Package: General Chemistry with Connect 2-year Access Card
Package: General Chemistry with Connect 2-year Access Card
7th Edition
ISBN: 9781259680458
Author: Raymond Chang Dr.
Publisher: McGraw-Hill Education
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Chapter 13, Problem 13.17QP

(a)

Interpretation Introduction

Interpretation: For the aqueous 1.22 M Sugar (C12H22O11) solution molality has to be calculated.

Concept introduction:

Molality: Molality is defined as number of moles of the solute present in the specified amount of the solvent in kilograms.

Molality =number of moles of the solutekg of solvent

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in 12g of 12C.

From given mass of substance moles could be calculated by using the following formula,

Molesofsubstance GivenmassofsubstanceMolecularmass

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The S.I. unit of mass is kg.

(a)

Expert Solution
Check Mark

Answer to Problem 13.17QP

Molality of 1.22 M Sugar (C12H22O11) solution is 1.74m.

Explanation of Solution

Given data: Strength of sugar solution =1.22 M

Density of sugar solution =1.12g /mL

Calculation of mass of sugar:

Substitute the value of strength of sugar and molecular mass of sugar in the formula to calculate mass of sugar.

Molecular weight of sugar =342.3g

mass of sugar = 1.22 mol ×342.3g1mol sugar= 418 g418g = 0.418kg[1kg =1000g]

Calculation of mass of sugar solution:

mass of solution =1000mL×1.12g1mL=1120 g1120 g = 1.120kg[1kg =1000g]

Calculation of molality of the solution:

Substitute the value of moles of sugar and amount of solvent (water) into molality formula to calculate molality of sugar solution.

molality =1.22 mol sugar(1.120-0.418)kg H2O= 1.74m

(b)

Interpretation Introduction

Interpretation: For the aqueous 0.87 M NaOH  solution molality has to be calculated.

Concept introduction:

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent. Molality is estimation of moles in relationship with solvent in the solution.

Molality (m) =Numberofmolesofsolute1kgofsolvent

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in 12g of 12C.

From given mass of substance moles could be calculated by using the following formula,

Molesofsubstance GivenmassofsubstanceMolecularmass

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The S.I. unit of mass is kg.

(b)

Expert Solution
Check Mark

Answer to Problem 13.17QP

Molality of 0.87 M NaOH  solution is 0.87 m.

Explanation of Solution

Given data: Strength of Sodium hydroxide solution = 0.87M

Density of Sodium hydroxide solution =1.04 g /mL

Calculation of mass ofNaOH:

Substitute the value of strength of NaOH and molecular mass of NaOH in the formula to calculate mass of NaOH.

Molecular weight of NaOH= 40g

mass of NaOH = 0.87 molNaOH ×40g NaOH1mol NaOH= 35g NaOH

Calculation of mass of H2Osolution:

mass of solvent =1040g - 35g =1005g =1.005kg1kg =1000g

Calculation of molality of the solution:

Substitute the value of moles of Sodium hydroxide and amount of solvent (water) into molality formula, to calculate molality.

molality =0.87 mol 1.005kg H2O= 0.87m

(c)

Interpretation Introduction

Interpretation: For the aqueous 5.24 M NaHCO3 solution molality has to be calculated.

Concept introduction:

Molality (m): Molality is the number of moles of solute present in one kilogram of solvent. Molality is estimation of moles in relationship with solvent in the solution.

Molality (m) =Numberofmolesofsolute1kgofsolvent

Moles: One mole is equivalent to the mass of the substance consists same number of units equal to the atoms present in 12g of 12C.

From given mass of substance moles could be calculated by using the following formula,

Molesofsubstance GivenmassofsubstanceMolecularmass

Mass: It is the quantitative measure of a substance. The amount of matter present in substance is expressed as mass. The S.I. unit of mass is kg.

(c)

Expert Solution
Check Mark

Answer to Problem 13.17QP

Molality of 5.24 M NaHCO3 solution is 6.99m.

Explanation of Solution

Given data: Strength of Sodium bicarbonate solution =5.24 M

Density of Sodium bicarbonate solution =1.19g /mL

Calculation of mass for NaHCO3 and solvent:

Substitute the value of molecular mass and strength of Sodium bicarbonate to calculate mass of NaHCO3.

mass of NaHCO3 = 5.24 mol ×84.01g1mol NaHCO3= 440g NaHCO3

mass of solvent (H2O) =1190g - 440g = 750g = 0.750kg1kg =1000g

Calculation of molality for given solution:

Substitute the value of moles of Sodium bicarbonate and amount of solvent (water) into molality formula to calculate molality of the given solution.

molality =5.24 mol NaHCO3 0.750 kg H2O= 6.99 m

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Chapter 13 Solutions

Package: General Chemistry with Connect 2-year Access Card

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