± Convert between Units of Concentration Fortunately, there are several other wa

Question

± Convert between Units of Concentration

Fortunately, there are several other ways of expressing concentration that do not involve volume and are therefore temperature independent.

A 2.350×102M 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.

Part A

Calculate the molality of the salt solution.

Express your answer to four significant figures and include the appropriate units.

.0235m

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Your answer has units of mol/L of water. You need to convert the volume of water to kilograms using the density provided.

Part B

Calculate the mole fraction of salt in this solution.

Express the mole fraction to four significant figures.

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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.

%

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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.

ppm

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± Convert between Units of Concentration

Chemists often use molarity M, in moles/liter, to measure the concentration of solutions. Molarity is a common unit of concentration because the volume of a liquid is very easy to measure. However, the drawback of using molarity is that volume is a temperature-dependent quantity. As temperature changes, density changes, which affects volume. Volume markings for most laboratory glassware are calibrated for room temperature, about 20C.

Fortunately, there are several other ways of expressing concentration that do not involve volume and are therefore temperature independent.

A 2.350×102M 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.

Part A

Calculate the molality of the salt solution.

Express your answer to four significant figures and include the appropriate units.

mNaCl =

.0235m

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Incorrect; Try Again; no points deducted

Your answer has units of mol/L of water. You need to convert the volume of water to kilograms using the density provided.

Part B

Calculate the mole fraction of salt in this solution.

Express the mole fraction to four significant figures.

NaCl =

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Incorrect; Try Again; no points deducted

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.

percent by mass NaCl =

%

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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.

parts per million NaCl =

ppm

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Explanation / Answer

no of moles of NaCl = molarity * volume in L

                               = 2.35*10-2 *1 = 2.35*10-2 moles of NaCl

mass of water = volume * density

                        = 999.4*0.9982 = 997.6 g   = 0.9976KG

molality = no of moles of solute / mass of solvent in kG

             = 2.35*10-2/0.9976   = 0.0235m

no of moles of H2o   = W/G.M.Wt

                                = 997.6/18 = 55.42 moles of H2O

mole fraction of NaCl   = no of moles of NaCl/total no of moles(no of moles of naCl + no of moles of H2O)

                               = 0.0235/0.0235+55.42

                              = 0.0235/55.4435

                             = 0.000424

mass of NaCl = no of moles of NaCl* gram molar mass

                      = 0.0235*58.5 = 1.37475g

percentage of NaCl   = mass of naCl*100/total mass

                                 = 1.37475*100/1.37475+997.6

                                 = 1.37475*100/998.97475

                                 = 0.1376%

         part-D

                                 = 0.001376g   = 1376*10-6 g = 1376ppm of NaCl

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