Question

# A 2.100×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by...

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.

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

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