$\begin{array}{l}\text{=}\text{0}\text{.2}\text{kg×}\frac{\text{500}\text{candy}\text{pieces}}{\text{1}\text{kg}}\\ \text{=}\text{100}\text{candy}\text{pieces}\end{array}$

Give weight of candy and number of candies present in per kg are plugged in above equation to give number of candy pieces in $\text{0}\text{.2 kg}$ of candy.

Number of candy pieces in $\text{0}\text{.2 kg}$ of candy is $\text{100}\text{candy}\text{pieces}$.

Conclusion

The number of candy pieces in $\text{0}\text{.2 kg}$ of candy was calculated.

(1-b)

Interpretation Introduction

Interpretation:

The mass of 10 dozen candies should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.17QP

The mass of 10 dozen candies is $\text{0}\text{.24}\text{kg}$.

Explanation of Solution

To calculate the mass of 10 dozen candies.

Given,

$\text{1}\text{.0 kg}$ of candy has $\text{500}$ pieces.

From the above mole concept, 1 mole of elemental sample contains $6.022136\times {10}^{23}$ number of atoms, same as $\text{1}\text{.0 kg}$ of candy has $\text{500}$ pieces.

The mass of 10 dozen candies is,

1 dozen contains 12 pieces means 10 dozen contains 120 candy pieces.

$\begin{array}{l}\text{=}\text{10}\text{dozen}\text{×}\frac{\text{12}\text{candy}\text{pieces}}{\text{1 dozen}}\text{×}\frac{\text{1}\text{kg}}{\text{500}\text{candy}\text{pieces}}\\ \text{=}\text{0}\text{.24}\text{kg}\end{array}$

The number of candies present in per dozen and number of candies present in $\text{1}\text{.0 kg}$ of candy are plugged in above equation to give the mass of 10 dozen candies.

The mass of 10 dozen candies $\text{0}\text{.24}\text{kg}$.

Conclusion

The mass of 10 dozen candies was calculated.

(1-c)

Interpretation Introduction

Interpretation:

The mass of 1 candy piece should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.17QP

The mass of 1 candy piece is $\text{0}\text{.0020}\text{kg}$.

Explanation of Solution

To calculate the mass of 1 candy piece.

Given,

$\text{1}\text{.0 kg}$ of candy has $\text{500}$ pieces.

From the above mole concept, 1 mole of elemental sample contains $6.022136\times {10}^{23}$ number of atoms, same as $\text{1}\text{.0 kg}$ of candy has $\text{500}$ pieces.

The mass of 1 candy piece is,

$\begin{array}{l}\text{=}\text{1}\text{candy}\text{×}\frac{\text{1}\text{kg}}{\text{500}\text{candy}\text{pieces}}\\ \text{=}\text{0}\text{.0020}\text{kg}\end{array}$

Number of candies present in $\text{1}\text{.0 kg}$ of candy are plugged in above equation to give the mass of 1 candy piece.

The mass of 1 candy piece is $\text{0}\text{.0020}\text{kg}$.

Conclusion

The mass of 1 candy piece was calculated.

(1-d)

Interpretation Introduction

Interpretation:

The mass of 2.0 moles of candy should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.17QP

The mass of 2.0 mole candy is ${\text{2×10}}^{\text{21}}\text{kg}$.

Explanation of Solution

To calculate the mass of 2.0 moles of candies.

Given,

$\text{1}\text{.0 kg}$ of candy has $\text{500}$ pieces.

From the above mole concept, 1 mole of elemental sample contains $6.022136\times {10}^{23}$ number of atoms, same as $\text{1}\text{.0 kg}$ of candy has $\text{500}$ pieces.

1 mole of candy contains $6.022136\times {10}^{23}$ number of candies.

The mass of 1 mole candy is,

$\begin{array}{l}\text{=}\text{2}\text{.0}\text{mole}\text{candy}\text{×}\frac{\text{6}{\text{.02×10}}^{\text{23}}\text{candy}\text{pieces}}{\text{1}\text{mole}}\text{×}\frac{\text{1}\text{kg}}{\text{500}\text{candy}\text{pieces}}\\ \text{=}{\text{2×10}}^{\text{21}}\text{kg}\end{array}$

Number of candies present in $\text{1}\text{.0 kg}$ of candy are plugged in above equation to give the mass of 1 candy piece.

The mass of 2.0 mole candy is ${\text{2×10}}^{\text{21}}\text{kg}$.

Conclusion

The mass of 2.0 moles of candies were calculated.

(2-a)

Interpretation Introduction

Interpretation:

The number of Helium atoms present in $\text{0}\text{.2 kg of Helium}$ should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.17QP

The number of Helium atoms present in $\text{0}\text{.2 kg of Helium}$ is $\text{3}{\text{.0×10}}^{\text{25}}\text{atoms}$.

Explanation of Solution

To calculate the number of Helium atoms present in $\text{0}\text{.2 kg of Helium}$.

Given,

Molar mass of Helium is $\text{4}\text{.003 g}$

From the above mole concept, 1 mole of elemental sample contains $6.022136\times {10}^{23}$ number of atoms.

The number of Helium atoms present in $\text{0}\text{.2 kg of Helium}$ is,

$\begin{array}{l}\text{=}\text{0}\text{.2kg×}\frac{\text{1}\text{mole}}{\text{4}\text{.003}}\text{×}\frac{\text{6}{\text{.02×10}}^{\text{23}}\text{atoms}}{\text{1}\text{mole}}\\ \text{=}\text{3}{\text{.0×10}}^{\text{25}}\text{atoms}\end{array}$

Molar mass of Helium and given weight of Helium sample are plugged in above equation to give the number of Helium atoms present in $\text{0}\text{.2 kg of Helium}$

The number of Helium atoms present in $\text{0}\text{.2 kg of Helium}$ is $\text{3}{\text{.0×10}}^{\text{25}}\text{atoms}$.

Conclusion

The number of Helium atoms present in $\text{0}\text{.2 kg of Helium}$ were calculated.

(2-b)

Interpretation Introduction

Interpretation:

The mass of 10 dozen Helium atoms should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.17QP

The mass of 10 dozen Helium atoms is $\text{7}{\text{.98×10}}^{\text{-22}}\text{g}$.

Explanation of Solution

To calculate the mass of 10 dozen Helium atoms

Given,

Molar mass of Helium is $\text{4}\text{.003 g}$

From the above mole concept, 1 mole of elemental sample contains $6.022136\times {10}^{23}$ number of atoms.

1 dozen contains 12 numbers so 10 dozen contains 120 number of Helium atoms.

The mass of 10 dozen Helium atoms is,

$\begin{array}{l}\text{=}\text{10}\text{dozen}\text{×}\frac{\text{12}\text{atoms}}{\text{1}\text{dozen}}\text{×}\frac{\text{1}\text{mole}}{\text{6}{\text{.02×10}}^{\text{23}}\text{atoms}}\text{×}\frac{\text{4}\text{.003}\text{g}}{\text{1}\text{mole}}\\ \text{=}\text{7}{\text{.98×10}}^{\text{-22}}\text{g}\end{array}$

Molar mass of Helium and number of atoms that are present in 10 dozen are plugged in above equation to give mass of 10 dozen Helium atoms.

The mass of 10 dozen Helium atoms is $\text{7}{\text{.98×10}}^{\text{-22}}\text{g}$.

Conclusion

The mass of 10 dozen Helium atoms calculated.

(2-c)

Interpretation Introduction

Interpretation:

The mass of one Helium atom should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.17QP

The mass of one Helium atom is $6.646{\text{×10}}^{\text{-24}}\text{g}$.

Explanation of Solution

To calculate the mass of one Helium atom.

Given,

Molar mass of Helium is $\text{4}\text{.003 g}$

From the above mole concept, 1 mole of elemental sample contains $6.022136\times {10}^{23}$ number of atoms.

The mass of one Helium atom is,

$\begin{array}{l}\text{=}\text{1}\text{atom}\text{×}\frac{\text{1}\text{mole}}{\text{6}{\text{.02×10}}^{\text{23}}\text{atoms}}\text{×}\frac{\text{4}\text{.003}\text{g}}{\text{1}\text{mole}}\\ \text{=}6.646{\text{×10}}^{\text{-24}}\text{g}\end{array}$

Molar mass of Helium and is plugged in above equation to give the mass of one Helium atom.

The mass of one Helium atom is $6.646{\text{×10}}^{\text{-24}}\text{g}$.

Conclusion

The mass of one Helium atom was calculated.

(2-d)

Interpretation Introduction

Interpretation:

The mass of 2.0 mole of Helium should be calculated.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Answer to Problem 3.17QP

The mass of 2.0 mole Helium is $8.0\text{g}$.

Explanation of Solution

To calculate the mass of 2.0 mole Helium.

Given,

Molar mass of Helium is $\text{4}\text{.003 g}$

From the above mole concept, 1 mole of elemental sample contains $6.022136\times {10}^{23}$ number of atoms.

The mass of 2.0 mole Helium is,

$\begin{array}{l}\text{=}\text{2}\text{.0}\text{mole}\text{×}\frac{\text{4}\text{.003}\text{g}}{\text{1}\text{mole}}\\ \text{=}\text{8}\text{.0}\text{g}\end{array}$

Molar mass of Helium and is plugged in above equation to give the mass of 2.0 mole Helium.

The mass of 2.0 mole Helium is $8.0\text{g}$.

Conclusion

The mass of 2.0 mole of Helium was calculated.

(3-a)

Interpretation Introduction

Interpretation:

$\text{1}\text{.0}\text{kg}$ nails and $\text{1}\text{.0}\text{kg}$ Helium is expected to have the same number of Binkles has to be explained.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Binkles:

$3\times {10}^{12}$ number of objects present in a sample is known as one Binkle.

Answer to Problem 3.17QP

The masses of nails and Helium atom are different so the $\text{1}\text{.0}\text{kg}$ of nail and Helium are does not contains equal number of binkles.

Explanation of Solution

To explain the binkles of $\text{1}\text{.0}\text{kg}$ nails and $\text{1}\text{.0}\text{kg}$ Helium.

Given,

Molar mass of Helium is $\text{4}\text{.003 g}$

One binkle has $3\times {10}^{12}$ number of objects

The masses of nails and Helium atom are different so the $\text{1}\text{.0}\text{kg}$ of nail and Helium are does not contains equal number of binkles.

Conclusion

Binkles of $\text{1}\text{.0}\text{kg}$ nails and $\text{1}\text{.0}\text{kg}$ Helium was explained.

(3-b)

Interpretation Introduction

Interpretation:

The higher mass containing option should be given from the $\text{3}\text{.5}$ binkles of nails and $\text{3}\text{.5}$ binkles of Helium.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Binkles:

The $3\times {10}^{12}$ number of objects present in a sample is known as one Binkle.

Answer to Problem 3.17QP

The masses of $\text{3}\text{.5}$ binkle nail is higher than $\text{3}\text{.5}$ binkles of Helium atom so the $\text{3}\text{.5}$ binkle nail has more mass than $\text{3}\text{.5}$ binkles of Helium.

Explanation of Solution

To explain the mass of $\text{3}\text{.5}$ binkles of nails and $\text{3}\text{.5}$ binkles of Helium.

Given,

Molar mass of Helium is $\text{4}\text{.003 g}$

One binkle has $3\times {10}^{12}$ number of objects

The masses of $\text{3}\text{.5}$ binkle nail is higher than $\text{3}\text{.5}$ binkles of Helium atom so the $\text{3}\text{.5}$ binkle nail has more mass than $\text{3}\text{.5}$ binkles of Helium.

Conclusion

The higher mass containing option should be given from $\text{3}\text{.5}$ binkles of nails and $\text{3}\text{.5}$ binkles of Helium.

(3-c)

Interpretation Introduction

Interpretation:

The more number of atom containing option should be given from $\text{3}\text{.5}$ binkles of Helium and $\text{3}\text{.5}$ binkles of Lithium.

Concept introduction:

Mole:

Number of atoms present in gram atomic mass of element is known as Avogadro number.

Avogadro number is $6.022136\times {10}^{23}$

One mole equal of atom equal to Avogadro number $(6.022136\times {10}^{23})$ hence, 1 mole of Iron $\text{(55}\text{.9 g)}$ contain atom $6.022136\times {10}^{23}$ $\text{Fe}$ atoms.

The mole of taken gram mass of compound is given by ratio between taken mass of compound to molar mass of compound.

$\text{Mole}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Binkles:

The $3\times {10}^{12}$ number of objects present in a sample is known as one Binkle.

Answer to Problem 3.17QP

The masses of $\text{3}\text{.5}$ binkle of Lithium is higher than $\text{3}\text{.5}$ binkles of Helium atom so the $\text{3}\text{.5}$ binkle Helium has more number of atoms than $\text{3}\text{.5}$ binkles of Lithium.

Explanation of Solution

To explain the mass of $\text{3}\text{.5}$ binkles of nails and $\text{3}\text{.5}$ binkles of Helium.

Given,

Molar mass of Helium is $\text{4}\text{.003 g}$

Molar mass of Lithium is $\text{6}\text{.941}\text{g}$

One binkle has $3\times {10}^{12}$ number of objects

The masses of $\text{3}\text{.5}$ binkle of Lithium is higher than $\text{3}\text{.5}$ binkles of Helium atom so the $\text{3}\text{.5}$ binkle Helium has more number of atoms than $\text{3}\text{.5}$ binkles of Lithium.

Conclusion

The more number of atom containing option was given from $\text{3}\text{.5}$ binkles of Helium and $\text{3}\text{.5}$ binkles of Lithium.