Archived Calculating molality given density and molarity

[LakeMountD]

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 I am given how many grams are in the solution but the way this one is worded I am confused. Not looking for the answer, just a little help on where to start.

 

[MexChemE]

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
\textrm{molality} = \frac{2.32 \ mol}{0.91028 \ kg \ water} = 2.549 \ m