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diff --git a/827/CH9/EX9.1.a/9_1a.sce b/827/CH9/EX9.1.a/9_1a.sce
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+// CHEMISTRY FOR ENVIRONMENTAL AND ENGINEERING SCIENCE FIFTH EDITION
+disp(' Chapter 9: Introduction to Water and Wastewater analysis')
+
+//To calcluate the TCE concentration 
+
+disp('What is the TCE concentration in g/m^3 for each of the following')
+
+disp('example 9.1 a')
+ 
+
+disp( 'A soil having a density of of 2 g/cm^3 with TCE concentration of 4 ppm')
+
+
+
+// Since density is given in cm^3 it has to be converted into m^3 
+
+density=2*1000
+
+//Since 1ppm= 1/10^6 g, we get
+
+ppm = 4/10^6
+
+disp(' TCE concentration in g/cm^3 is')
+
+TCE= ppm*density
+
+disp(TCE)
+
diff --git a/827/CH9/EX9.1.b/9_1b.sce b/827/CH9/EX9.1.b/9_1b.sce
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+// CHEMISTRY FOR ENVIRONMENTAL AND ENGINEERING SCIENCE FIFTH EDITION
+disp(' Chapter 9: Introduction to Water and Wastewater analysis')
+
+//To calcluate the TCE concentration 
+
+disp('What is the TCE concentration in g/m^3 for each of the following')
+ 
+disp('example 9.1 b') 
+
+disp(' An air sample at 20 degree celsius and 1 atm pressure with TCE concentration of 4 ppm')
+
+
+// Since in air 4ppm equals 4mL gaseous TCE/10^6 mL air 
+
+TCE = 4/10^6
+Giventemp= 20
+
+// here the given temperature is in degree celsius and has to be converted into kelvin in standard terms
+
+ Actualtemp= 273+ Giventemp
+
+// Since at standard temperature and pressure the volume is 22.4L 
+
+disp('The answer is')
+
+ans = (TCE*273*131.5*10^3)/( Actualtemp*22.4)
+
+disp(ans)
+
+//We see that the concentrations in g/m^3 are quite different for the soil and air samples, even thought the concentrations when expressed in ppm are the same.This illustrates that we must understand well the general basis for mass expressions in different media
+