A 2.150*10^-2 M solution of NaCl in water is at 20.0C. The sample was created by

Question

A 2.150*10^-2 M solution of NaCl in water is at 20.0C. The sample was created by dissolving a sample of NaClbin water and then bringing the volume up to 1.000L. It was determined that the water needed to do this was 999.4 mL. The density of water at 20.0C is 0.9982 g/mL.
Calculate the molality of the salt solution. Calculate the mole fraction of salt in this solution. Calculate the concentration of the salt solution in percent by mass. Calculate the concentration of the salt solution in parts per million.
A 2.150*10^-2 M solution of NaCl in water is at 20.0C. The sample was created by dissolving a sample of NaClbin water and then bringing the volume up to 1.000L. It was determined that the water needed to do this was 999.4 mL. The density of water at 20.0C is 0.9982 g/mL.
Calculate the molality of the salt solution. Calculate the mole fraction of salt in this solution. Calculate the concentration of the salt solution in percent by mass. Calculate the concentration of the salt solution in parts per million.

Calculate the molality of the salt solution. Calculate the mole fraction of salt in this solution. Calculate the concentration of the salt solution in percent by mass. Calculate the concentration of the salt solution in parts per million.

Explanation / Answer

1)

Molarity = 2.150 x 10^-2 M

moles = molarity x volume

          = 2.150 x 10^-2 x 1

          = 2.150 x 10^-2 mol

mass of water = 999.4 x 0.9982 = 997.6 g

molality = moles / mass of solvent

            = 2.150 x 10^-2 / 0.9976

molality = 0.02155 m

2)

moles of NaCl = 2.150 x 10^-2 mol

moles of water = 997.6 / 18.02 = 55.36 mol

mole fraction of salt = 2.150 x 10^-2 / 2.150 x 10^-2 + 55.36

mole fraction of salt    = 3.882 x 10^-4

3)

mass of NaCl = 0.0215 x 58.5 = 1.258 g

mass of solution = 1.258 + 997.6 = 998.86 g

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

            = (1.258 / 998.86) x 100

           = 0.126 %

4)

concentration of salt = 2.15 x 10^-2 mol / L

                                = 1.258 g / L

                                = 1258 mg / L

                                = 1258 ppm

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