Question

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

Answer 1

`"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.