Question

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

Answer 1

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?