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

A 2.500×10⁻² 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 999.3 mL . The density of water at 20.0∘C is 0.9982 g/mL.

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

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

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 = ???

Solutions

Expert Solution

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)

queen_honey_blossom answered 1 year ago

Next > < Previous

Related Solutions

A 2.800×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by...

A 2.800×10−2 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 999.2 mL . The density of water at 20.0∘C is 0.9982 g/mL. 1a) Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units. 1b) Calculate the...

A 2.250×10−2 M solution of NaCl in water is at 20.0∘C. The sample was created by...

A 2.250×10−2 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. A. Calculate the molality of the salt solution. Express your answer to four significant figures and include the appropriate units. B. Calculate the...

A 2.750×10−2M solution of NaCl in water is at 20.0∘C. The sample was created by dissolving...

A 2.750×10−2M 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. Part A Calculate the molality of the salt solution. Part B Calculate the mole fraction of salt in this solution. Express the mole fraction...

A 2.400×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created...

A 2.400×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 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 999.0 mL . The density of water at 20.0∘C is 0.9982 g/mL. Part A: Calculate the molality of the glycerol solution. Part B: Calculate the mole fraction of glycerol in this solution. Part C:...

A 2.250×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created...

A 2.250×10−2 M solution of glycerol (C3H8O3) in water is at 20.0∘C. The sample was created by dissolving a sample of C3H8O3 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 999.0 mL . The density of water at 20.0∘C is 0.9982 g/mL. Part A: Calculate the molality of the glycerol solution. Express your answer to four significant figures and include the appropriate units. Definitely...

A 0.380 m aqueous solution of NaCl is prepared at 20.0∘C. Assume that the density of...

A 0.380 m aqueous solution of NaCl is prepared at 20.0∘C. Assume that the density of the solution at 20.0∘C is 1.082 g/mL. Calculate the molarity of the salt solution.

A water sample is prepared by adding 10-4M NaCl, 10-3 M NaHCO3 and 10-4 M CaCl2...

A water sample is prepared by adding 10-4M NaCl, 10-3 M NaHCO3 and 10-4 M CaCl2 to water. Assume that when these salts dissolve, they dissociate to Na+, Cl- , Ca2+, and HCO3 - and then do not react further. a. Compute the ionic strength of the solution. b. What is the activity of each dissolved ion if the infinitely solution approximation applies and the infinite dilution scale is used? What is the activity of each ion if the extended...

A 1.00 L solution contains 2.00×10-4 M Cu(NO3)2 and 2.500×10-3 M ethylenediamine (en). The Kf for...

A 1.00 L solution contains 2.00×10-4 M Cu(NO3)2 and 2.500×10-3 M ethylenediamine (en). The Kf for Cu(en)22+ is 1.00×1020. What is the concentration of Cu2+(aq ) in the solution?

An aqueous solution of 0.0190g of a protein in 10.0 mL of water at 20.0 ̊C...

An aqueous solution of 0.0190g of a protein in 10.0 mL of water at 20.0 ̊C shows a 5.22 cm rise inthe apparatus shown in the figure. Assume the density of the solution to be 0.998g/mL, andthe density of mercury to be 13.6 g/cm3. What is the molar mass of the protein?

A 20.0 mL sample of 0.432 M HBr is titrated with a 0.314 M NaOH solution....

A 20.0 mL sample of 0.432 M HBr is titrated with a 0.314 M NaOH solution. The pH after the addition of 35.0 mL of NaOH is __________. A.0.04 B.1.37 C.12.53 D.12.63 E.13.33

Subjects