If the saturation concentration of potassium nitrate is #3.0*mol*L^-1#, is a solution prepared by dissolving a #67.05*g# mass of this salt in water to make a volume of #250*mL# solution, saturated?

1 Answer
anor277
Feb 12, 2018

#"Solubility"="Moles of solute"/"Volume of solution"#

Explanation:

And here..for the dissolution reaction....

#KNO_3(s) stackrel(H_2O)rarrK^(+) + NO_3^(-)#

#"solubility"="moles of solute"/"volume of solution"=((67.05*g)/(101.1*g*mol^-1))/(250*gxx10^-3*L*g^-1)=2.65*mol*L^-1#

And while including the units in the quotient is a major pfaff, it does give me a chance to review my reasoning and arithmetic. I got an answer in #mol*L^-1#..as I require for a solubility measurement. This persuades me that I got the order of operations right (for once!).

And since here the calculated solubility is LESS than #3.00*mol*L^-1-=3.00*mol*dm^-3# #(1*L-=1*dm^3)#, the solution is #"UNSATURATED"#.

By the way what does #"saturation"# mean in this context? Does it represent the maximum concentration of solute that the solvent could hold?

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