The vapor pressure of water is to be calculated above given solution at 338K. Concept introduction: a) Lowering of vapor pressure and elevation in boiling point are colligative property and depends upon only number of molecules and not nature. b) The boiling point increases on addition of a non volatile solute. c) The ebullioscopic constant of water is 0 .512 0 C .kg m . Given: Boiling point of solution with non-volatile solute = 375.3 K Vapor pressure of pure water = 0.2467 atm,at 338 K

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Answer 1

Question

Chapter 14, Problem 111E

Interpretation Introduction

Interpretation:

The vapor pressure of water is to be calculated above given solution at 338K.

Concept introduction:

a) Lowering of vapor pressure and elevation in boiling point are colligative property and depends upon only number of molecules and not nature.

b) The boiling point increases on addition of a non volatile solute.

c) The ebullioscopic constant of water is $0 .512 \frac{^{0} C .kg}{m}$ .

Given:

Boiling point of solution with non-volatile solute = 375.3 K

Vapor pressure of pure water = 0.2467 atm,at 338 K

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Answer 2

Question

Chapter 14, Problem 111E

Interpretation Introduction

Interpretation:

The vapor pressure of water is to be calculated above given solution at 338K.

Concept introduction:

a) Lowering of vapor pressure and elevation in boiling point are colligative property and depends upon only number of molecules and not nature.

b) The boiling point increases on addition of a non volatile solute.

c) The ebullioscopic constant of water is $0 .512 \frac{^{0} C .kg}{m}$ .

Given:

Boiling point of solution with non-volatile solute = 375.3 K

Vapor pressure of pure water = 0.2467 atm,at 338 K

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Answer 3

Question

Chapter 14, Problem 111E

Interpretation Introduction

Interpretation:

The vapor pressure of water is to be calculated above given solution at 338K.

Concept introduction:

a) Lowering of vapor pressure and elevation in boiling point are colligative property and depends upon only number of molecules and not nature.

b) The boiling point increases on addition of a non volatile solute.

c) The ebullioscopic constant of water is $0 .512 \frac{^{0} C .kg}{m}$ .

Given:

Boiling point of solution with non-volatile solute = 375.3 K

Vapor pressure of pure water = 0.2467 atm,at 338 K

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Answer 4

Question

Chapter 14, Problem 111E

Interpretation Introduction

Interpretation:

The vapor pressure of water is to be calculated above given solution at 338K.

Concept introduction:

a) Lowering of vapor pressure and elevation in boiling point are colligative property and depends upon only number of molecules and not nature.

b) The boiling point increases on addition of a non volatile solute.

c) The ebullioscopic constant of water is $0 .512 \frac{^{0} C .kg}{m}$ .

Given:

Boiling point of solution with non-volatile solute = 375.3 K

Vapor pressure of pure water = 0.2467 atm,at 338 K

Expert Solution & Answer

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Answer 5

Chapter 14, Problem 111E

Interpretation Introduction

Interpretation:

The vapor pressure of water is to be calculated above given solution at 338K.

Concept introduction:

a) Lowering of vapor pressure and elevation in boiling point are colligative property and depends upon only number of molecules and not nature.

b) The boiling point increases on addition of a non volatile solute.

c) The ebullioscopic constant of water is $0 .512 \frac{^{0} C .kg}{m}$ .

Given:

Boiling point of solution with non-volatile solute = 375.3 K

Vapor pressure of pure water = 0.2467 atm,at 338 K

Answer 6

Chapter 14, Problem 111E

Interpretation Introduction

Interpretation:

The vapor pressure of water is to be calculated above given solution at 338K.

Concept introduction:

a) Lowering of vapor pressure and elevation in boiling point are colligative property and depends upon only number of molecules and not nature.

b) The boiling point increases on addition of a non volatile solute.

c) The ebullioscopic constant of water is $0 .512 \frac{^{0} C .kg}{m}$ .

Given:

Boiling point of solution with non-volatile solute = 375.3 K

Vapor pressure of pure water = 0.2467 atm,at 338 K

Answer 7

Chapter 14, Problem 111E

Interpretation Introduction

Interpretation:

The vapor pressure of water is to be calculated above given solution at 338K.

Concept introduction:

a) Lowering of vapor pressure and elevation in boiling point are colligative property and depends upon only number of molecules and not nature.

b) The boiling point increases on addition of a non volatile solute.

c) The ebullioscopic constant of water is $0 .512 \frac{^{0} C .kg}{m}$ .

Given:

Boiling point of solution with non-volatile solute = 375.3 K

Vapor pressure of pure water = 0.2467 atm,at 338 K

Answer 8

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Answer 9

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Answer 10

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Answer 11

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Solution Summary: The author explains that the vapor pressure of water above solution at 338K is 0.229 atm.

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Chapter 14 Solutions