Question

A 2.100×10−2 M 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.

Answer 1

a)

molality of salt

mol = (2.1*10^-2 )* 1 =2.1*10^-2

molality = mol/kg = MV / kg = (2.1*10^-2 )* 1 / 0.9994 = 0.021012 molal

b)

mole frac of salt in solution

mol frac = mole Salt / total mol

mol water = mass/MW = 998/18 = 55.4444

total mol = mol water + mol salt = 2.1*10^-2 + 55.4444

mol frac = (2.1*10^-2)/(55.4444+ 2.1*10^-2 ) = 0.00037861441

c)

calculate % by mass

mass aslt = mol*MW = (2.1*10^-2)(58) =1.218

% mass = mass of salt / total mass = 1.218/998.2 * 100 = 0.122019 %

d)

ppm = mg /kg

mg of salt = 1.218 *1000 = 1218 mg of salt

kw = 0.998

ppm = 1218/0.998 = 1220.4408 ppm