(a) Interpretation: A concept map is to be drawn and the grams of CO 2 gas dissolved in the solution when 1 .0 L of carbon dioxide dissolves in 500 .0 mL of solution is to be calculated. Concept introduction: A mole is a basic unit used in the International system of units (SI). It is abbreviated as mol Mole is defined that the amount of substance that contains molecules or atoms equals to 12 g of C − 12 molecule. The concept map of a mole is a diagram used to relate the different concepts of mole chemistry.

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Answer 1

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Chapter 15, Problem 19E

Interpretation Introduction

(a)

Interpretation:

A concept map is to be drawn and the grams of $CO2$ gas dissolved in the solution when $1.0 L$ of carbon dioxide dissolves in $500.0 mL$ of solution is to be calculated.

Concept introduction:

A mole is a basic unit used in the International system of units (SI). It is abbreviated as $mol$ Mole is defined that the amount of substance that contains molecules or atoms equals to $12 g$ of $C−12$ molecule. The concept map of a mole is a diagram used to relate the different concepts of mole chemistry.

Expert Solution

Answer to Problem 19E

The concept map is shown below.

EP INTRODUCTORY CHEM.-MOD.MASTERINGCHEM, Chapter 15, Problem 19E , additional homework tip 1

The grams of $CO2$ gas dissolved in the solution when $1.0 L$ of carbon dioxide dissolves in $500.0 mL$ of solution is $1.96 g$ .

Explanation of Solution

When $1.0 L$ of carbon dioxide dissolves in $500.0 mL$ of solution, mole concept map looks like as shown below.

EP INTRODUCTORY CHEM.-MOD.MASTERINGCHEM, Chapter 15, Problem 19E , additional homework tip 2

Figure 1

The volume occupied by $1 mol$ of $CO2$ gas at STP is $22.4 L$ .

Therefore, the number of moles which occupy $1.0 L$ is calculated below.

$Number of moles=1.00 L CO2×1 mol CO222.4 L CO2=0.04464 mol$

The molar mass of $CO2$ gas is $44.01 g mol−1$ .

Therefore, the mass of $1 mol$ of $CO2$ gas is $44.01 g$ .

The formula to calculate the mass of $0.04464 mol$ of $CO2$ gas is shown below.

$Mass= 0.04464 mol CO2×Mass of 1 mol CO21 mol CO2$

Substitute the mass of $1 mol$ of $CO2$ gas in the above equation.

$Mass= 0.04464 mol CO2×44.01 g CO2 1 mol CO2=1.96 g$

Therefore, the grams of $CO2$ gas dissolved in the solution is $1.96 g$ .

Conclusion

The grams of $CO2$ gas is $1.96 g$ .

Interpretation Introduction

(b)

Interpretation:

A concept map is to be drawn and the molecules of $CO2$ gas dissolved in the solution when $1.0 L$ of carbon dioxide dissolves in $500.0 mL$ of solution is to be calculated.

Concept introduction:

A mole is a basic unit used in the International system of units (SI). It is abbreviated as $mol$ Mole is defined that the amount of substance that contains molecules or atoms equals to $12 g$ of $C−12$ molecule. The concept map of a mole is a diagram used to relate the different concepts of mole chemistry.

Expert Solution

Answer to Problem 19E

The concept map is shown below.

EP INTRODUCTORY CHEM.-MOD.MASTERINGCHEM, Chapter 15, Problem 19E , additional homework tip 3

The molecules of $CO2$ gas dissolved in the solution when $1.0 L$ of carbon dioxide dissolves in $500.0 mL$ of solution is $2.69×1022 moleules$ .

Explanation of Solution

When $1.0 L$ of carbon dioxide dissolves in $500.0 mL$ of solution, mole concept map looks like as shown below.

EP INTRODUCTORY CHEM.-MOD.MASTERINGCHEM, Chapter 15, Problem 19E , additional homework tip 4

Figure 1

The volume occupied by $1 mol$ of $CO2$ gas at STP is $22.4 L$ .

Therefore, the number of moles which occupy $1.0 L$ of is calculated below.

$Number of moles=1.00 L CO2×1 mol CO222.4 L CO2=0.04464 mol$

The molecules present in $1 mol$ of $CO2$ gas at STP are $6.02×1023 moleules$ .

The formula to calculate the molecules occupied by $0.04464 mol$ of $CO2$ gas is shown below.

$Number of molecules= 0.04464 mol CO2×Molecules in 1 mol CO21 mol CO2$

Substitute the molecules in $1 mol$ of $CO2$ gas in the above equation.

$Number of molecules= 0.04464 mol CO2×6.02×1023 moleules1 mol CO2=2.69×1022 moleules$

Therefore, the molecules of $CO2$ gas dissolved in the solution is $2.69×1022 moleules$ .

Conclusion

The molecules of $CO2$ gas is $2.69×1022 moleules$ .

Interpretation Introduction

(c)

Interpretation:

A concept map is to be drawn and the molar concentration of the carbonic acid solution when $1.0 L$ of carbon dioxide dissolves in $500.0 mL$ of solution is to be calculated.

Concept introduction:

A mole is a basic unit used in the International system of units (SI). It is abbreviated as $mol$ Mole is defined that the amount of substance that contains molecules or atoms equals to $12 g$ of $C−12$ molecule. The concept map of a mole is a diagram used to relate the different concepts of mole chemistry.

Expert Solution

Answer to Problem 19E

The concept map is shown below.

EP INTRODUCTORY CHEM.-MOD.MASTERINGCHEM, Chapter 15, Problem 19E , additional homework tip 5

The molar concentration of the carbonic acid solution is $0.0893 M$ .

Explanation of Solution

When $1.0 L$ of carbon dioxide dissolves in $500.0 mL$ of solution, mole concept map looks like as shown below.

EP INTRODUCTORY CHEM.-MOD.MASTERINGCHEM, Chapter 15, Problem 19E , additional homework tip 6

Figure 1

The volume occupied by $1 mol$ of $CO2$ gas at STP is $22.4 L$ .

Therefore, the number of moles which occupy $1.0 L$ of is calculated below.

$Number of moles=1.00 L CO2×1 mol CO222.4 L CO2=0.04464 mol$

The number of moles in $500.0 mL$ of solution is $0.04464 mol$ .

The relation between $L$ and $mL$ is shown below.

$1 L=1000 mL$

The probable unit factors are given below.

$1 L1000 mL and 1000 mL1 L$

The unit factor to determine $L$ from $mL$ is given below.

$1 L1000 mL$

Therefore, the volume in $L$ is calculated below.

$Volume=500.0 mL×1 L1000 mL=0.5 L$

The formula to determine molarity is shown below.

$M=nV$ …(1)

Where

• $M$ is the molarity of a solution.

• $n$ is the number of moles of solute.

• $V$ is the volume of the solution.

Substitute the value of number of moles and volume in equation (1).

$Molar concentration of CO2=0.04464 mol0.5 L=0.0893 mol/L$

The relation between $M$ and $mol/L$ as shown below.

$1 M=1 mol/L$

The unit factors are given below.

$1 M1 mol/L and 1 mol/L1 M$

The unit factor to determine $M$ from $mol/L$ is given below.

$1 M1 mol/L$

Therefore, $0.0893 mol/L$ can be written as shown below.

$Molarity=0.0893 mol/L×1 M1 mol/L=0.0893 M$

Therefore, the molar concentration of $H2CO3$ , carbonic acid solution is $0.0893 M$ .

Conclusion

The molar concentration of carbonic acid solution is $0.0893 M$ .

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