Chapter 8, Problem 109AP

Interpretation Introduction

(a)

Interpretation:

The number of atoms present in the sample of gold should be calculated.

Concept Introduction:

Moles of an element are calculated by dividing the given mass by its molar mass.

Moles of an element $=\frac{\text{Mass}}{\text{MolarMass}}$

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is $6.023\times {10}^{23}$.

Answer to Problem 109AP

$9.03\times {10}^{21}$ atoms are present in 2.89g of gold.

Explanation of Solution

Mass of gold = 2.89g

Molar mass of gold = 196.96 g/mol

Moles of gold $=\frac{\text{M}\text{a}\text{s}\text{s}}{\text{M}\text{o}\text{l}\text{a}\text{r}\text{M}\text{a}\text{s}\text{s}}=\frac{\text{2}\text{.89 g}}{\text{196}\text{.96 g/mol}}=0.015\text{ mol}$

Number of atoms present in 0.015 mole of gold $\left(\text{A}\text{u}\right)=\left(0.015\times 6.023\times {10}^{23}\right)=9.03\times {10}^{21}\text{ atoms}$.

Interpretation Introduction

(b)

Interpretation:

The number of atoms present in the sample of platinum should be calculated.

Concept Introduction:

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is $6.023\times {10}^{23}$.

Answer to Problem 109AP

$1.56\times {10}^{20}$ atoms are present in 0.000259 mole of platinum.

Explanation of Solution

Moles of platinum present in the sample = 0.000259

Number of atoms present in 0.000259 mole of platinum $\left(\text{P}\text{t}\right)=\left(0.000259\times 6.023\times {10}^{23}\right)=1.56\times {10}^{20}$.

Interpretation Introduction

(c)

Interpretation:

The number of atoms present in the sample of platinum should be calculated.

Concept Introduction:

Moles of an element are calculated by dividing the given mass by its molar mass.

Moles of an element $=\frac{\text{Mass}}{\text{MolarMass}}$

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is $6.023\times {10}^{23}$.

Answer to Problem 109AP

$8.01\times {10}^{17}$ atoms are present in 0.000259 g of platinum.

Explanation of Solution

Mass of platinum = 0.000259g

Molar mass of platinum = 195.08g/mol

Moles of platinum $=\frac{\text{M}\text{a}\text{s}\text{s}}{\text{M}\text{o}\text{l}\text{a}\text{r}\text{M}\text{a}\text{s}\text{s}}=\frac{\text{0}\text{.000259 g}}{\text{195}\text{.08 g/mol}}=1.33\times {10}^{-6}\text{ mol}$

Number of atoms present in $1.33\times {10}^{-6}$ mole of platinum $\left(\text{P}\text{t}\right)=\left(1.33\times {10}^{-6}\times 6.023\times {10}^{23}\right)=8.01\times {10}^{17}$.

Interpretation Introduction

(d)

Interpretation:

The number of atoms present in the sample of magnesium should be calculated.

Concept Introduction:

Moles of an element are calculated by dividing the given mass by its molar mass.

Moles of an element $=\frac{\text{Mass}}{\text{MolarMass}}$

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is $6.023\times {10}^{23}$.

Answer to Problem 109AP

$2.25\times {10}^{25}$ atoms are present in 2.0lb of platinum.

Explanation of Solution

Mass of magnesium = 2.0lb = 907.184g [Since, 1lb = 453.592g]

Molar mass of magnesium = 24.3g/mol

Moles of magnesium $=\frac{\text{Mass}}{\text{MolarMass}}=\frac{\text{907}\text{.184 g}}{\text{24}\text{.3 g/mol}}=37.33\text{ mol}$

Number of atoms present in 37.33 moles of magnesium $\left(\text{M}\text{g}\right)=\left(37.33\times 6.023\times {10}^{23}\right)=2.25\times {10}^{25}$.

Interpretation Introduction

(e)

Interpretation:

The number of atoms present in the sample of mercury should be calculated.

Concept Introduction:

Mass can be calculated by multiplying the density by the volume.

$\text{M}\text{a}\text{s}\text{s}=\text{D}\text{e}\text{n}\text{s}\text{i}\text{t}\text{y}\times \text{V}\text{o}\text{l}\text{u}\text{m}\text{e}$

Moles of an element are calculated by dividing the given mass by its molar mass.

Moles of an element $=\frac{\text{Mass}}{\text{MolarMass}}$

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is $6.023\times {10}^{23}$.

Answer to Problem 109AP

$7.7\times {10}^{22}$ atoms are present in 1.90 mL of mercury.

Explanation of Solution

Volume of mercury = 1.90mL

Density of mercury = 13.6g/mL

Mass of mercury $=\left(\text{Density×Volume}\right)=\left(13.6\text{g}/\text{m}\text{L}\times 1.90\text{m}\text{L}\right)=25.84\text{ g}$

Molar mass of mercury = 200.59g/mol

Moles of mercury $=\frac{\text{Mass}}{\text{Molar Mass}}=\frac{\text{25}\text{.84 g}}{\text{200}\text{.59 g/mol}}=0.128\text{ mol}$

Number of atoms present in 37.33 moles of

Mercury $\left(\text{H}\text{g}\right)=\left(0.128\times 6.023\times {10}^{23}\right)=7.7\times {10}^{22}$.

Interpretation Introduction

(f)

Interpretation:

The number of atoms present in the sample of tungsten should be calculated.

Concept Introduction:

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is $6.023\times {10}^{23}$.

Answer to Problem 109AP

$2.58\times {10}^{23}$ atoms are present in 4.30 moles of tungsten.

Explanation of Solution

Moles of tungsten = 4.30

Number of atoms present in 4.30 moles of

Tungsten $\left(\text{W}\right)=\left(4.30\times 6.023\times {10}^{23}\right)=2.58\times {10}^{23}$.

Interpretation Introduction

(f)

Interpretation:

The number of atoms present in the sample of tungsten should be calculated.

Concept Introduction:

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is $6.023\times {10}^{23}$.

Answer to Problem 109AP

$2.58\times {10}^{23}$ atoms are present in 4.30 moles of tungsten.

Explanation of Solution

Moles of tungsten = 4.30

Number of atoms present in 4.30 moles of

Tungsten $\left(\text{W}\right)=\left(4.30\times 6.023\times {10}^{23}\right)=2.58\times {10}^{23}$.

Interpretation Introduction

(g)

Interpretation:

The number of atoms present in the sample of tungsten should be calculated.

Concept Introduction:

Moles of an element are calculated by dividing the given mass by its molar mass.

Moles of an element $=\frac{\text{Mass}}{\text{MolarMass}}$

According to Avogadro’s number, the number of atoms present in one mole of a monoatomic element is $6.023\times {10}^{23}$.

Answer to Problem 109AP

$1.38\times {10}^{22}$ atoms are present in 4.30g of tungsten.

Explanation of Solution

Mass of tungsten = 4.30g

Molar mass of tungsten = 183.84g/mol

Moles of tungsten $=\frac{\text{Mass}}{\text{Molar Mass}}=\frac{\text{4}\text{.30 g}}{\text{183}\text{.84 g/mol}}=0.023\text{ mol}$

Number of atoms present in 0.023 moles of

Tungsten $\left(\text{W}\right)=\left(0.023\times 6.023\times {10}^{23}\right)=1.38\times {10}^{22}$.