A 2.350×102 M solution of NaCl in water is at 20.0C. The sample was created by d

Question

A 2.350×102 M solution of NaCl in water is at 20.0C. 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 20.0C 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.

Explanation / Answer

Part-A

no of moles of NaCl   = molarity* volume in L

                                = 2.35*10^-2*1    = 2.35*10^-2   = 0.0235 moles

mass of water            = volume * density

                                 = 999.4*0.9982   = 997.6g

molality                   = no of moles/weight of solvent in kg

                                =0.0235/0.9976   = 0.0235m

part-B

no of moles of H2O   = W/G.M.Wt

                               = 997.6/18   = 55.42moles

no of moles of NaCl   = 0.0235moles

mole fraction of NaCl   = no of moles of NaCl/no of moles of NaCl + no of moles of H2O

                                  = 0.0235/0.0235+55.42

                                   = 0.0235/55.4435   = 0.000424

part-C

mass of NaCl   = no of moles * gram molar mass

                     = 0.0235*58.5   = 1.37475g

mass of H2O    = 997.6

mass percent of NaCl   = 1.37475*100/1.37475+997.6

                                 = 1.37475*100/998.97475   = 0.138% NaCl by mass

part-D

1.37475g NaCl/1000ml    = 0.001375g/ml   = 1375*10^-6g/ml   = 1375ppm NaCl

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