What is the osmolality of total ions in an aqueous solution prepared by dissolving 0.400 moles of lead(IV) nitrate in 750 grams of water? Alright. So, we're going to say here that our total ionic molality equals the number of ions times the molality of the solution. So, we're going to say here that our total number of ions, this breaks up into 5 ions because it's ionic. It breaks up into 1 lead(IV) ion plus 4 nitrate ions.

Okay. So, 5 total ions. So, it's going to be 5 times. Now we need the molality of the solution. Remember, molality equals moles of solute, which in this case would be the lead(IV) nitrate.

So, 0.400 moles of lead(IV) nitrate divided by kilograms of solvent. Our solvent here is water, and we're going to say 750 grams is 0.750 kilograms. So, here, we plug that in, that gives me 0.5333 as my molality for the solution. So, plug that in here, and when we do that we're going to get our osmolality equal to 2.6665 molal. Here, this has 3 significant figures and 4 significant figures, so we'll want our answer to have 3 significant figures.

So, it comes out to 2.67 molal as our osmolality of the total number of ions.