Question

What is the `"molality"` of a `10*g` mass of salt, whose formula mass was `36.46*g*mol^-1`, that is dissolved in a `100*g` mass of solvent?

Answer 1

`"Molality"<3*mol*kg^-1.`

Explanation:

By definition, `"Molality"="Moles of solute"/"Kilograms of solvent"`.

Here you have a 10% solution, and here I ASSUME that you mean

`w/wxx100%`, i.e. `"Mass of solute"/"Mass of solvent"xx100%`

And so we have a `10*g` mass of solute, in a `100*g` mass of solvent (which is reasonably assumed to have a `100*mL` volume; this is another unspoken assumption; the solvent is clearly water).

And finally `"molality"-=((10*g)/(36.46*g*mol^-1))/(0.100*kg)=2.74*mol*kg^-1`.

Note that at these concentrations, `"molality"` is equivalent to `"molarity"`.