Calculate the concentration in % ( v / v ) of the following solutions: a. 250 . mL of solution contains 15 .0 mL of acetone. b. 250 . mL of solution contains 15 .0 mL of any soluble liquid solute. c. 1 .0 quart of acetic acid is put into a 5- gallon container, and enough water is added to fill the container. d. A solution of acetone and water is separated by distillation. A 300 .-mL sample gives 109 mL of acetone.

Question

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Answer 1

Textbook Question

Chapter 7, Problem 7.35E

Calculate the concentration in % ( v / v ) of the following solutions:

a. 250 . mL of solution contains 15 .0 mL of acetone.

b. 250 . mL of solution contains 15 .0 mL of any soluble liquid solute.

c. 1 .0 quart of acetic acid is put into a 5- gallon container, and enough water is added to fill the container.

d.A solution of acetone and water is separated by distillation. A 300 .-mL sample gives 109 mL of acetone.

Expert Solution

Interpretation Introduction

(a)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of acetone and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

%(v/v)=volume of acetoneSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Conclusion

The concentration in %(v/v) of 200. mL of solution containing 15 mL of alcohol is 6.0 %v/v.

Expert Solution

Interpretation Introduction

(b)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of solute and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=volume of soluteSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Conclusion

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Expert Solution

Interpretation Introduction

(c)

Interpretation:

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is 5 %(v/v).

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The 1- gallon container contains 4.0 quart. Thus, the 5- gallon container contains 4×5=20 quart.

The volume of solute and volume of solution is 1.0 quart and 20.0 quart respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=120×100=5 %

Thus, the concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Conclusion

The concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Expert Solution

Interpretation Introduction

(d)

Interpretation:

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of acetone and volume of solution is 109 mL and 300.-mL respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

%(v/v)=109 mL300 mL×100=36.33 %v/v

Thus, the concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Conclusion

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

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Answer 2

Textbook Question

Chapter 7, Problem 7.35E

Calculate the concentration in % ( v / v ) of the following solutions:

a. 250 . mL of solution contains 15 .0 mL of acetone.

b. 250 . mL of solution contains 15 .0 mL of any soluble liquid solute.

c. 1 .0 quart of acetic acid is put into a 5- gallon container, and enough water is added to fill the container.

d.A solution of acetone and water is separated by distillation. A 300 .-mL sample gives 109 mL of acetone.

Expert Solution

Interpretation Introduction

(a)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of acetone and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

%(v/v)=volume of acetoneSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Conclusion

The concentration in %(v/v) of 200. mL of solution containing 15 mL of alcohol is 6.0 %v/v.

Expert Solution

Interpretation Introduction

(b)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of solute and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=volume of soluteSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Conclusion

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Expert Solution

Interpretation Introduction

(c)

Interpretation:

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is 5 %(v/v).

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The 1- gallon container contains 4.0 quart. Thus, the 5- gallon container contains 4×5=20 quart.

The volume of solute and volume of solution is 1.0 quart and 20.0 quart respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=120×100=5 %

Thus, the concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Conclusion

The concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Expert Solution

Interpretation Introduction

(d)

Interpretation:

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of acetone and volume of solution is 109 mL and 300.-mL respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

%(v/v)=109 mL300 mL×100=36.33 %v/v

Thus, the concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Conclusion

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

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Answer 3

Textbook Question

Chapter 7, Problem 7.35E

Calculate the concentration in % ( v / v ) of the following solutions:

a. 250 . mL of solution contains 15 .0 mL of acetone.

b. 250 . mL of solution contains 15 .0 mL of any soluble liquid solute.

c. 1 .0 quart of acetic acid is put into a 5- gallon container, and enough water is added to fill the container.

d.A solution of acetone and water is separated by distillation. A 300 .-mL sample gives 109 mL of acetone.

Expert Solution

Interpretation Introduction

(a)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of acetone and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

%(v/v)=volume of acetoneSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Conclusion

The concentration in %(v/v) of 200. mL of solution containing 15 mL of alcohol is 6.0 %v/v.

Expert Solution

Interpretation Introduction

(b)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of solute and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=volume of soluteSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Conclusion

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Expert Solution

Interpretation Introduction

(c)

Interpretation:

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is 5 %(v/v).

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The 1- gallon container contains 4.0 quart. Thus, the 5- gallon container contains 4×5=20 quart.

The volume of solute and volume of solution is 1.0 quart and 20.0 quart respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=120×100=5 %

Thus, the concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Conclusion

The concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Expert Solution

Interpretation Introduction

(d)

Interpretation:

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of acetone and volume of solution is 109 mL and 300.-mL respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

%(v/v)=109 mL300 mL×100=36.33 %v/v

Thus, the concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Conclusion

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

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Answer 4

Textbook Question

Chapter 7, Problem 7.35E

Calculate the concentration in % ( v / v ) of the following solutions:

a. 250 . mL of solution contains 15 .0 mL of acetone.

b. 250 . mL of solution contains 15 .0 mL of any soluble liquid solute.

c. 1 .0 quart of acetic acid is put into a 5- gallon container, and enough water is added to fill the container.

d.A solution of acetone and water is separated by distillation. A 300 .-mL sample gives 109 mL of acetone.

Expert Solution

Interpretation Introduction

(a)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of acetone and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

%(v/v)=volume of acetoneSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Conclusion

The concentration in %(v/v) of 200. mL of solution containing 15 mL of alcohol is 6.0 %v/v.

Expert Solution

Interpretation Introduction

(b)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of solute and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=volume of soluteSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Conclusion

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Expert Solution

Interpretation Introduction

(c)

Interpretation:

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is 5 %(v/v).

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The 1- gallon container contains 4.0 quart. Thus, the 5- gallon container contains 4×5=20 quart.

The volume of solute and volume of solution is 1.0 quart and 20.0 quart respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=120×100=5 %

Thus, the concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Conclusion

The concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Expert Solution

Interpretation Introduction

(d)

Interpretation:

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of acetone and volume of solution is 109 mL and 300.-mL respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

%(v/v)=109 mL300 mL×100=36.33 %v/v

Thus, the concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Conclusion

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

Interpretation Introduction

(a)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of acetone and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

%(v/v)=volume of acetoneSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Conclusion

The concentration in %(v/v) of 200. mL of solution containing 15 mL of alcohol is 6.0 %v/v.

Expert Solution

Interpretation Introduction

(b)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of solute and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=volume of soluteSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Conclusion

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Expert Solution

Interpretation Introduction

(c)

Interpretation:

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is 5 %(v/v).

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The 1- gallon container contains 4.0 quart. Thus, the 5- gallon container contains 4×5=20 quart.

The volume of solute and volume of solution is 1.0 quart and 20.0 quart respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=120×100=5 %

Thus, the concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Conclusion

The concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Expert Solution

Interpretation Introduction

(d)

Interpretation:

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of acetone and volume of solution is 109 mL and 300.-mL respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

%(v/v)=109 mL300 mL×100=36.33 %v/v

Thus, the concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Conclusion

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Answer 8

Expert Solution

Interpretation Introduction

(a)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of acetone and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

%(v/v)=volume of acetoneSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Conclusion

The concentration in %(v/v) of 200. mL of solution containing 15 mL of alcohol is 6.0 %v/v.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

(a)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer 13

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Answer 14

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of acetone and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

%(v/v)=volume of acetoneSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of acetone is 6.0 %v/v.

Answer 15

Conclusion

The concentration in %(v/v) of 200. mL of solution containing 15 mL of alcohol is 6.0 %v/v.

Answer 16

Expert Solution

Interpretation Introduction

(b)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of solute and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=volume of soluteSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Conclusion

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Interpretation Introduction

(b)

Interpretation:

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer 21

Answer to Problem 7.35E

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Answer 22

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of solute and volume of solution is 15.0 mL and 250. mL respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=volume of soluteSolution volume×100=15 mL250 mL×100=6.0 %v/v

Thus, the concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Answer 23

Conclusion

The concentration in %(v/v) of 250. mL of solution that contains 15.0 mL of any soluble liquid solute is 6.0 %v/v.

Answer 24

Expert Solution

Interpretation Introduction

(c)

Interpretation:

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is 5 %(v/v).

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The 1- gallon container contains 4.0 quart. Thus, the 5- gallon container contains 4×5=20 quart.

The volume of solute and volume of solution is 1.0 quart and 20.0 quart respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=120×100=5 %

Thus, the concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Conclusion

The concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Interpretation Introduction

(c)

Interpretation:

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer 29

Answer to Problem 7.35E

The concentration in %(v/v) of 1.0 quart of acetic acid that is put into a 5- gallon container is 5 %(v/v).

Answer 30

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The 1- gallon container contains 4.0 quart. Thus, the 5- gallon container contains 4×5=20 quart.

The volume of solute and volume of solution is 1.0 quart and 20.0 quart respectively.

Substitute the value of volume of solute and volume of solution in the above equation as follows.

%(v/v)=120×100=5 %

Thus, the concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Answer 31

Conclusion

The concentration in %(v/v) of 1.0 quart of acetic acid is put into a 5- gallon container is 5 %(v/v).

Answer 32

Expert Solution

Interpretation Introduction

(d)

Interpretation:

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).

Answer to Problem 7.35E

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Explanation of Solution

The formula to calculate concentration in %(v/v) is given below as,

%(v/v)=Solute volumeSolution volume×100

The volume of acetone and volume of solution is 109 mL and 300.-mL respectively.

Substitute the value of volume of acetone and volume of solution in the above equation as follows.

%(v/v)=109 mL300 mL×100=36.33 %v/v

Thus, the concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Conclusion

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is 36.33 %v/v.

Answer 33

Expert Solution

Answer 34

Expert Solution

Answer 35

Expert Solution

Answer 36

Interpretation Introduction

(d)

Interpretation:

The concentration in %(v/v) of a 300.-mL sample giving 109 mL of acetone on distillation is to be calculated.

Concept introduction:

A solution is prepared by mixing proper amount of solute and solvent. The concentration of solution when measured on the basis of solution volume, is expressed in terms of percent weight by volume (% w/v), percent volume by volume (% v/v) and molarity (M). When concentration is measured on the basis of mass of solvent, it is expressed in terms of percent weight by weight (% w/w).