Question

A 2.650×10?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 999.3 mL . The density of water at 20.0?C is 0.9982 g/mL. 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. Part D Calculate the concentration of the salt solution in parts per million.

Answer 1

For the given aqueous solution of NaCl

Part A)

moles of NaCl = 2.65 x 10^-2 mol

mass of water = 999.3 ml x 0.9982 g/mol/1000 = 0.9975 kg

molality of solution = 2.65 x 10^-2 mol/0.9975 kg = 0.0266 m

Part B)

moles water = 997.5 g/18 g/mol = 55.42 mol

total moles = 55.42 + 0.0265 = 55.45 mol

mole fraction of salt in solution = 0.0265/55.45 = 0.0005

Part C)

mass NaCl in solution = 2.65 x 10^-2 mol x 58.44 g/mol = 1.55 g

Total mass of solution = 1.55 + 997.5 g = 999.05 g

mass% NaCl in solution = 1.55 x 100/999.05 = 0.155%

Part D)

NaCl concentration in ppm = 1.55 g x 1000/1 L = 1550 ppm