What is the mole fraction, x, of solute and the molality for an aqueous solution that is 16.0 % naoh by mass

Step 1) identify the solvent and solute
      solvent= H2O 
      solute= NaOH
Step 2) Assume that there are 100 grams total. Assume that the percent of the compound present is the number of grams present as well
           100 g total
          16g NaOH 
            84g H20 
Step 3) convert grams to moles 
          84g H2O =4.64 mol H2O 
          19 g NaOH = 0.400 mol NaOH 
Step 4) mole fraction = moles A/(mol A + mol B +) 
          X= 0.400/(0.400 + 4.64) = 0.079  (no units) 
Step 5) convert g of solvent to kg
        84 g H2O = 0.084 kg 
Step 6) molality = moles of solute/ kg of solvent 
           molality = 0.400 mol NaOH/ 0.084 kg H2O 
           molality=  4.76 m
  
          

Ibelieve y y even though i've only taken chemistry for one year