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Archived Calculating molality given density and molarity

  • Thread starter LakeMountD
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I haven't taken chemistry in over 3 years so I have forgotten a lot of the very basic concepts (even while understanding the higher level stuff) and am having problems with this simple thing! Please help :(.

Question 1- At 20°C, a 2.32 M aqueous solution of ammonium chloride has a density of 1.0344 g/mL What is the molality of ammonium chloride in the solution? The formula weight of NH4Cl is 53.5 g/mol.

I don't think I have to use temperature here. I know how to calculate molality when im given how many grams are in the solution but the way this one is worded im confused. Not looking for the answer, just a little help on where to start.
 
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Alright, molarity is defined as moles of solute per liters of solution and molality is defined as moles of solute per kilograms of solvent. That said, we must define a basis to work with, let's say we have 1 liter of solution, and because we know the molarity of this solution is 2.32 M, then we know there's 2.32 moles of NH4Cl in the solution. And we know 2.32 moles of NH4Cl are equal to 124.12 grams. Next, we have the density of the solution, which is defined as mass of solution per volume of solution. So, 1 liter of solution contains 1034.4 grams, of which 124.12 grams are of NH4Cl, then we have 910.28 grams of solvent. Now we can calculate the molality of the solution
[tex]\textrm{molality} = \frac{2.32 \ mol}{0.91028 \ kg \ water} = 2.549 \ m[/tex]
 

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