So, here it says what is the ionic molality of nitrate ions in 0.305 molal of lead(IV) nitrate? Alright, so here we're just looking for the molality of our nitrate ions. Here, lead(IV) nitrate breaks up into 1 lead(IV) ion plus 4 nitrate ions. So, the ionic molality or osmolality would equal the 4 nitrate ions times the molality of the entire compound. So that comes out to being 1.22 molal. So just remember, with ionic molality, they can ask you for the total ionic molality which takes into account all the ions, so that would be 5 ions total times the molality of the compound or they can ask for the ionic molality of individual ions. In this case, there are 4 nitrate ions, so we'd multiply the total molality by 4 to get our final answer.

Now that we've seen examples 1 and 2, tackle the practice question left on the bottom of the page.