In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture: 5 mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I 2 + 25 mL H 2 O a. How many moles of acetone were in the reaction mixture? Recall that, for a component A , moles A = M A × V , where M A is the molarity of A and V is the volume in liters of the solution of A that was used. moles acetone b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL, 0.050 L, and the number of moles of acetone was found in Part (a). Again, M A = moles of A V of soln. in liters M acetone c. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentrations of H + ion and I 2 as in the original mixture?

Question

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Answer 1

Textbook Question

Chapter 20, Problem 1ASA

In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture:

5 mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I 2 + 25 mL H 2 O

a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, moles A = M A × V , where M A is the molarity of A and V is the volume in liters of the solution of A that was used.

__________ moles acetone

b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL, 0.050 L, and the number of moles of acetone was found in Part (a). Again,

M A = moles of A V of soln. in liters

__________ M acetone

c. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentrations of H + ion and I 2 as in the original mixture?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The moles of acetone are present in the reaction mixture should be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

Moles of acetone = 0.02 mol

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of moles of acetone is shown below:

moles = molarity × volume= 4 molL-1 × 0.005 L= 0.02 mol

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The molarity of acetone in the reaction mixture should be determined when volume of mixture is 50 mL

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

The ratio of moles to the volume in liters is known as molarity.

Answer to Problem 1ASA

Molarity of acetone = 0.4 M

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of molarity of acetone is shown below:

Molarity = molesvolume(L)= 0.02 mol0.05 L= 0.4 M

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

Keeping the total volume and concentration of H+ and I₂ remains same; predict how the molarity of aceteone should be doubled.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O

Explanation of Solution

If concentration of H+ and I₂ remains the same then decreases the volume of H₂O by 15 mL and increases the initial concentration of acetone with volume by 15 mL.

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O

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Answer 2

Textbook Question

Chapter 20, Problem 1ASA

In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture:

5 mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I 2 + 25 mL H 2 O

a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, moles A = M A × V , where M A is the molarity of A and V is the volume in liters of the solution of A that was used.

__________ moles acetone

b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL, 0.050 L, and the number of moles of acetone was found in Part (a). Again,

M A = moles of A V of soln. in liters

__________ M acetone

c. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentrations of H + ion and I 2 as in the original mixture?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The moles of acetone are present in the reaction mixture should be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

Moles of acetone = 0.02 mol

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of moles of acetone is shown below:

moles = molarity × volume= 4 molL-1 × 0.005 L= 0.02 mol

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The molarity of acetone in the reaction mixture should be determined when volume of mixture is 50 mL

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

The ratio of moles to the volume in liters is known as molarity.

Answer to Problem 1ASA

Molarity of acetone = 0.4 M

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of molarity of acetone is shown below:

Molarity = molesvolume(L)= 0.02 mol0.05 L= 0.4 M

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

Keeping the total volume and concentration of H+ and I₂ remains same; predict how the molarity of aceteone should be doubled.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O

Explanation of Solution

If concentration of H+ and I₂ remains the same then decreases the volume of H₂O by 15 mL and increases the initial concentration of acetone with volume by 15 mL.

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O

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Answer 3

Textbook Question

Chapter 20, Problem 1ASA

In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture:

5 mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I 2 + 25 mL H 2 O

a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, moles A = M A × V , where M A is the molarity of A and V is the volume in liters of the solution of A that was used.

__________ moles acetone

b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL, 0.050 L, and the number of moles of acetone was found in Part (a). Again,

M A = moles of A V of soln. in liters

__________ M acetone

c. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentrations of H + ion and I 2 as in the original mixture?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The moles of acetone are present in the reaction mixture should be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

Moles of acetone = 0.02 mol

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of moles of acetone is shown below:

moles = molarity × volume= 4 molL-1 × 0.005 L= 0.02 mol

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The molarity of acetone in the reaction mixture should be determined when volume of mixture is 50 mL

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

The ratio of moles to the volume in liters is known as molarity.

Answer to Problem 1ASA

Molarity of acetone = 0.4 M

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of molarity of acetone is shown below:

Molarity = molesvolume(L)= 0.02 mol0.05 L= 0.4 M

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

Keeping the total volume and concentration of H+ and I₂ remains same; predict how the molarity of aceteone should be doubled.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O

Explanation of Solution

If concentration of H+ and I₂ remains the same then decreases the volume of H₂O by 15 mL and increases the initial concentration of acetone with volume by 15 mL.

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O

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Answer 4

Textbook Question

Chapter 20, Problem 1ASA

In a reaction involving the iodination of acetone, the following volumes were used to make up the reaction mixture:

5 mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I 2 + 25 mL H 2 O

a. How many moles of acetone were in the reaction mixture? Recall that, for a component A, moles A = M A × V , where M A is the molarity of A and V is the volume in liters of the solution of A that was used.

__________ moles acetone

b. What was the molarity of acetone in the reaction mixture? The volume of the mixture was 50 mL, 0.050 L, and the number of moles of acetone was found in Part (a). Again,

M A = moles of A V of soln. in liters

__________ M acetone

c. How could you double the molarity of the acetone in the reaction mixture, keeping the total volume at 50 mL and keeping the same concentrations of H + ion and I 2 as in the original mixture?

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The moles of acetone are present in the reaction mixture should be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

Moles of acetone = 0.02 mol

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of moles of acetone is shown below:

moles = molarity × volume= 4 molL-1 × 0.005 L= 0.02 mol

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The molarity of acetone in the reaction mixture should be determined when volume of mixture is 50 mL

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

The ratio of moles to the volume in liters is known as molarity.

Answer to Problem 1ASA

Molarity of acetone = 0.4 M

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of molarity of acetone is shown below:

Molarity = molesvolume(L)= 0.02 mol0.05 L= 0.4 M

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

Keeping the total volume and concentration of H+ and I₂ remains same; predict how the molarity of aceteone should be doubled.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O

Explanation of Solution

If concentration of H+ and I₂ remains the same then decreases the volume of H₂O by 15 mL and increases the initial concentration of acetone with volume by 15 mL.

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The moles of acetone are present in the reaction mixture should be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

Moles of acetone = 0.02 mol

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of moles of acetone is shown below:

moles = molarity × volume= 4 molL-1 × 0.005 L= 0.02 mol

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The molarity of acetone in the reaction mixture should be determined when volume of mixture is 50 mL

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

The ratio of moles to the volume in liters is known as molarity.

Answer to Problem 1ASA

Molarity of acetone = 0.4 M

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of molarity of acetone is shown below:

Molarity = molesvolume(L)= 0.02 mol0.05 L= 0.4 M

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

Keeping the total volume and concentration of H+ and I₂ remains same; predict how the molarity of aceteone should be doubled.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O

Explanation of Solution

If concentration of H+ and I₂ remains the same then decreases the volume of H₂O by 15 mL and increases the initial concentration of acetone with volume by 15 mL.

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O

Answer 8

(a)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The moles of acetone are present in the reaction mixture should be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

Moles of acetone = 0.02 mol

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of moles of acetone is shown below:

moles = molarity × volume= 4 molL-1 × 0.005 L= 0.02 mol

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The moles of acetone are present in the reaction mixture should be determined.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer 13

Answer to Problem 1ASA

Moles of acetone = 0.02 mol

Answer 14

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of moles of acetone is shown below:

moles = molarity × volume= 4 molL-1 × 0.005 L= 0.02 mol

Answer 15

(b)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The molarity of acetone in the reaction mixture should be determined when volume of mixture is 50 mL

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

The ratio of moles to the volume in liters is known as molarity.

Answer to Problem 1ASA

Molarity of acetone = 0.4 M

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of molarity of acetone is shown below:

Molarity = molesvolume(L)= 0.02 mol0.05 L= 0.4 M

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

The molarity of acetone in the reaction mixture should be determined when volume of mixture is 50 mL

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

The ratio of moles to the volume in liters is known as molarity.

Answer 20

Answer to Problem 1ASA

Molarity of acetone = 0.4 M

Answer 21

Explanation of Solution

Given information:

Molarity of acetone = 4 M

Molarity of HCl = 1M

Molarity of I₂ = 0.0050 M

Volume of acetone = 5 mL

Volume of HCl = 10 mL

Volume of I₂ = 10 mL

Volume of H₂O = 25 mL

Total volume = 5 + 10 + 10 + 25 = 50 mL = 0.05 L

The calculation of molarity of acetone is shown below:

Molarity = molesvolume(L)= 0.02 mol0.05 L= 0.4 M

Answer 22

(c)

Expert Solution

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

Keeping the total volume and concentration of H+ and I₂ remains same; predict how the molarity of aceteone should be doubled.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer to Problem 1ASA

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O

Explanation of Solution

If concentration of H+ and I₂ remains the same then decreases the volume of H₂O by 15 mL and increases the initial concentration of acetone with volume by 15 mL.

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Interpretation Introduction

Interpretation:

The reaction of iodination of acetone, forms a reaction mixture;

5mL 4.0 M acetone + 10 mL 1.0 M HCl + 10 mL 0.0050 M I2+ 25 mL H2O

Keeping the total volume and concentration of H+ and I₂ remains same; predict how the molarity of aceteone should be doubled.

Concept Introduction:

Mole is the amount of the substance that contains the same number of particles or atoms or molecules. Molar mass is defined as an average mass of atoms present in the chemical formula. It is the sum of the atomic masses of all the atoms present in the chemical formula of any compound.

Answer 27

Answer to Problem 1ASA

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O

Answer 28

Explanation of Solution

If concentration of H+ and I₂ remains the same then decreases the volume of H₂O by 15 mL and increases the initial concentration of acetone with volume by 15 mL.

20 mL of 4M acetone + 10 mL of 1M HCl + 10 mL of 0.005 M I2 + 10 mL H2O