A 2.900×10 2 M solution of NaCl in water is at 20.0C. The sample was created by

Question

A 2.900×102M 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 999.2 mL . The density of water at 20.0C is 0.9982 g/mL.

Part A

Calculate the molality of the salt solution.

Express your answer to four significant figures and include the appropriate units.

5M

Use either an integer, decimal number, or scientific notation for the numeric portion of your answer. Do not use calculations or functions.

Part B

Calculate the mole fraction of salt in this solution.

Express the mole fraction to four significant figures.

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Part C

Calculate the concentration of the salt solution in percent by mass.

Express your answer to four significant figures and include the appropriate units.

Submit

Part D

Calculate the concentration of the salt solution in parts per million.

Express your answer as an integer to four significant figures and include the appropriate units.

5M

Explanation / Answer

Part A)

mass of solvent = 999.2 x 0.9982 = 997.4 g

                         = 0.9974 kg

moles of NaCl = 2.90 x 10^-2 mol

molality = moles / mass of solvent in kg

              = 2.90 x 10^-2 / 0.9974

molality = 0.02908 m

Part B)

moles of water = 997.4 / 18.02 = 55.35 mol

moles of salt = 2.90 x 10^-2 mol

total moles = 55.38 mol

mole fraction of salt = 2.90 x 10^-2 / 55.38

mole fraction of salt = 5.237 x 10^-4

Part C)

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

                        = 1.6965 / 999.09) x 100

% mass of salt = 0.1698 %

part D)

concentration of salt = 1.6965 / 1.000

                                = 1696 ppm

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