Question

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

Answer 1

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