Question

# A 2.050×10−2M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving...

A 2.050×10−2M 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 999.4 mL . The density of water at 20.0∘C 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.

Part B

Calculate the mole fraction of salt in this solution.

Express the mole fraction to four significant figures.

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.

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.

A) Molality = mol of solute / mass of solvent.
Mol of solute = Molarity of solute x Volume of solution = (0.02050 M) x (1.000 L) = 0.02050 mol of solute
Mass of solvent = 999.4 mL x (0.9983 g / 1 mL) = 997.7 g = 0.9977 kg
Molality = 0.02050 mol / 0.9977 kg = 0.02055 mol / kg

B) Mol fraction of salt = mol of salt / mol total
Mol of water = 997.7 g x (1 mol H2O / 18.0154 g H2) = 55.38 mol H2O
Mol fraction of salt = 0.02050 mol / (0.02050 mol + 55.38 mol) = 0.0003700

C) Concentration by percent mass = (mass of salt / mass of solution) x 100%
Mass of salt = 0.02050 mol NaCl x (58.44 g NaCl / 1 mol NaCl) = 1.198 g NaCl
Mass of water = 997.7 g H2O
Mass of solution = 1.198 g + 997.7 g = 998.90 g
Therefore, (1.198 g / 998.90 g) x 100% = 0.1199 %

D)Parts per million = (g of solute / g of solution)*1000000 = (1.198 g / 998.90 g)*1000000 = 1199 ppm

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