Mole fraction, weight %, Molarity

Question

6.0m solution of NAOH has a density of 1.39 g/cm3 at 20°C. Calculate mole fraction, weight percent, the molarity of NaOH

Answer 1

Starting with 6.0 m NaOH, tells us that we have 6.0 moles NaOH/kg H2O.

Moles NaOH = 6.0

Moles H2O = 1 kg x 1000 g/kg x 1 mol/18 g = 55.6 moles 

Total moles = 6.0 + 55.6 = 61.6 moles

Mole fraction NaOH = 6.0/61.6 = 0.0974 = 0.10 (to 2 significant figures)

Weight percent NaOH = mass NaOH/total mass of solution (x100%)

mass NaOH = 6.0 moles x 40 g/mol = 240 g

mass H2O = 1000 g

total mass = 1240 g

Weight percent NaOH = 240 g/1240 g (x100%) = 19% (to 2 significant figures)

Molarity NaOH = moles NaOH/liter solution

moles NaOH = 6.0

liters of solution = (1000 g + 240 g) x 1 ml/1.39 g = 1240 g x 1 ml/1.39 g = 892 ml = 0.892 L

Molarity of NaOH = 6.0 moles/0.892 L = 6.7 M (to 2 significant figures)