Question

What is the mole fraction, X, of solute and the molality, m (or b), for an aqueous solution that is 16.0% NaOH by mass?

Please show work to help me learn how to work it out! Thanks

Answer 1

16.0% NaOH by mass means 16gm of NaOH present in 100gm of solution

weight of solvent (water ) = weight of solution -weight of solute(NaOH)

                                   = 100-16 = 84gm

no of moles of NaOH(n1) = 16/40 = 0.4 moles

no of moles of water(n2) = 84/18 = 4.7 moles

mole fraction of NaOH (X1) = n1/n1+n2

                                         = 0.4/0.4+4.7

                                        = 0.4/5.1 = 0.078

mole fraction of water X2 = n2/n1+n2

                                     = 4.7/0.4+4.7

                                     = 4.7/5.1

                                    = 0.92

molality     = weight of solute*1000/gram molar mass of solute*weight of solvent in gm

             = 16*1000/40*84 = 4.76m