Asked by CHAIMA
4. A 500 g of ammonium chloride solution is 3 % by mass with a density of 1.024 g/mL. Calculate molarity and molality of a solution.
Answers
Answered by
DrBob222
3% by mass means 3 g NH4Cl in 100 g solution or 15 g NH4Cl in 500 g solution. With a density of 1.024 g/mL that 500 g will have a volume = mass/density = 500/1.024 = 488 mL.
mols NH4Cl in 15.0 g = g/molar mass = 15.0/53.5 = 0.28
So M = mols/L = 0.28 mols/488 mL = ? M
m = molality = mols/kg solvent = 0.28/0.500 = ? m.
mols NH4Cl in 15.0 g = g/molar mass = 15.0/53.5 = 0.28
So M = mols/L = 0.28 mols/488 mL = ? M
m = molality = mols/kg solvent = 0.28/0.500 = ? m.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.