 I have this electrolytic cell here. So, there is a power source here, let us say a battery and we have two plates dipped in some electrolytic solution and let us say the length between both of these plates is L and the area of each of these plates is A and we know that the circuit here is complete. So, if we think of some electrons leaving from this negative terminal we know that they are going back and completing the circuit. So, then how is the conduction taking place in the solution? How can we describe that? So, we have seen something very similar when we have looked at resistance and resistivity before. Let us go back to that idea. So, if we have this metallic wire with a cross section area A and a length L, we know that in this case the resistance R is defined as rho times L by A where R is the resistance and rho is the resistivity. So, if I just rearrange this I can write it as 1 by R is equal to 1 by rho times A by L. I have just taken each of these terms to the other side of the equal to sign. So, since R was our resistance I can write 1 by R or the inverse of resistance as conductance and let us say I denote this by the symbol G and similarly for this 1 by rho rho was resistivity. So, I am going to call 1 by rho that is the inverse of resistivity as conductivity. So, if this conductivity is denoted by this kappa we can write the conductance G is equal to kappa times A by L. Now, in the case of this wire A was the cross section area and L was the length of this wire. But what will be the area and the length for our electrolytic cell? You can probably guess what it is going to be. So, the length L is the distance between both of these plates and the area A is the common area in the solution between these two plates. So, you can see that both of these plates are dipped in the solution and both of them each have a area which is equal to A. But the important point is that this A is the common area or the effective area. So, if I pull this plate that is that only half of it is dipped in the solution the effective area will now become A by 2 and if I keep pulling this plate out so that it is completely out of the solution the circuit breaks and we do not have any conduction. So, the A here is the common area of both of these plates and for simplicity I have taken both of them to be equal. So, I know that the common area is A. Now, if we look at the units of the terms in this equation G which is the conductance is 1 over resistance and we know that resistance is measured in ohms. So, the units of G will be ohm inverse and similarly if you look at the resistivity because this is length so the units will be meters and area will be meter square. The SI unit of resistivity will be ohm meters. So, conductivity or kappa which is inverse of resistivity will have units of ohm inverse meter inverse. Now, in SI units this ohm inverse is referred to the unit Siemens which is denoted by S. So, we can write the units of conductance as Siemens and the units of conductivity as Siemens per meter. Now, let's say you want to find the conductance of different solutions. So, we use the same cell we just drain out the solution and use a different one. So, in that case the length and the area will remain the same because we are using the same cell. So, sometimes in problems this ratio of L by A is given as cell constant which is denoted by this G star. So, if you are using the same cell the length between these plates and the common area is going to be the same. So, for that particular cell the cell constant will be defined as L over A and so we can rewrite this equation as G times G star is equal to kappa where G is the conductance in Siemens and G star or the cell constant has a unit of 1 over meter or meter inverse which is the same as we saw before. So, kappa was ohm inverse meter inverse ohm inverse can be called Siemens and kappa here is Siemens per meter. So, when we started out we wanted to better understand the flow of electrons in this electrolytic solution and to do that we used what we knew from conduction in a metallic wire and we drew some parallels and based on a very similar idea we came to these equations. But if we go back to this wire in this case when we write the relationship between resistance and resistivity, resistivity is a fundamental property of the material. So, let's say if I have a copper wire I know that the resistivity of copper is some value and if instead of copper let's say I want to use a silver wire or a gold wire I can just compare these values of resistivity to find out which one will be a better conductor. But this was in the case of a metallic wire for the electrolytic cell if we use this form of the equation and we know that the L by A or the cell constant is a fixed value and we use this kappa or conductivity in the same way to understand how the conduction changes when we change the solution and the key word here is solution because unlike the case of this solid metallic wire if we are comparing solutions we need to think about how much of the solute are we adding and into how much of the solvent so we probably need to think in terms of concentration. Let's see how we can do that. So, we looked at this way of defining conductivity before. So, let's say I want to standardize this conductivity in some way and the idea is if you look at this cell constant which is G star we saw how the cell constant was equal to L over A where L is this distance between the plates and A is the common area of these plates. So, what if I take this common area to be 1 square centimeter and the length between the plates to be 1 centimeter. So, essentially what I am doing is if I think of this cube I am saying that this face has an area of 1 centimeter square and this length is 1 centimeter. So, the volume of this cube will be 1 centimeter cube which can also be written as 1 cc or 1 ml. So, if I take the plates of areas 1 centimeter square and if I keep them 1 centimeter apart effectively the volume inside this space is equal to 1 ml but the problem here is that when I talk of the conductivity of this unit volume I'm not including the ions anywhere. So, unlike the case of the metallic wire it's the ions here that actually take part in the charge transfer and where are the ions coming from. The ions are determined by how much of the solute you added in this solvent to get this solution and so it would be useful to rethink this value of conductivity on a per mole basis. So, let's define a new term called molar conductivity and we're going to define it as this kappa divided by molar concentration or molarity c. If the symbol looks new to you this is actually the upper case lambda. So, the other lambda that you would have seen is small case lambda which is written like this. This is the upper case lambda that we are using to define this quantity called molar conductivity. I know what you're thinking, why do we have to define this molar quantity? Can we not directly use this value of kappa from here? So, the problem with looking at this kappa for solutions is that if we look at this unit volume this is the unit volume of the solution. So, when you talk of a solution you will have some fixed amount of solvent and some amount of solute and the amount of solute is what is determining the amount of ions in the solution and the transfer of charges and therefore the conductivity of this solution will depend on how many ions are there in the solution and so we use the molar concentration or the molarity to take this into account. Now, if we look at the units of molar conductivity kappa in SI units is given as C-minus per meter and concentration can be written in the form of mol per meter cube. So, if we cancel out this term the units of molar conductivity then become C-minus per mol per meter square but usually the kappa is given in C-minus per centimeter and the concentration is given in moles per liter. So, let's adjust these units here since such a way that if we are given this concentration in mol per liter and the kappa in C-minus per centimeter we can directly plug them in and not worry about adjusting these units and we can work our way backwards. We know that one centimeter cube is the same as one ml. So, I can write this as 1 instead of centimeters I can write it as 10 to the power minus 2 meters and this I can write as 1 times 10 to the power minus 3 liters. So, I have 10 to the power minus 6 meter cube is equal to 10 to the power minus 3 liters. I can cancel this out on both sides and so I can write 10 to the power minus 3 meter cube is equal to 1 liter. Now, if we take a reciprocal on both sides we can write this as 10 to the power 3 per meter cube is equal to per liter. So, we can already see that we are getting there. We need a mol per liter on the right hand side. So, we can write this as 10 to the power 3 mol per meter cube is equal to 1 mol per liter. Now, if we take this 10 to the power 3 to the other side, we have 1 mol per meter cube equal to 10 to the power minus 3 mol per liter. Now, why we did all of this unit conversion exercise is because we want to be able to plug in this mol per meter cube here to get the molar conductivity in CGS units. So, if we do that we can write the molar conductivity lambda m as kappa in Siemens per centimeter divided by the concentration given in moles per liter times 1000 and the 1000 here is when you take this 10 to the power minus 3 from the denominator to the numerator. So, if a numerator is equal to 1, then we can write this as 10 to the power minus 3 per liter. So, we can write this as 10 to the power minus 3 times C in mol per liter. Now, I can just take this 10 to the power minus 3 above. So, I can write this as molar conductivity is kappa which is expressed in Siemens per centimeter divided by the concentration given in moles per liter If in numerical problems you are given the kappa value in these units and the concentration in mol per liter you can directly plug them in into this formula to get the molar conductivity in Siemens centimeter square per mol.