Asked by Ambachew Abata
Commercial Concentrated solution of ammonia and a density of 0.898 g/ml a) what is the percent of ammonia in such solution? b) what is the molality
Answers
Answered by
DrBob222
I don't think you have enough information given. However, I remember that commercial NH3 is 15.3 M so let x = % NH3, then
0.898 g/mL * 1000 mL * (x) * (1/17) = 15.3*100
Solve for %.
molality = m = mols/Kg solvent
0.898 g/mL * 1000 mL * (x) * (1/17) = 15.3*100
Solve for %.
molality = m = mols/Kg solvent
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