The freezing temperature of 1.50 m urea has to be calculated. Concept Introduction: Depression in freezing point: At freezing point solid phase of the solvent will be in equilibrium with liquid phase. When solute molecule is added, it will interfere with the rate at which the liquid molecules associate to form solid state, hence lower the freezing point of the solution. Δ T f = k f ( m p a r t i c l e s ) Where, Δ T f difference in temperature of pure solvent and solution, k f is freezing point depression constant.

Question

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Answer 1

Question

Chapter 6, Problem 6.73QP

(a)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m urea has to be calculated.

Concept Introduction:

Depression in freezing point:

At freezing point solid phase of the solvent will be in equilibrium with liquid phase. When solute molecule is added, it will interfere with the rate at which the liquid molecules associate to form solid state, hence lower the freezing point of the solution.

ΔTf = kf (m particles)

Where, ΔTf difference in temperature of pure solvent and solution, kf is freezing point depression constant.

(a)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m urea solution is -2.79 oC.

Explanation of Solution

kf for aqueous solution is 1.86oCm.

So freezing point temperature of 1.50 m urea can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(1.5 m)= 2.79 oCT1−T2 = 2.79 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the urea solution.

Therefore freezing point of the solution is given below.

T1−T2 = 2.79 oC0 oC−T2= 2.79 oC T2=0 oC−2.79 oC = -2.79 oC.

Freezing temperature of the solution is -2.79 oC.

(b)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m LiBr has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m LiBr is -5.58 oC.

Explanation of Solution

kf for aqueous solution is 1.86oCm.

LiBr being ionic compound, dissociate to form two particles. So 1.50 m LiBr is 3 m (1.5 m × 2) particles.

So freezing point temperature of 1.50 m LiBr can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(3 m)= 5.58 oCT1−T2 = 5.58 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the solution.

Therefore freezing point of the solution is given below.

T1−T2 = 5.58 oC0 oC−T2= 5.58 oCT2=0 oC−5.58 oC= -5.58 oC.

Freezing point temperature of the solution is -5.58 oC.

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Answer 2

Question

Chapter 6, Problem 6.73QP

(a)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m urea has to be calculated.

Concept Introduction:

Depression in freezing point:

At freezing point solid phase of the solvent will be in equilibrium with liquid phase. When solute molecule is added, it will interfere with the rate at which the liquid molecules associate to form solid state, hence lower the freezing point of the solution.

ΔTf = kf (m particles)

Where, ΔTf difference in temperature of pure solvent and solution, kf is freezing point depression constant.

(a)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m urea solution is -2.79 oC.

Explanation of Solution

kf for aqueous solution is 1.86oCm.

So freezing point temperature of 1.50 m urea can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(1.5 m)= 2.79 oCT1−T2 = 2.79 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the urea solution.

Therefore freezing point of the solution is given below.

T1−T2 = 2.79 oC0 oC−T2= 2.79 oC T2=0 oC−2.79 oC = -2.79 oC.

Freezing temperature of the solution is -2.79 oC.

(b)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m LiBr has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m LiBr is -5.58 oC.

Explanation of Solution

kf for aqueous solution is 1.86oCm.

LiBr being ionic compound, dissociate to form two particles. So 1.50 m LiBr is 3 m (1.5 m × 2) particles.

So freezing point temperature of 1.50 m LiBr can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(3 m)= 5.58 oCT1−T2 = 5.58 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the solution.

Therefore freezing point of the solution is given below.

T1−T2 = 5.58 oC0 oC−T2= 5.58 oCT2=0 oC−5.58 oC= -5.58 oC.

Freezing point temperature of the solution is -5.58 oC.

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Answer 3

Question

Chapter 6, Problem 6.73QP

(a)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m urea has to be calculated.

Concept Introduction:

Depression in freezing point:

At freezing point solid phase of the solvent will be in equilibrium with liquid phase. When solute molecule is added, it will interfere with the rate at which the liquid molecules associate to form solid state, hence lower the freezing point of the solution.

ΔTf = kf (m particles)

Where, ΔTf difference in temperature of pure solvent and solution, kf is freezing point depression constant.

(a)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m urea solution is -2.79 oC.

Explanation of Solution

kf for aqueous solution is 1.86oCm.

So freezing point temperature of 1.50 m urea can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(1.5 m)= 2.79 oCT1−T2 = 2.79 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the urea solution.

Therefore freezing point of the solution is given below.

T1−T2 = 2.79 oC0 oC−T2= 2.79 oC T2=0 oC−2.79 oC = -2.79 oC.

Freezing temperature of the solution is -2.79 oC.

(b)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m LiBr has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m LiBr is -5.58 oC.

Explanation of Solution

kf for aqueous solution is 1.86oCm.

LiBr being ionic compound, dissociate to form two particles. So 1.50 m LiBr is 3 m (1.5 m × 2) particles.

So freezing point temperature of 1.50 m LiBr can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(3 m)= 5.58 oCT1−T2 = 5.58 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the solution.

Therefore freezing point of the solution is given below.

T1−T2 = 5.58 oC0 oC−T2= 5.58 oCT2=0 oC−5.58 oC= -5.58 oC.

Freezing point temperature of the solution is -5.58 oC.

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Answer 4

Question

Chapter 6, Problem 6.73QP

(a)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m urea has to be calculated.

Concept Introduction:

Depression in freezing point:

At freezing point solid phase of the solvent will be in equilibrium with liquid phase. When solute molecule is added, it will interfere with the rate at which the liquid molecules associate to form solid state, hence lower the freezing point of the solution.

ΔTf = kf (m particles)

Where, ΔTf difference in temperature of pure solvent and solution, kf is freezing point depression constant.

(a)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m urea solution is -2.79 oC.

Explanation of Solution

kf for aqueous solution is 1.86oCm.

So freezing point temperature of 1.50 m urea can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(1.5 m)= 2.79 oCT1−T2 = 2.79 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the urea solution.

Therefore freezing point of the solution is given below.

T1−T2 = 2.79 oC0 oC−T2= 2.79 oC T2=0 oC−2.79 oC = -2.79 oC.

Freezing temperature of the solution is -2.79 oC.

(b)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m LiBr has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m LiBr is -5.58 oC.

Explanation of Solution

kf for aqueous solution is 1.86oCm.

LiBr being ionic compound, dissociate to form two particles. So 1.50 m LiBr is 3 m (1.5 m × 2) particles.

So freezing point temperature of 1.50 m LiBr can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(3 m)= 5.58 oCT1−T2 = 5.58 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the solution.

Therefore freezing point of the solution is given below.

T1−T2 = 5.58 oC0 oC−T2= 5.58 oCT2=0 oC−5.58 oC= -5.58 oC.

Freezing point temperature of the solution is -5.58 oC.

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Answer 5

Chapter 6, Problem 6.73QP

(a)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m urea has to be calculated.

Concept Introduction:

Depression in freezing point:

At freezing point solid phase of the solvent will be in equilibrium with liquid phase. When solute molecule is added, it will interfere with the rate at which the liquid molecules associate to form solid state, hence lower the freezing point of the solution.

ΔTf = kf (m particles)

Where, ΔTf difference in temperature of pure solvent and solution, kf is freezing point depression constant.

(a)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m urea solution is -2.79 oC.

Explanation of Solution

kf for aqueous solution is 1.86oCm.

So freezing point temperature of 1.50 m urea can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(1.5 m)= 2.79 oCT1−T2 = 2.79 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the urea solution.

Therefore freezing point of the solution is given below.

T1−T2 = 2.79 oC0 oC−T2= 2.79 oC T2=0 oC−2.79 oC = -2.79 oC.

Freezing temperature of the solution is -2.79 oC.

(b)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m LiBr has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m LiBr is -5.58 oC.

Explanation of Solution

kf for aqueous solution is 1.86oCm.

LiBr being ionic compound, dissociate to form two particles. So 1.50 m LiBr is 3 m (1.5 m × 2) particles.

So freezing point temperature of 1.50 m LiBr can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(3 m)= 5.58 oCT1−T2 = 5.58 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the solution.

Therefore freezing point of the solution is given below.

T1−T2 = 5.58 oC0 oC−T2= 5.58 oCT2=0 oC−5.58 oC= -5.58 oC.

Freezing point temperature of the solution is -5.58 oC.

Answer 6

Chapter 6, Problem 6.73QP

(a)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m urea has to be calculated.

Concept Introduction:

Depression in freezing point:

At freezing point solid phase of the solvent will be in equilibrium with liquid phase. When solute molecule is added, it will interfere with the rate at which the liquid molecules associate to form solid state, hence lower the freezing point of the solution.

ΔTf = kf (m particles)

Where, ΔTf difference in temperature of pure solvent and solution, kf is freezing point depression constant.

(a)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m urea solution is -2.79 oC.

Explanation of Solution

kf for aqueous solution is 1.86oCm.

So freezing point temperature of 1.50 m urea can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(1.5 m)= 2.79 oCT1−T2 = 2.79 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the urea solution.

Therefore freezing point of the solution is given below.

T1−T2 = 2.79 oC0 oC−T2= 2.79 oC T2=0 oC−2.79 oC = -2.79 oC.

Freezing temperature of the solution is -2.79 oC.

Answer 7

Chapter 6, Problem 6.73QP

(a)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m urea has to be calculated.

Concept Introduction:

Depression in freezing point:

At freezing point solid phase of the solvent will be in equilibrium with liquid phase. When solute molecule is added, it will interfere with the rate at which the liquid molecules associate to form solid state, hence lower the freezing point of the solution.

ΔTf = kf (m particles)

Where, ΔTf difference in temperature of pure solvent and solution, kf is freezing point depression constant.

Answer 8

(a)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m urea solution is -2.79 oC.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

kf for aqueous solution is 1.86oCm.

So freezing point temperature of 1.50 m urea can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(1.5 m)= 2.79 oCT1−T2 = 2.79 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the urea solution.

Therefore freezing point of the solution is given below.

T1−T2 = 2.79 oC0 oC−T2= 2.79 oC T2=0 oC−2.79 oC = -2.79 oC.

Freezing temperature of the solution is -2.79 oC.

Answer 13

(b)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m LiBr has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m LiBr is -5.58 oC.

Explanation of Solution

kf for aqueous solution is 1.86oCm.

LiBr being ionic compound, dissociate to form two particles. So 1.50 m LiBr is 3 m (1.5 m × 2) particles.

So freezing point temperature of 1.50 m LiBr can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(3 m)= 5.58 oCT1−T2 = 5.58 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the solution.

Therefore freezing point of the solution is given below.

T1−T2 = 5.58 oC0 oC−T2= 5.58 oCT2=0 oC−5.58 oC= -5.58 oC.

Freezing point temperature of the solution is -5.58 oC.

Answer 14

(b)

Interpretation Introduction

Interpretation:

The freezing temperature of 1.50 m LiBr has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 15

(b)

Expert Solution

Answer to Problem 6.73QP

The freezing temperature of 1.50 m LiBr is -5.58 oC.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

kf for aqueous solution is 1.86oCm.

LiBr being ionic compound, dissociate to form two particles. So 1.50 m LiBr is 3 m (1.5 m × 2) particles.

So freezing point temperature of 1.50 m LiBr can be calculated as follows,

ΔTf = kf (m particles)=(1.86oCm)(3 m)= 5.58 oCT1−T2 = 5.58 oC

Where T1 is the freezing point of pure water and T2 is the freezing point of the solution.

Therefore freezing point of the solution is given below.

T1−T2 = 5.58 oC0 oC−T2= 5.58 oCT2=0 oC−5.58 oC= -5.58 oC.

Freezing point temperature of the solution is -5.58 oC.

Answer 20

Ch. 6.1 - Describe how you would distinguish experimentally...Ch. 6.1 - Prob. 6.2Q Ch. 6.1 - Explain why, over time, a bottle of soft drink...Ch. 6.1 - Prob. 6.4Q Ch. 6.1 - Prob. 6.5Q Ch. 6.1 - Prob. 6.6Q Ch. 6.2 - Calculate the % (m/V) of 0.0600 L of solution...Ch. 6.2 - Prob. 6.2PP Ch. 6.2 - Prob. 6.3PP Ch. 6.2 - Prob. 6.4PP

Concept explainers

Videos

Answer to Problem 6.73QP

Explanation of Solution

Interpretation:

The freezing temperature of 1.50 m LiBr has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 6.73QP

Explanation of Solution

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Chapter 6 Solutions

Concept explainers

Videos

Answer to Problem 6.73QP

Explanation of Solution

Interpretation: The freezing temperature of 1.50 m LiBr has to be calculated. Concept Introduction: Refer to part (a).

Answer to Problem 6.73QP

Explanation of Solution

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Chapter 6 Solutions

Interpretation:

The freezing temperature of 1.50 m LiBr has to be calculated.

Concept Introduction:

Refer to part (a).