Question

± 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 20∘C. Fortunately, there are several other ways of expressing concentration that do not involve volume and are therefore temperature independent.

A 2.400×10−2M 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 999.4 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. mNaCl = SubmitHintsMy AnswersGive UpReview Part Part B Calculate the mole fraction of salt in this solution. Express the mole fraction to four significant figures. χNaCl = SubmitHintsMy AnswersGive UpReview Part 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 = SubmitHintsMy AnswersGive UpReview Part 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 = SubmitHintsMy AnswersGive UpReview Part Provide FeedbackContinue

Answer 1

NaCl = 0.02 M , vol = 1000ml , water vol = 999.4 ml , density of water = 0.9982 g/ml

water mass = vol x density = 999.4 ml x 0.9982 g/ml = 997.6 ml = 0.9976 kg

NaCl Molarity = moles / vol of solution in L

0.024 = moles of NaCl / 1 L

moles of NaCl = 0.024 , mass of NaCl = moles x molar mass of NaCl = 0.024 x 58.44 = 1.40256 g

A) molality = moles of NaCl / solvent mass in Kg

               = 0.024 /0.9976   = 0.02406

B) water moles = mass/molar mass = 997.6 /18.015 = 55.376

mol fraction of NaCl = ( NaCl moles ) / ( NaCl moles + water moles)

               = ( 0.024 ) / ( 0.024+55.376)   = 0.0004332

C) NaCl % = ( 100 x NaCl mass/ solution mass)   wher solution mass = NaCl mass + water mass

             = ( 100 x 1.40256) / ( 1.40256+997.6)   

                  = 0.1404 %

D) NaCl in ppm =10^ 6 x ( NaCl mass) / ( solution mass) = 140400 ppm