What is the #"molality"# of a #0.030*g# mass of sugar dissolved in #300*mL# of water?

1 Answer
May 3, 2017

Answer:

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

Explanation:

And so #"Molality"=((0.03*g)/(180.16*g*mol^-1))/(0.3*kg)=??mol*kg^-1#.

At these concentrations the #"molality"# would be identical to the #"molarity"#.

Note that #1*dm^3# is a fancy way of saying #1*L#, #1*dm^3# #=# #"1 decimetre"#, where the prefix #"deci"-=10^-1#.

And thus #1*dm^3"-=(10^-1*m)^3=10^-3*m^3=1*L#, because there are #1000*L*m^-3#. A #"cubic metre"# is a huge volume.