Hello, dear friends, we are studying the chapter, Solution and Colligative Properties. So in this module I will take you through the next colligative property which is elevation in boiling point, so let us see that.

So when I say elevation in boiling point, the first thing we need to know is what is boiling point. It is the temperature at which the vapour pressure of the liquid becomes equal to the external pressure. So that is what is boiling point.

Next, on addition of a non volatile solute the vapour pressure of the solvent will always decrease. That is what we learnt from the colligative properties definition itself earlier and which is in fact giving us the first one which is relative lowering of vapour pressure, right.

So, next thing, therefore to boil the solution, the required temperature will always be higher so it will take more amount of temperature or more amount of energy to boil the same solution. So let me show you this more graphically, instead of reading it out more like a point wise, let me show you the same thing graphically here. So we know that at boiling point vapour pressure of the solution will be equal to external pressure so taking that into consideration let us start with the graph here.

So, let me draw the graph between vapour pressure versus temperature, right. So now closely observe the vapour pressure line, I have drawn a line here which is the P external that is 1 Atm. Now for volatile solvent at equilibrium we know that solvent and solvent which is at liquid and which is at vapour, there is always an equilibrium established and we can use Clausius-Clapeyron Equation which is lnP is equal to delta H vap by RT. So therefore lnP versus T you can see how it is going to be. It is more like a logarithmic graph versus 1 by T. So, if at all I draw that between vapour pressure and T, the graph will look something like this. So this is that solvent graph, got it. So using that equation which we just mentioned.

Now observe it very carefully, if at all I draw this particular line and it hits at that point X axis and that point Tb0 is the boiling point of the solvent. And as I said earlier the first point, so vapour pressure of the liquid has just become equal to external pressure and that’s why it is called the boiling point.

Now, the next point is now on adding the non volatile solute I told you that the vapour pressure will decrease. Now if the vapour pressure will decrease we learnt this earlier the curve was supposed to shift onto this side, so now the curve will look more like this for solution, got it. Now observe again carefully, if at all I draw this line it will touch at Tb. Now this is a temperature which is higher, right. So to reach that 1 Atm which is the external pressure I am requiring more temperature here. So the temperature will be higher, so that is what I said. Now there is a certain difference which is established and that difference, delta Tb is what is the elevation in the boiling point, and therefore this point becomes the boiling point of solution. So to reach that 1 Atm, so we need Tb. So what I can conclude here that Delta Tb is equal to Tb minus Tb0 and that is what we call it as the elevation in boiling point.

So we will see how we can write it in the mathematical way. So elevation in the boiling point is directly related to molality, so I can write delta Tb is directly proportional to molality and molality will always have the unit of mol per kg. So therefore I can write delta Tb is equal to some constant Kb into m. And this constant Kb is called Ebullioscopic constant or Molal Elevation constant. Please remember very, very important constant using problem solving, got it.

So now, m depends upon number of mols of solute particles, that is more the number of solute higher will be the elevation in the boiling point, right, because it is directly related. So therefore, if m is equal to 1 I can say delta Tb will be equal to Kb. And units of Kb will be Kb is equal to delta Tb by m, so therefore unit will be Kelvin kg per mol or degree centigrade kg per mol. So units are also very important, students, so please remember that carefully, it is Kelvin kg per mol.

And Kb for water usually it is always given in the problems but still it is better to remember, it is 0.52 Kelvin kg per mol.

Now, let me take you ahead in determining Kb and this can be derived from enthalpy and boiling point. Now in this equation delta Tb is equal to Kb into m. We know that m is responsible for the property of solute, while Kb will be responsible for the property of the pure solvent. So that means Kb is all related with the solvent properties. So therefore, Kb is given as R Tb square M by delta H vap. What is this Tb, M and delta H vap, well these are all values for the solvent only. So what are those, let me write down that.

So R is the gas constant, we always take it as 8.314 joules per mol per Kelvin.

Tb is the boiling point temperature in Kelvin.

M is the molar mass in kg per mol.

Again please remember we have to write that in kg per mol and this is for solvent, not for solution.

And then delta H vaporization which is the enthalpy of vaporization in joules per mol or cal per mol as per whatever be the values of Kb.

So this is how we have to calculate Kb.

Kb can also be represented in another form which is R Tb square by 1000 into Lv. Now what is this Lv, we know that R is gas constant, Tb is boiling point, M is molar mass, but Lv is the latent heat of vaporization which is usually represented in joules per gram. So either the question might give you delta H vaporization or Lv. Using any of these we have to calculate Kb and accordingly we have to go for further applications.

As we go for the next modules we will see the applications of elevation in boiling point.

Thank you