Molality(m) = moles solute/Kilograms Solvent
For a 10.0% w/w aqueous solution of EtOH(f.wt. = 46g/mol) => 10gms EtOH/100gms Soln. Using this ratio as solute-in-solution base solution, then the data needed for molality would be 10gms EtOH in (100-90)gms solvent => molality = ((10/46)mole EtOH)/0.09Kg Solvent = 2.42 molal in EtOH...
Molarity(M) = Moles Solute/Liter Solution
For the same 10% w/w EtOH solution => 10gm EtOH/100gm Solution:
=> [EtOH]=[(10/46)mole EtOH]/[(100gm soln/0.984g/ml)/(1000ml/L)]
=(0.2174 mole EtOH/0.1016L soln)
=2.10 Molar in EtOH
You probably already know this, but molality(m) is used in solution type problems where the solution goes through a temperature change history, whereas, Molarity(M) based solutions are bound to a constant temperature. Since Molarity = moles solute/Liters of Solution, heating or cooling a 'Molar Solution' changes the solute concentration b/c volume expands or contracts changing the concentration. The value of Molality(m) = (moles solute/Kg Solvent) is not the same if the temperature of the solution changes.Boiling point elevation and Freezing Pt depression type problems are molality type cases in point. Just a polite FYI if you needed it. Enjoy.
the density of an aqueous solution containing 10.0% by mass ethanol is 0.984 g/ml. calculate molality and molarity. what volume of the solution would contain 1.15 g ethanol ?
2 answers
Correction... Molality(m) remains the same if temp is changed... (next to last sentence). remove the 'not'. If there's an edit feature on this site, I don't know how to use it. sorry bout that.
2 answers