What are the #"molarities"# and #"molalities"# of a #180xx10^3*g# mass of glucose dissolved in #1*kg# of water?

1 Answer
Jul 26, 2017

Answer:

Unreasonably high.......

Explanation:

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

While we can find the molar quantity, i.e. #"moles of solute"=(180xx10^3*g)/(180.16*g*mol^-1)=1000*mol#, the presumption that such a quantity would dissolve in approx. one litre of solution is unreasonable.

If the question specified a #180.16*g# quantity, then I doubt that the volume of the aqueous solution would change much......

And so #"molality"=((180.16*g)/(180.16*g*mol^-1))/(1*kg*"water")=1*mol*kg^-1#. Would this solution be denser or less dense than pure water?