Question

What is the mole fraction, X, of solute and the molality, m (or b), for an aqueous solution that is 19.0% NaOH by mass?

Answer 1

mass of NaOH = 19 g

mass of solvent = 81 g = 0.081 kg

moles of solvent = 81 / 18 = 4.5

moles of NaOH = 19 / 40 = 0.475

total moles = 4.5 + 0.475 = 4.975

mole fraction of NaOH = moles of NaOH / total moles = 0.475 / 4.975

                                     = 0.0955

molality = moles of solute / mass of solvent = 0.475 / 0.081

              = 5.86 m

molality = 5.86m