Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 13, Problem 116QRT
Interpretation Introduction

Interpretation:

The boiling point, the freezing point and osmotic pressure of NaCl solution have to be calculated.

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 at temperature 100oC.

  ΔTb= Tb- Tb°.

Where,

  ΔTb Change in boiling point.

  Tb Boiling point of the solution.

  Tb° Boiling point of pure solvent.

TheBoiling point elevation(ΔTb) is indicated as 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.

TheFreezing point depression(ΔTf) is indicated as ΔTf= Kfm

Where,

ΔTf Change in freezing point.

Kf Molal freezing point constant.

m Molality of the solution.

Expert Solution & Answer
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Answer to Problem 116QRT

The Boiling point of the solution is 100.0957oC. and the freezing point of the solution is -0.348oC and the osmotic pressure is 4.62atm.

Explanation of Solution

Given data as follows:

Mole of NaCl= 0.100mole; molality of NaCl solution = 0.100mol/kg;

‘i’ is = 1.87.

Mass of water = 1 kg.

Value of Kb=0.512oC kg /mol.

Value of Kf=1.86oC kg/mol.

Calculate Freezing point of the solution:

Substituting the values of molality, KfandKb the freezing point was calculated as shown below,

    ΔT= Kfm =(-1.86°C kg/mol)(= 0.100mol/kg)(1.87) = -0.348CΔT= T°-T -T=ΔTT°.

    T=T°ΔT=0°C +(-0.348C)=-0.348oC.

Calculate Boiling point of the solution:

Substituting the values of molality, KfandKb the boiling point was calculated as shown below,

  ΔT= Kbm = (0.512°C kg/mol)(0.100m) (1.87)= 0.0957oCΔTb= Tb- Tb°.

  Tb(solution)=Tb°(solvent)+ΔTb=100+0.0957oC=100.0957oC.

Therefore, the Boiling point of the solution is 100.0957oC. and the freezing point of the solution is -0.348oC.

Osmotic pressure calculation:

The molarity is calculated as follows:

  Molarity=MolesofsoluteVolume of solution in L=(0.100mol NaCl1kgsolvent)(1kgsolvent1kgsolution)(1.01 kg solution1L solution)=0.101MNaCl.

Hence, molarity obtained is 0.101M.

Temperature = 25oC+273=298K.

The formula of osmotic pressure is as shown below,

Osmotic pressure (π) = iMRT.

  (π) = iMRT=(1.87)(0.101M)(0.08206L.atmmol.K)(298K)=4.62 atm.

Therefore, the osmotic pressure of solution is 4.62atm.

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Chapter 13 Solutions

Chemistry: The Molecular Science

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