#"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?