Question

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`?

Answer 1

`"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.