CHM 152
posted by Car on .
At 20oC, a 2.32 M aqueous solution of Ammonium Chloride has a density of 1.0344g/ml. WHat is the molality of Ammonium Chloride in the solution? The formula weight of NH4Cl is 53.50 g/mol??
I am not sure how to incorporate the density into the problem to come up with how to start. This is a sample of a question that will appear on CHM 152 test I have on MOnday. I just need to know where to begin in my steps please......HELP!!!!!

2.32 M = 2.32 moles NH4Cl/L solution.
2.32 moles NH4Cl = 2.32 x (53.5g/mol) = 124.12 g NH4Cl
mass of 1000 mL of the solution is 1000 x 1.0344 g/mL = 1034.4 g and 123.12 g is NH4Cl; therefore, the amount of water is
1034.4123.12 = 910.28 g H2O or 0.91028 kg solvent.
Then 2.32 moles NH4Cl/0.91028 kg solvent =2.549 m which rounds to 2.55 m.
Check my thinking carefully, 
A solution is prepared by dissolving 23.7 g of Ca in 375 g of water. The density of the resulting solution is
density of 1.05 g/ml