A 2.250×10 2 M solution of NaCl in water is at 20.0C. The sample was created by

Question

A 2.250×102M 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 999.4 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.

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0.4150.415MM

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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.

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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.

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0.4150.415MM

Explanation / Answer

Ans. #A. Given, molality of salt solution = 2.250 x 10-2 M = 0.0225 M

Both the values (2.250 x 10-2 M, 0.0225 M) have 4 significant figures.

#B. Moles of NaCl = Molarity x Volume of solution in liters

                                    = 2.250 x 10-2 M x 1.000 L

                                    = 0.0225 mol

# Mass of NaCl = Moles x Molar mass = 0.0225 mol x (58.442468 g/ mol) = 1.3150 g

# Mass of water (solvent) = Volume x Density

= 999.4 mL x (0.9982 g/mL) = 997.60108 g

Moles of H2O = Mass / Molar mass = 997.60108 g / (18.01528 g/ mol) = 55.3753 mol

# Total moles of Solution = Moles of NaCl + Moles of H2O

                                                = 0.0225 mol + 55.3753 mol

                                                = 55.3978 mol

# Mole fraction of NaCl = Moles of NaCl / Total moles of solution

                                                = 0.0225 mol / 55.3978 mol

                                                = 4.061 x 10-4 = 0.0004 (Prefer scientific notation)

#C. Mass of solution = Mass of NaCl + Mass of Water

                                    = 1.3150 g + 997.60108 g

                                    = 998.91608 g

% NaCl = (Mass of NaCl / Mass of Solution) x 100

                        = (1.3150 g / 998.91608 g) x 100

                        = 1.316 x 10-1 %

#D. [NaCl], ppm = Mass of NaCl in mg / Mass of solution in kg

                        = 1315.0 mg / 0.99760108 g

                        = 1318.162 mg/ kg                                                   ; [1 mg/ kg = 1 ppm]

                        = 1.318 x 103 ppm

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