Question

A 2.550×10⁻²M 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 999.3 mL . The density of water at 20.0∘C is 0.9982 g/mL.

A) Calculate the molality of the salt solution.

B) Calculate the mole fraction of salt in this solution.

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

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

Answer 1

(A) molarity of NaCl = 2.550 x 10^-2 M

volume of solution = 1.000 L

moles NaCl = (molarity of NaCl) * (volume of solution)

moles NaCl = (2.550 x 10^-2 M) * (1.000 L)

moles NaCl = 2.550 x 10^-2 mol

volume water = 999.3 mL

mass water = (volume water) * (density of water)

mass water = (999.3 mL) * (0.9982 g/mL)

mass water = 997.5 g

mass water = 997.5 g * (1 kg / 1000 g)

mass water = 0.9975 kg

molality NaCl = (moles NaCl) / (mass water in kg)

molality NaCl = (2.550 x 10^-2 mol) / (0.9975 kg)

molality NaCl = 2.556 x 10^-2 m

(B) moles NaCl = 2.550 x 10^-2 mol

mass water = 997.5 g

moles water = (mass water) / (molar mass water)

moles water = (997.5 g) / (18.0 g/mol)

moles water = 55.37 mol

Total moles = (moles NaCl) + (moles water)

Total moles = (2.550 x 10^-2 mol) + (55.37 mol)

Total moles = 55.39550 mol

mole fraction NaCl = (moles NaCl) / (Total moles)

mole fraction NaCl = (2.550 x 10^-2 mol) / (55.39550 mol)

mole fraction NaCl = 4.60 x 10^-4

(C) moles NaCl = 2.550 x 10^-2 mol

mass NaCl = (moles NaCl) * (molar mass NaCl)

mass NaCl = (2.550 x 10^-2 mol) * (58.44 g/mol)

mass NaCl = 1.49 g

Total mass = (mass NaCl) + (mass water)

Total mass = (1.49 g) + (997.5 g)

Total mass = 998.99 g

percent by mass NaCl = (mass NaCl / Total mass) * 100

percent by mass NaCl = (1.49 g / 998.99 g) * 100

percent by mass NaCl = 0.149 %

(D) ppm NaCl = (mass NaCl / Total mass) * 10⁶

ppm NaCl = (1.49 g / 998.99 g) * 10⁶

ppm NaCl = 1490 ppm