14. Solutions
Molality
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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? Alright. 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 gonna say here that our total number of ions, this breaks up into, 5 ions because it's ionic, it breaks up into 1 lead 4 ion plus 4 nitrate ions. Okay. So 5 total ions. So it's gonna 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 4 nitrate, so 0.400 moles of lead 4 nitrate divided by kilograms of solvent. Our solvent here is water and we're gonna say 750 grams is 0.75 0 kilograms. So here, we plug that in, that gives me 0.533 3 as my molality for solution. So plug that in here, and when we do that, we're gonna get our osmolality equal to 2.6665 modal. Here, this has 3 sig figs and 4 sig figs. So we'll want our answer to have 3 significant figures. So it comes out to 2.6 7 mole as our osmolality of the total number of ions.
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