Volume of solution that is prepared using 2 .111 g of sodium carbonate in 250 .0 mL solution that student need to transfer in order to obtain 2 .15 mmol Na + has to be given. Concept Introduction: Molarity of the solution is defined as the number of moles of solute that is dissolved in the volume of solution in liters. Unit of molarity is mol ⋅ L − 1 . The equation for molarity is given as follows; Molarity (M) = Amount of solute in moles Volume of solution in liters

Question

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Answer 1

Question

Chapter F, Problem G.5E

(a)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution that student need to transfer in order to obtain $2.15 mmol Na+$ has to be given.

Concept Introduction:

Molarity of the solution is defined as the number of moles of solute that is dissolved in the volume of solution in liters. Unit of molarity is $mol⋅L−1$ . The equation for molarity is given as follows;

$Molarity (M)=Amount of solute in molesVolume of solution in liters$

(a)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $13.44 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $2.15 mmol$ of $Na+$ ion. One mole of sodium carbonate gives two moles of $Na+$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $2.15 mmol$ of $Na+$ ion is shown below;

$n(NaCO3)=2.15×10−3 mol2=1.075×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $2.15 mmol$ of $Na+$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=1.075×10−3 mol0.08 M=13.44 mL$

Therefore, the volume of sodium carbonate solution is $13.44 mL$ .

(b)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $4.98 mmol CO32−$ has to be given.

Concept Introduction:

Refer part (a).

(b)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $62.25 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $4.98 mmol$ of $CO32−$ ion. One mole of sodium carbonate gives one mole of $CO32−$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $4.98 mmol$ of $CO32−$ ion is shown below;

$n(NaCO3)=4.98×10−3 mol1=4.98×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $4.98 mmol$ of $CO32−$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=4.98×10−3 mol0.08 M=62.25 mL$

Therefore, the volume of sodium carbonate solution is $62.25 mL$ .

(c)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $50.0 mg$ of $Na2CO3$ has to be given.

Concept Introduction:

Refer part (a).

(c)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $5.89 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Mass of sodium carbonate that has to be present in the solution is given as $50.0 mg$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Number of moles of sodium carbonate in $50.0 mg$ is calculated as shown below;

$n=mM=50 mg106 g⋅mol−1=0.471×10−3 mol$

The volume of solution that is needed to obtain $50.0 mg$ of sodium carbonate is given as follows;

$Volume of solution(V)=n(Na2CO3)Molarity=0.471×10−3 mol0.08 M=5.89 mL$

Volume of sodium carbonate solution is $5.89 mL$ .

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Answer 2

Question

Chapter F, Problem G.5E

(a)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution that student need to transfer in order to obtain $2.15 mmol Na+$ has to be given.

Concept Introduction:

Molarity of the solution is defined as the number of moles of solute that is dissolved in the volume of solution in liters. Unit of molarity is $mol⋅L−1$ . The equation for molarity is given as follows;

$Molarity (M)=Amount of solute in molesVolume of solution in liters$

(a)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $13.44 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $2.15 mmol$ of $Na+$ ion. One mole of sodium carbonate gives two moles of $Na+$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $2.15 mmol$ of $Na+$ ion is shown below;

$n(NaCO3)=2.15×10−3 mol2=1.075×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $2.15 mmol$ of $Na+$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=1.075×10−3 mol0.08 M=13.44 mL$

Therefore, the volume of sodium carbonate solution is $13.44 mL$ .

(b)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $4.98 mmol CO32−$ has to be given.

Concept Introduction:

Refer part (a).

(b)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $62.25 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $4.98 mmol$ of $CO32−$ ion. One mole of sodium carbonate gives one mole of $CO32−$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $4.98 mmol$ of $CO32−$ ion is shown below;

$n(NaCO3)=4.98×10−3 mol1=4.98×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $4.98 mmol$ of $CO32−$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=4.98×10−3 mol0.08 M=62.25 mL$

Therefore, the volume of sodium carbonate solution is $62.25 mL$ .

(c)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $50.0 mg$ of $Na2CO3$ has to be given.

Concept Introduction:

Refer part (a).

(c)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $5.89 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Mass of sodium carbonate that has to be present in the solution is given as $50.0 mg$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Number of moles of sodium carbonate in $50.0 mg$ is calculated as shown below;

$n=mM=50 mg106 g⋅mol−1=0.471×10−3 mol$

The volume of solution that is needed to obtain $50.0 mg$ of sodium carbonate is given as follows;

$Volume of solution(V)=n(Na2CO3)Molarity=0.471×10−3 mol0.08 M=5.89 mL$

Volume of sodium carbonate solution is $5.89 mL$ .

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Answer 3

Question

Chapter F, Problem G.5E

(a)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution that student need to transfer in order to obtain $2.15 mmol Na+$ has to be given.

Concept Introduction:

Molarity of the solution is defined as the number of moles of solute that is dissolved in the volume of solution in liters. Unit of molarity is $mol⋅L−1$ . The equation for molarity is given as follows;

$Molarity (M)=Amount of solute in molesVolume of solution in liters$

(a)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $13.44 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $2.15 mmol$ of $Na+$ ion. One mole of sodium carbonate gives two moles of $Na+$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $2.15 mmol$ of $Na+$ ion is shown below;

$n(NaCO3)=2.15×10−3 mol2=1.075×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $2.15 mmol$ of $Na+$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=1.075×10−3 mol0.08 M=13.44 mL$

Therefore, the volume of sodium carbonate solution is $13.44 mL$ .

(b)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $4.98 mmol CO32−$ has to be given.

Concept Introduction:

Refer part (a).

(b)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $62.25 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $4.98 mmol$ of $CO32−$ ion. One mole of sodium carbonate gives one mole of $CO32−$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $4.98 mmol$ of $CO32−$ ion is shown below;

$n(NaCO3)=4.98×10−3 mol1=4.98×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $4.98 mmol$ of $CO32−$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=4.98×10−3 mol0.08 M=62.25 mL$

Therefore, the volume of sodium carbonate solution is $62.25 mL$ .

(c)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $50.0 mg$ of $Na2CO3$ has to be given.

Concept Introduction:

Refer part (a).

(c)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $5.89 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Mass of sodium carbonate that has to be present in the solution is given as $50.0 mg$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Number of moles of sodium carbonate in $50.0 mg$ is calculated as shown below;

$n=mM=50 mg106 g⋅mol−1=0.471×10−3 mol$

The volume of solution that is needed to obtain $50.0 mg$ of sodium carbonate is given as follows;

$Volume of solution(V)=n(Na2CO3)Molarity=0.471×10−3 mol0.08 M=5.89 mL$

Volume of sodium carbonate solution is $5.89 mL$ .

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Answer 4

Question

Chapter F, Problem G.5E

(a)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution that student need to transfer in order to obtain $2.15 mmol Na+$ has to be given.

Concept Introduction:

Molarity of the solution is defined as the number of moles of solute that is dissolved in the volume of solution in liters. Unit of molarity is $mol⋅L−1$ . The equation for molarity is given as follows;

$Molarity (M)=Amount of solute in molesVolume of solution in liters$

(a)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $13.44 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $2.15 mmol$ of $Na+$ ion. One mole of sodium carbonate gives two moles of $Na+$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $2.15 mmol$ of $Na+$ ion is shown below;

$n(NaCO3)=2.15×10−3 mol2=1.075×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $2.15 mmol$ of $Na+$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=1.075×10−3 mol0.08 M=13.44 mL$

Therefore, the volume of sodium carbonate solution is $13.44 mL$ .

(b)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $4.98 mmol CO32−$ has to be given.

Concept Introduction:

Refer part (a).

(b)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $62.25 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $4.98 mmol$ of $CO32−$ ion. One mole of sodium carbonate gives one mole of $CO32−$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $4.98 mmol$ of $CO32−$ ion is shown below;

$n(NaCO3)=4.98×10−3 mol1=4.98×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $4.98 mmol$ of $CO32−$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=4.98×10−3 mol0.08 M=62.25 mL$

Therefore, the volume of sodium carbonate solution is $62.25 mL$ .

(c)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $50.0 mg$ of $Na2CO3$ has to be given.

Concept Introduction:

Refer part (a).

(c)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $5.89 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Mass of sodium carbonate that has to be present in the solution is given as $50.0 mg$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Number of moles of sodium carbonate in $50.0 mg$ is calculated as shown below;

$n=mM=50 mg106 g⋅mol−1=0.471×10−3 mol$

The volume of solution that is needed to obtain $50.0 mg$ of sodium carbonate is given as follows;

$Volume of solution(V)=n(Na2CO3)Molarity=0.471×10−3 mol0.08 M=5.89 mL$

Volume of sodium carbonate solution is $5.89 mL$ .

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Answer 5

Chapter F, Problem G.5E

(a)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution that student need to transfer in order to obtain $2.15 mmol Na+$ has to be given.

Concept Introduction:

Molarity of the solution is defined as the number of moles of solute that is dissolved in the volume of solution in liters. Unit of molarity is $mol⋅L−1$ . The equation for molarity is given as follows;

$Molarity (M)=Amount of solute in molesVolume of solution in liters$

(a)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $13.44 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $2.15 mmol$ of $Na+$ ion. One mole of sodium carbonate gives two moles of $Na+$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $2.15 mmol$ of $Na+$ ion is shown below;

$n(NaCO3)=2.15×10−3 mol2=1.075×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $2.15 mmol$ of $Na+$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=1.075×10−3 mol0.08 M=13.44 mL$

Therefore, the volume of sodium carbonate solution is $13.44 mL$ .

(b)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $4.98 mmol CO32−$ has to be given.

Concept Introduction:

Refer part (a).

(b)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $62.25 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $4.98 mmol$ of $CO32−$ ion. One mole of sodium carbonate gives one mole of $CO32−$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $4.98 mmol$ of $CO32−$ ion is shown below;

$n(NaCO3)=4.98×10−3 mol1=4.98×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $4.98 mmol$ of $CO32−$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=4.98×10−3 mol0.08 M=62.25 mL$

Therefore, the volume of sodium carbonate solution is $62.25 mL$ .

(c)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $50.0 mg$ of $Na2CO3$ has to be given.

Concept Introduction:

Refer part (a).

(c)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $5.89 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Mass of sodium carbonate that has to be present in the solution is given as $50.0 mg$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Number of moles of sodium carbonate in $50.0 mg$ is calculated as shown below;

$n=mM=50 mg106 g⋅mol−1=0.471×10−3 mol$

The volume of solution that is needed to obtain $50.0 mg$ of sodium carbonate is given as follows;

$Volume of solution(V)=n(Na2CO3)Molarity=0.471×10−3 mol0.08 M=5.89 mL$

Volume of sodium carbonate solution is $5.89 mL$ .

Answer 6

Chapter F, Problem G.5E

(a)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution that student need to transfer in order to obtain $2.15 mmol Na+$ has to be given.

Concept Introduction:

Molarity of the solution is defined as the number of moles of solute that is dissolved in the volume of solution in liters. Unit of molarity is $mol⋅L−1$ . The equation for molarity is given as follows;

$Molarity (M)=Amount of solute in molesVolume of solution in liters$

(a)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $13.44 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $2.15 mmol$ of $Na+$ ion. One mole of sodium carbonate gives two moles of $Na+$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $2.15 mmol$ of $Na+$ ion is shown below;

$n(NaCO3)=2.15×10−3 mol2=1.075×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $2.15 mmol$ of $Na+$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=1.075×10−3 mol0.08 M=13.44 mL$

Therefore, the volume of sodium carbonate solution is $13.44 mL$ .

Answer 7

Chapter F, Problem G.5E

(a)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution that student need to transfer in order to obtain $2.15 mmol Na+$ has to be given.

Concept Introduction:

Molarity of the solution is defined as the number of moles of solute that is dissolved in the volume of solution in liters. Unit of molarity is $mol⋅L−1$ . The equation for molarity is given as follows;

$Molarity (M)=Amount of solute in molesVolume of solution in liters$

Answer 8

(a)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $13.44 mL$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $2.15 mmol$ of $Na+$ ion. One mole of sodium carbonate gives two moles of $Na+$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $2.15 mmol$ of $Na+$ ion is shown below;

$n(NaCO3)=2.15×10−3 mol2=1.075×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $2.15 mmol$ of $Na+$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=1.075×10−3 mol0.08 M=13.44 mL$

Therefore, the volume of sodium carbonate solution is $13.44 mL$ .

Answer 13

(b)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $4.98 mmol CO32−$ has to be given.

Concept Introduction:

Refer part (a).

(b)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $62.25 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $4.98 mmol$ of $CO32−$ ion. One mole of sodium carbonate gives one mole of $CO32−$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $4.98 mmol$ of $CO32−$ ion is shown below;

$n(NaCO3)=4.98×10−3 mol1=4.98×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $4.98 mmol$ of $CO32−$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=4.98×10−3 mol0.08 M=62.25 mL$

Therefore, the volume of sodium carbonate solution is $62.25 mL$ .

Answer 14

(b)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $4.98 mmol CO32−$ has to be given.

Concept Introduction:

Refer part (a).

Answer 15

(b)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $62.25 mL$ .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Required solution has $4.98 mmol$ of $CO32−$ ion. One mole of sodium carbonate gives one mole of $CO32−$ . Therefore, the number of moles of sodium carbonate that has to be taken in order to obtain $4.98 mmol$ of $CO32−$ ion is shown below;

$n(NaCO3)=4.98×10−3 mol1=4.98×10−3 mol$

Molarity of the sodium carbonate solution is calculated as $0.08 M$ . Therefore, the volume of sodium carbonate solution that is required to obtain $4.98 mmol$ of $CO32−$ is calculated as shown below;

$Volume of solution(V)=n(Na2CO3)Molarity=4.98×10−3 mol0.08 M=62.25 mL$

Therefore, the volume of sodium carbonate solution is $62.25 mL$ .

Answer 20

(c)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $50.0 mg$ of $Na2CO3$ has to be given.

Concept Introduction:

Refer part (a).

(c)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $5.89 mL$ .

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Mass of sodium carbonate that has to be present in the solution is given as $50.0 mg$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Number of moles of sodium carbonate in $50.0 mg$ is calculated as shown below;

$n=mM=50 mg106 g⋅mol−1=0.471×10−3 mol$

The volume of solution that is needed to obtain $50.0 mg$ of sodium carbonate is given as follows;

$Volume of solution(V)=n(Na2CO3)Molarity=0.471×10−3 mol0.08 M=5.89 mL$

Volume of sodium carbonate solution is $5.89 mL$ .

Answer 21

(c)

Interpretation Introduction

Interpretation:

Volume of solution that is prepared using $2.111 g$ of sodium carbonate in $250.0 mL$ solution, that student need to transfer in order to obtain $50.0 mg$ of $Na2CO3$ has to be given.

Concept Introduction:

Refer part (a).

Answer 22

(c)

Expert Solution

Answer to Problem G.5E

Volume of sodium carbonate solution is $5.89 mL$ .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Mass of sodium carbonate dissolved is given as $2.111 g$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Formula of sodium carbonate is $Na2CO3$ . Number of moles of sodium carbonate in $2.111 g$ is calculated as shown below;

$n=mM=2.111 g106 g⋅mol−1=0.020 mol$

Volume of solution is given as $250 mL$ . Number of moles of sodium carbonate is calculated as $0.020 mol$ . Therefore, Molarity of the sodium carbonate solution is calculated as shown below;

$Molarity (M)=0.020 mol0.250 L=0.08 M$

Therefore, the molarity of the sodium carbonate solution is $0.08 M$ .

Mass of sodium carbonate that has to be present in the solution is given as $50.0 mg$ . Molar mass of sodium carbonate is $106 g⋅mol−1$ . Number of moles of sodium carbonate in $50.0 mg$ is calculated as shown below;