Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781133949640
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
Question
Chapter 13, Problem 37PS

(a)

Interpretation Introduction

Interpretation: The freezing point of the given solution has to be determined.

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

  • Decrease in the vapor pressure
  • Increase in the boiling point
  • Decline in the freezing point
  • Osmotic pressure

Change in freezing point is calculated by using the equation,

  ΔTfp=Kfpmsolute

  where,

  Kfp is the molal freezing point elevation constant.

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

  Molality (m) =Numberofmolesofsolute1kgofsolvent

(a)

Expert Solution
Check Mark

Answer to Problem 37PS

  1. a) The freezing point of the solution is 0.3480C

Explanation of Solution

Given,

Mass of phenylalanine is 3%=3100×100g=3g

Mass of solvent is 100g3g=97g=0.097kg

Molar mass of phenylalanine is 165.19g/mol

Molal freezing point elevation constant of water is 1.860Cm1

  Numberofmole=GivenmassofthesubstanceMolarmass=3gphenylalanine165.19g/mol=0.018mol

Molality of the solute is calculated,

  Molality (m) =Numberofmolesofsolute1kgofsolvent=0.018160.097kg=0.187m

Hence,

Change in freezing point is,

  ΔTfp=Kfpmsolute=(1.860Cm1)×0.187m=0.34820C

Therefore,

The freezing point of the solution is 0.3480C

(b)

Interpretation Introduction

Interpretation: The boiling point of the given solution has to be determined.

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

  • Decrease in the vapor pressure
  • Increase in the boiling point
  • Decline in the freezing point
  • Osmotic pressure

Change in boiling point is calculated by using the equation,

  ΔTbp=Kbpmsolute

  where,

  Kbp is the molal boiling point elevation constant.

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

  Molality (m) =Numberofmolesofsolute1kgofsolvent

(b)

Expert Solution
Check Mark

Answer to Problem 37PS

  1. b) The boiling point of the solution is 100.09590C

Explanation of Solution

Given,

Mass of phenylalanine is 3%=3100×100g=3g

Mass of solvent is 100g3g=97g=0.097kg

Molar mass of phenylalanine is 165.19g/mol

Molal boiling point elevation constant of water is 0.51210Cm1

  Numberofmole=GivenmassofthesubstanceMolarmass=3gphenylalanine165.19g/mol=0.018mol

Molality of the solute is calculated,

  Molality (m) =Numberofmolesofsolute1kgofsolvent=0.018160.097kg=0.187m

Hence,

Change in boiling point is,

  ΔTbp=Kbpmsolute=0.51210Cm1×0.187m=0.095880C

Therefore,

The boiling point of the solution is,

  bp=1000C0.09590C=100.09590C

The boiling point of the solution is 100.09590C

(c)

Interpretation Introduction

Interpretation: The osmotic pressure of the given solution has to be determined. The values that are easily measureable in laboratory have to be identified among these values.

Concept introduction:

Colligative properties: Properties of solutions which having influence on the concentration of the solute in it. Colligative properties are,

  • Decrease in the vapor pressure
  • Increase in the boiling point
  • Decline in the freezing point
  • Osmotic pressure

Osmotic pressure is calculated by using the equation,

  π=cRT

  where,

  c is the molar concentration.

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

  Molality (m) =Numberofmolesofsolute1kgofsolvent

(c)

Expert Solution
Check Mark

Answer to Problem 37PS

  1. c) The osmotic pressure of the solution is 4.58atm

It is found that osmotic pressure can be easily measured experimentally in laboratories.

Explanation of Solution

Given,

Mass of phenylalanine is 3%=3100×100g=3g

Mass of solvent is 100g3g=97g=0.097kg

Molar mass of phenylalanine is 165.19g/mol

R=0.08206L.atm K1mol1T=250C=(25+273.15)K=298.15K

  Numberofmole=GivenmassofthesubstanceMolarmass=3gphenylalanine165.19g/mol=0.018mol

Molality of the solute is calculated,

  Molarity (M) =Numberofmolesofsolute1literofsolution=0.018160.1L=0.1816M

Therefore,

The osmotic pressure of the solution is,

  π=cRT=0.1816M×0.08206L.atm K1mol1×298.15K=4.58atm

The osmotic pressure of the solution is 4.58atm

  • Comparing to other colligative properties, osmotic pressure is the measurement with large values. Therefore, it can be easily measured experimentally in laboratories with less error.

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Chapter 13 Solutions

Chemistry & Chemical Reactivity

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Similar questions
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    Calculate the freezing point of 525 g of water that contains 25.0 g of NaCl. Assume i, the vant Hoff factor, is 1.85 for NaCl.
    (a) If you dissolve 10.0 g (about one heaping teaspoonful) of sugar (sucrose, C12H22O11) in a cup of water (250. g), what are the mole fraction, molality, and weight percent of sugar? (b) Seawater has a sodium ion concentration of 1.08 104 ppm. If the sodium is present in the form of dissolved sodium chloride, what mass of NaCl is in each liter of seawater? Seawater is denser than pure water because of dissolved salts. Its density is 1.05 g/mL
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