A 2.100×102 M solution of NaCl in water is at 20.0C. The sample was created by d

Question

A 2.100×102 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.

A) Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units.

B) Calculate the mole fraction of salt in this solution. Express the mole fraction to four significant figures.

C) Calculate the concentration of the salt solution in percent by mass. Express your answer to four significant figures and include the appropriate units.

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.

Explanation / Answer

a) Molality = mol of solute / mass of solvent.
Mol of solute = Molarity of solute x Volume of solution = (0.021 M) x (1.000 L) = 0.021 mol of solute
Mass of solvent = 999.4 mL x (0.9982 g / 1 mL) = 997.6 g = 0.9976 kg
Molality = 0.021 mol / 0.9976 kg = 0.02105 mol / kg

b) Mol fraction of salt = mol of salt / mol total
Mol of water = 997.6 g x (1 mol H2O / 18.0154 g H2) = 55.37 mol H2O
Mol fraction of salt = 0.021 mol / (0.021 mol + 55.37 mol) = 0.0003791

c) Concentration by percent mass = (mass of salt / mass of solution) x 100%
Mass of salt = 0.021 mol NaCl x (58.44 g NaCl / 1 mol NaCl) = 1.227 g NaCl
Mass of water = 997.6 g H2O
Mass of solution = 1.227 g + 997.6 g = 998.82 g
Therefore, (1.227 g / 998.82 g) x 100% = 0.1228 % by mass

d) Parts per million = (g of solute / g of solution)*1000000 = (1.227 g / 998.82 g)*1000000 = 1228 ppm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.