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?

1 Answer
Jul 26, 2017

Answer:

#"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"#.