Molality Example 2

Jules Bruno
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What is the osmolarity of total ions in an equal solution prepared by dissolving .400 moles of lead for nitrate in 750 g of water. All right, so we're going to say here that are total ionic morality equals number of ions, times the morality 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 for ion plus four nitrate ions. Okay, so five total ions, So it's gonna be five times now. We need the morality of the solution. Remember, morality equals moles of solute, which in this case, would be the lead for nitrate. So .400 moles of lead for nitrate Divided by kg of solvent are solvent ears, water and we're gonna say 750 g is .750 kg. So here we plug that in. That gives me .5333 as my morality for a solution. So plug that in here. And when we do that, we're gonna get our osmolarity equal to 2.6665 molo here This has three sig figs and four sig figs. So we want to answer to have three significant figures. So it comes out to 2.67 mol as our Oz modality of the total number of ions.
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