Question

25.0 mL of a KOH solution (D = 1.46 g/mL and 11.7 M KOH) was added to 25.0 mL of a NaOH solution (D = 1.54 g/mL and 50.5 % NaOH by mass). Determine the molality (in terms of total solute) of the final solution.

Answer 1

Mass of KOH solution = volume x density = 25 x 1.46 = 36.5 g

Moles of KOH = M xV = 11.7 x ( 25/1000) = 0.2925

mass of KOh = moles x molar mass = 0.2925 x 56.1 = 16.4 g

solvent mass = 36.5-16.4 = 20.1 g

Mass of NaOH solution = 25 x 1.54 = 38.5 g

NaOH is 50.5 % by mass , hence NaOH mass = ( 38.5 x 50.5/100) = 19.4425 g

solvent mass = 38.5-19.4425 =19 g

NaOH moles = mass of NaoH / Molar mass of NaoH = 19.4425/40 = 0.486

solution mass = 20.1+19 = 39.1 g = 0.0391 Kg

Molality of total solute = ( total moles / solvent mass in kg)

                  = ( 0.2925+0.486) / ( 0.0391) = 19.9