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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