<p>Don't bother unless you do chemistry, otherwise this post makes little sense. My chem class last year only touched on molality, but in the McGraw Hill book there's a bit of a contradiction.</p>
<p>When it calculates molality for an ionic substance, in one example it factors in all atoms in (AlCl(3)), and considers that to be 4 particles (so if there's one mol of AlCl(3), it's multiplied by 4 to give 4 mols when calculating molality). In another problem that is worked involving CaCl(2), however, where there are 3 atoms, it calculates the molality (which it later uses to calculate the boiling point increase of the water) without multiplying the initial molar value by the 3 CaCl(2)particles. Simply put, it's calculating moles for all four atoms in AlCl(3), but for CaCl(2), it only counts the mols of the molecule, not all four atoms when calculating molality. Which are you supposed to use?</p>
<p>maybe the context in the two situations was different:
the first one could've been asking for moles of ions in the solvent
while the other one could've been asking for moles of solute in the solvent</p>
<p>good luck on Chem, i'll be taking it too</p>
<p>For both questions, it was calculating the same thing, the temperature change based on molality, yet both situations factored in particles in different ways to determine the mols of solute involved . . .</p>
<p>arghh! now, i'm scared too.
Could you type up the Q?
Are we talking about freezing point depression here?</p>
<p>I'm pretty sure you have to multiply it by the number of particles, like in the first problem.</p>
<p>k</p>
<p>What matters is that I know what particles to use when calculating the molality to determine the factor by which there's an increase or depression in boiling or freezing point--they're derived by the same process. (but it's about increasing boiling points, to answer your question).</p>
<p>Yes, molality is total moles of solute particles per kilogram of solvent.</p>