Now the van Hoff factor which uses the variable I equals the number of ions produced from dissolving a soluble solute. Now when we talk about solutes, we group them as being either ionic or covalent in nature. Now remember ionic compounds are composed of a positive ion connected to a negative ion. That positive ion can be in the form of the ammonium ion or a metal, and then that negative ion will be in the form of nonmetal.
So here, if we take a look, we have sodium hydroxide, ammonium carbonate, and aluminum sulfate as our three ionic compounds. Each one, because they are ionic, can break up into ions. Here we'd have Na+ OH-. Here it breaks up into two ions. We just said that the Van Hoff factor is a number of ions produced when a soluble solute dissolves. So since there's two ions I = 2, ammonium carbonate breaks up into two ammonium ions plus one carbonate ion, for a total of 3 ions. So here I = 3. Aluminum sulfate breaks up into two aluminum ions and three sulfate ions, for a total of 5 ions, and because of that I = 5.
Now, covalent compounds are just compounds composed of only nonmetals together. OK, we're going to say here that they are. Because of this they are non volatile and you're going to say they are non ionizable or non electrolytes. OK, so here if they mention covalent solutes, if they mention that they're non volatile, if they mention that they're non electrolytes, we group them all under covalent solutes. So here we have glucose, we have chlorine, we have methanol, and here we have something called urea. So urine, that's one of the main components of it.
All of these are covalent in nature and therefore they are non volatile, which means they don't break up into ions and they're not electrolytes, which also means they don't break up into ions. Because of that, I = 1. Now technically 0 ions are formed, so they just stay in the form that they're in, so that still counts. They stay in the form that they're in, so we just count them as one. So I for them would equal to 1. So remember ionic compounds break up into ions. It's important you get the number correct to write to find the right number for I. For covalent solutes they don't break up into ions, their I will always just be one in its value.