   Chapter 13.5, Problem 3.3ACP

Chapter
Section
Textbook Problem

A 1.0 L sample of water is shaken in air to allow it to become saturated with both N2 and O2. What is the total amount of dissolved gases? If the solution was now heated and the dissolved gases expelled from solution and collected, what is the mole fraction of oxygen in the gas mixture?

Interpretation Introduction

Interpretation: The total amount of nitrogen and oxygen in water and the mole fraction of oxygen in the given gas mixture should be determined.

Concept introduction:

• Mole fraction is one of the concentrations expressing term which is equal to the number of moles of a component divided by the total number of a solution. The mole fraction of all components of a solution, when added together will equal to 1

The mole fraction of component a can be determined by using the equation,

Molefractionofa(Xa)=MoleofaTotalmoles

• Raoult’s law: the vapor pressure of a solvent above a solution (Psolution)is equal to the vapor pressure of the pure solvent(P0solvent ) at the same temperature scaled by the mole fraction of the solvent (XSolvent )present.

Psolution=Xsolvent×P0solvent

• Density=MassVolume
• Henry’s law: The solubility of a gas in liquid is directly proportional to the gas pressure.

Sg= kHPg,where,Sgisthegassolubility(inmol/kg)kHis Henry's constant Pg is the partial pressure of the gaseous solute.

• Pressureinatm=pressureinpascal×1atm101325Pascal
Explanation

Partial pressure of the two gases namely oxygen and nitrogen can be determined by using the Raoult’s law and the solubility of these gases in water can be determined by using Henry’s law as follows,

PartialpressureofN2 =XN2×(Ptotal) =0.7808(1.0atm)=0.781atmSolubility of N2 =kHPg =(6.0×10-4mol/kg.bar)(0.781atm)(1bar/0.98692atm) =4.75×10-4mol/kgsimilarly,PartialpressureofO2 =XO2×(Ptotal) =0.209(1.0atm)=0.209atmSolubility of O2 =kHPg =(1.3×10-3mol/kg

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