Question

A 2.300×10⁻² M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving a sample of NaCl in water and then bringing the volume up to 1.000 L. It was determined that the volume of water needed to do this was 999.4 mL . The density of water at 20.0∘C is 0.9982 g/mL. Express your answer(s) to four significant figures and include the appropriate units.

Part A) Calculate the molality of the salt solution.

Part B) Calculate the mole fraction of salt in this solution.

Part C) Calculate the concentration of the salt solution in percent by mass.

Answer 1

Part A)

Molarity = moles / volume

moles = Molarity x volume

          = 2.300 x 10^-2 x 1.000

moles = 2.300 x 10^-2 mol

mass of solvent = 999.4 x 0.9982 = 997.6 g

molality = moles / mass of solvent in kg

             = 2.300 x 10^-2 / 0.9976

molality = 0.02306 m

Part B)

moles of solute = 2.30 x 10^-2

moles of solvent = 997.6 / 18.02 = 55.36 mol

mole fraction of salt = 2.30 x 10^-2 / (55.36 + 0.023)

                                 = 4.153 x 10^-4

Part C)

mass of salt = 0.023 x 58.5 = 1.3455 g

mass % = (mass of salt / mass of solution ) x 100

               = 1.3455 / (997.6+ 1.3455) x 100

               = 0.1346 %

             =