A 2.500×10 2 M solution of NaCl in water is at 20.0C. The sample was created by

Question

A 2.500×102 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 999.3 mL . The density of water at 20.0C is 0.9982 g/mL.

Calculate the molality of the salt solution. XNaCl = 2.506 X 10-2 m

Calculate the mole fraction of salt in this solution. XNaCl = ???

Calculate the concentration of the salt solution in percent by mass. Percent by mass NaCl = ???

Calculate the concentration of the salt solution in parts per million. Parts per million NaCl = ???

Explanation / Answer

in 1.000L... you have
.. 2.500x10^-2 mol NaCl
.. 2.500x10^-2 mol NaCl x (58.44g / mol) = 1.461 g NaCl

.. 999.3mL x (0.9982g / mL) x (1kg / 1000g) = 0.99750kg H2O
.. 999.3mL x (0.9982g / mL) x (1 mol / 18.016g) = 55.367 mol H2O

so..

*** A ***
molality = moles solute / kg solvent = 2.500x10^-2 mol / 0.99750kg = 2.5062 x10^-2 mol/Kg

*** B ***
mole fraction NaCl = moles NaCl / total moles = 2.500x10^-2 mol / (2.500x10^-2 mol + 55.367 mol) = 4.51 x10^-4

*** C ***
mass % NaCl = (mass NaCl / total mass) x 100% = 1.461g NaCl / (1.461g NaCl + 997.50g H2O) x 100% = 0.1462%

*** D ***
here's an example.. ppm is parts per million parts..

0.1462% NaCl...that means 0.1462 g NaCl / 100g solution.. so I can do this
.. ppm NaCl = (0.1462g NaCl / 100g solution) x (10^6 / 10^6)
... .... ... ... .. = 1462g / 10^6 grams solution
... ... ... .. .. ..= 1462g / million grams solution
... .... ... ..... .= 1462 ppm (g / g)

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