What are the molarity AND molality of an ammoniacal solution composed of a mass of #30.6*g# with respect to ammonia, and #81.3*g# with respect to water, to give a solution whose density is #0.982*g*mL^-1#?

1 Answer
anor277
Apr 20, 2017

#"Molarity"="Moles of solute"/"Volume of solution"#.........

Explanation:

And thus..........................

#"Molarity"=((30.6*g)/(17.01*g*mol^-1))/((30.6+81.3*g)/(0.982*g*mL^-1))=(1.80*mol)/(0.114*L)=15.8*mol*L^-1#.

And #"Molality"="Moles of solute"/"Kilograms of solvent"#

#=((30.6*g)/(17.01*g*mol^-1))/(81.3xx10^-3*kg)# #=(1.80*mol)/(0.114*L)=22.1*mol*kg^-1#.

I am not overly happy with this question, because an ammoniacal solution of this concentration is almost unreasonable; open this in a lab and the #"pen and ink"# would clear the lab pdq.

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