EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
7th Edition
ISBN: 9780100257047
Author: Chang
Publisher: YUZU
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Chapter 13, Problem 13.71QP

(a)

Interpretation Introduction

Interpretation:

For following solutions normal freezing point and boiling point to be calculated.

  1. a) 21.2g NaCl in 135 mL water
  2. b)

Concept introduction:

  • Boiling point is the temperature at which liquid turns into a gas. Example: boiling point of water is 100°C.  That is water changes from liquid phase to gas phase.
  •  

    ΔTb= Tb- Tb°

Where,

ΔTb Change in boiling point

Tb -  Boiling point of the solution

Tb° -  Boiling point of pure solvent

  • Boiling point elevation (ΔTb) is distinction between boiling point of the pure solvent (Tb°) and the boiling point of the solution (Tb).
  •  

    ΔTb= Kbm

Where,

ΔTb Change in boiling point

Kb Molal boiling point constant

m- molality of the solution

  • Freezing point is the temperature at which liquid turns into solid.
  • Freezing point depression (ΔTf) is distinction between freezing point of the pure solvent (Tf°) and the freezing point of the solution (Tf).
  •  

    ΔTf= Kfm

Where,

ΔTf Change in freezing point

Kf Molal freezing point constant

m- Molality of the solution

(a)

Expert Solution
Check Mark

Answer to Problem 13.71QP

Freezing point of NaCl  solution = -10.0°C

Boiling point of NaCl  solution = 102.8°C

Explanation of Solution

To record the given data

Amount of water in which NaCl present = 135mL

Amount of Sodium chloride = 21.2g

Amount of water and Sodium chloride are recorded as shown.

Sodium chloride is strong electrolyte. The concentration of the particles is double this solution. Note that because the density of water is 1g/mL and mass of water 135g.

To calculate mole of Sodium chloride

21.2g ×1mol58.44g= 0.363mol

By plugging in the values of amount and molecular mass of Sodium chloride, a mole of Sodium chloride has calculated.

Calculation of molality of the solution

molality =0.363 mol0.135kg water= 2.70m

By plugging in the value of moles of the solute in kilogram water, molality of the solution has calculated.

Finding change in freezing point and boiling point of the solution

i=2

ΔT= iKfm = 2(1.86°C/m)(2.70m) =10.0°C

ΔT= iKbm = 2(0.52°C/m)(2.70m) =2.8°C

New boiling point = 100 + 2.8°C=102.8°C

By plugging in the value of molal freezing point constant and molality of the solution, change in freezing point and boiling point of the solution has calculated.

Conclusion

For given solutions normal freezing point and boiling point has been calculated.

Freezing point of NaCl  solution = -10.0°C

Boiling point of NaCl  solution = 102.8°C

(b)

Interpretation Introduction

Interpretation:

For following solutions normal freezing point and boiling point to be calculated.

(b)15.4g of Urea in 66.7mL of water

Concept introduction:

  • Boiling point is the temperature at which liquid turns into a gas. Example: boiling point of water is 100°C.  That is water changes from liquid phase to gas phase.

ΔTb= Tb- Tb°

Where,

ΔTb Change in boiling point

Tb -  Boiling point of the solution

Tb° -  Boiling point of pure solvent

  • Boiling point elevation (ΔTb)  is distinction between boiling point of the pure solvent (Tb°) and the boiling point of the solution (Tb).

ΔTb= Kbm

Where,

ΔTb Change in boiling point

Kb Molal boiling point constant

m- molality of the solution

  • Freezing point is the temperature at which liquid turns into solid.
  • Freezing point depression (ΔTf) is distinction between freezing point of the pure solvent (Tf°) and the freezing point of the solution (Tf).

ΔTf= Kfm

Where,

ΔTf Change in freezing point

Kf Molal freezing point constant

m- Molality of the solution

(b)

Expert Solution
Check Mark

Answer to Problem 13.71QP

Freezing point of Urea solution = -7.14°C

Boiling point of Urea solution = 102.0°C

Explanation of Solution

Given data

Amount of water in which Urea present = 66.7 mL

Amount of Urea = 15.4g

Amount of water and urea are recorded as shown.

Calculation of mole of Urea

Urea is non- electrolyte. The concentration of the particles is just equal to the concentration of urea.

Molecular mass of urea = 60.06g

15.4g ×1mol60.06g= 0.256 mol

By plugging in the values of amount and molecular mass of Urea, a mole of Urea has calculated.

Calculate\ion of molality of the solution

molality =0.256 mol0.0667kg water= 3.84 m

By plugging in the value of moles of the solute in kilogram water, molality of the solution has calculated.

Finding change in freezing point and boiling point of the solution

i=1

ΔT= iKfm = 1(1.86°C/m)(3.84m) =7.14°C

ΔT= iKbm =1(0.52°C/m)(3.84m) =2.0°C

 boiling point = 100 + 2.8°C=102.0°C

By plugging in the value of molal freezing point constant and molality of the solution, change in freezing point and boiling point of the solution has calculated.

Conclusion

For given solutions normal freezing point and boiling point has been calculated.

Freezing point of Urea solution was calculated as -7.14°C

Boiling point of Urea solution was calculated as 102.0°C

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Chapter 13 Solutions

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO

Ch. 13.6 - Prob. 2PECh. 13.6 - Prob. 2RCCh. 13.6 - Prob. 3PECh. 13.6 - Prob. 4PECh. 13.6 - Prob. 3RCCh. 13.6 - Prob. 5PECh. 13.6 - Prob. 4RCCh. 13 - Prob. 13.1QPCh. 13 - Prob. 13.2QPCh. 13 - Prob. 13.3QPCh. 13 - 13.4 As you know, some solution processes are...Ch. 13 - Prob. 13.5QPCh. 13 - Prob. 13.6QPCh. 13 - Prob. 13.7QPCh. 13 - Prob. 13.8QPCh. 13 - 13.9 Arrange these compounds in order of...Ch. 13 - Prob. 13.10QPCh. 13 - Prob. 13.11QPCh. 13 - Prob. 13.12QPCh. 13 - Prob. 13.13QPCh. 13 - 13.14 Calculate the amount of water (in grams)...Ch. 13 - Prob. 13.15QPCh. 13 - Prob. 13.16QPCh. 13 - Prob. 13.17QPCh. 13 - 12.20 For dilute aqueous solutions in which the...Ch. 13 - Prob. 13.19QPCh. 13 - 13.20 The concentrated sulfuric acid we use in the...Ch. 13 - 13.21 Calculate the molarity and the molality of...Ch. 13 - 13.22 The density of an aqueous solution...Ch. 13 - Prob. 13.23QPCh. 13 - Prob. 13.25QPCh. 13 - 13.26 The solubility of KNO3 is 155 g per 100 g of...Ch. 13 - Prob. 13.27QPCh. 13 - Prob. 13.28QPCh. 13 - Prob. 13.29QPCh. 13 - Prob. 13.30QPCh. 13 - Prob. 13.31QPCh. 13 - 13.32 A man bought a goldfish in a pet shop. Upon...Ch. 13 - Prob. 13.33QPCh. 13 - 13.34 A miner working 260 m below sea level opened...Ch. 13 - Prob. 13.35QPCh. 13 - 13.36 The solubility of N2 in blood at 37°C and at...Ch. 13 - Prob. 13.37QPCh. 13 - Prob. 13.38QPCh. 13 - Prob. 13.39QPCh. 13 - 13.40 How is the lowering in vapor pressure...Ch. 13 - Prob. 13.41QPCh. 13 - Prob. 13.42QPCh. 13 - Prob. 13.43QPCh. 13 - Prob. 13.44QPCh. 13 - Prob. 13.45QPCh. 13 - Prob. 13.46QPCh. 13 - Prob. 13.47QPCh. 13 - 13.48 How many grams of sucrose (C12H22O11) must...Ch. 13 - Prob. 13.49QPCh. 13 - 13.50 The vapor pressures of ethanol (C2H5OH) and...Ch. 13 - Prob. 13.51QPCh. 13 - Prob. 13.52QPCh. 13 - 13.53 What are the boiling point and freezing...Ch. 13 - 13.54 An aqueous solution contains the amino acid...Ch. 13 - 13.55 Pheromones are compounds secreted by the...Ch. 13 - 12.58 The elemental analysis of an organic solid...Ch. 13 - Prob. 13.57QPCh. 13 - 13.58 A solution is prepared by condensing 4.00 L...Ch. 13 - Prob. 13.59QPCh. 13 - 13.60 A solution of 2.50 g of a compound of...Ch. 13 - Prob. 13.61QPCh. 13 - 13.62 A solution containing 0.8330 g of a protein...Ch. 13 - Prob. 13.63QPCh. 13 - 13.64 A solution of 6.85 g of a carbohydrate in...Ch. 13 - 13.65 Define ion pairs. What effect does ion-pair...Ch. 13 - Prob. 13.66QPCh. 13 - Prob. 13.67QPCh. 13 - Prob. 13.68QPCh. 13 - Prob. 13.69QPCh. 13 - Prob. 13.70QPCh. 13 - Prob. 13.71QPCh. 13 - 13.72 At 25°C the vapor pressure of pure water is...Ch. 13 - 13.73 Both NaCl and CaCl2 are used to melt ice on...Ch. 13 - Prob. 13.74QPCh. 13 - Prob. 13.75QPCh. 13 - Prob. 13.76QPCh. 13 - Prob. 13.77QPCh. 13 - Prob. 13.78QPCh. 13 - Prob. 13.79QPCh. 13 - Prob. 13.80QPCh. 13 - Prob. 13.81QPCh. 13 - Prob. 13.82QPCh. 13 - Prob. 13.83QPCh. 13 - Prob. 13.84QPCh. 13 - Prob. 13.85QPCh. 13 - Prob. 13.86QPCh. 13 - Prob. 13.87QPCh. 13 - Prob. 13.88QPCh. 13 - Prob. 13.89QPCh. 13 - Prob. 13.90QPCh. 13 - 13.91 Hydrogen peroxide with a concentration of...Ch. 13 - 13.92 Before a carbonated beverage bottle is...Ch. 13 - Prob. 13.93QPCh. 13 - 13.94 Explain each of these statements: (a) The...Ch. 13 - Prob. 13.95QPCh. 13 - Prob. 13.96QPCh. 13 - Prob. 13.97QPCh. 13 - Prob. 13.98QPCh. 13 - Prob. 13.99QPCh. 13 - Prob. 13.100QPCh. 13 - Prob. 13.101QPCh. 13 - Prob. 13.102QPCh. 13 - Prob. 13.103QPCh. 13 - Prob. 13.104QPCh. 13 - Prob. 13.105SPCh. 13 - Prob. 13.106SPCh. 13 - Prob. 13.107SPCh. 13 - Prob. 13.108SPCh. 13 - Prob. 13.109SPCh. 13 - Prob. 13.110SPCh. 13 - 13.111 A student carried out the following...Ch. 13 - Prob. 13.112SPCh. 13 - Prob. 13.113SPCh. 13 - Prob. 13.114SP
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Solutions: Crash Course Chemistry #27; Author: Crash Course;https://www.youtube.com/watch?v=9h2f1Bjr0p4;License: Standard YouTube License, CC-BY