What is the osmolality of total ions in an aqueous solution prepared by dissolving 0.400 moles of lead 4 nitrate in 750 grams of water? All right, so we're going to say here that our total ionic molality equals 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 five ions. Because it's ionic, it breaks up into one lead 4 ion plus 4 nitrate ions. OK, 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 for nitrate.
So 0.400 moles of lead, 4 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 solution.
So plug that in here, and when we do that, we're going to get our osmolality equal to 2.6665 mol here. This has three sig figs and four sig figs. So we'll want to answer to have three significant figures. So it comes out to 2.67 mole as our osmolality of the total number of ions.
