A 2.400×10 2 M solution of NaCl in water is at 20.0C. The sample was created by

Question

A 2.400×102M 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.

Calculate the molality of the salt solution.

Calculate the mole fraction of salt in this solution.

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

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

Explanation / Answer

M = mol/L

Calculate the molality of the salt solution.

M is kept, therefore M = 2.4*10^-2 M (given data)

Calculate the mole fraction of salt in this solution.

mol frac = mol of NaCl / total mol

mol of W = mass/MW = 1000/18 = 55.56 mol

mol of NaCl = 2.4*10^-2

mol frac = mol of NaCl / total mol = (2.4*10^-2) /(55.56 +2.4*10^-2) =0.000431

x = 0.000431

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

C = mass/Total mass

mass = mol*MW = (2.4*10^-2)(58) = 1.392 g

C = 1.392 /(0.9982*1000) *100% = 0.1394%

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

ppm = mg of salt / kg of solution

g = 1.392 ---> 1392 mg of salt

kg of solution = 1

1392/1 = 1392 ppm

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