Question

Week 7 Homework ± Convert between Units of Concentration 25 of 29 Review | Constants | Periodic Table Chemists often use molarity MM, in moles/litermoles/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∘C. Fortunately, there are several other ways of expressing concentration that do not involve volume and are therefore temperature independent. A 2.200×10−2 MM solution of glycerol (C3H8O3C3H8O3) in water is at 20.0∘C∘C. The sample was created by dissolving a sample of C3H8O3C3H8O3 in water and then bringing the volume up to 1.000 LL. It was determined that the volume of water needed to do this was 999.1 mLmL . The density of water at 20.0∘C∘C is 0.9982 g/mLg/mL. Part A Calculate the molality of the glycerol solution. Express your answer to four significant figures and include the appropriate units. View Available Hint(s) mC3H8O3mC3H8O3 = SubmitPrevious Answers Incorrect; Try Again Part B Calculate the mole fraction of glycerol in this solution. Express the mole fraction to four significant figures. View Available Hint(s) χC3H8O3χC3H8O3chi_C_3H_8O_3 = SubmitPrevious Answers Incorrect; Try Again Part C Calculate the concentration of the glycerol solution in percent by mass. Express your answer to four significant figures and include the appropriate units. View Available Hint(s) percent by mass C3H8O3C3H8O3 = nothingnothing Submit Part D Calculate the concentration of the glycerol solution in parts per million. Express your answer as an integer to four significant figures and include the appropriate units. View Available Hint(s) parts per million C3H8O3C3H8O3 = SubmitPrevious Answers Incorrect; Try Again Provide Feedback Next

Answer 1

@ Giren he have prepared a solution by a dissolving Glycerol (chita Og a water. so here Isolute - C3H₂O) Tatto @ Most ose no = 2.200x16 PM INS ☺ Mow qanday manal sedali fesolved Podji Icon seal rabim of the glyceril sem S 2 -200x1072 omel Gweg, volline of solutoon taleq= 1.00 so moles of solute presenta conc. xual = 2.2 X10-?na IL 2-2 X10-2 me] Again ginant volume of water usedz 999.1 ml density of water taken 0: 9982 Sonh So man of water taken - densich xual 20.99827 x999.1 and on

- 1997:39) molality of sal moles of salute S he Man of solvent eskg. yli 2-2710-2 meel 997.39 = 0.022. mal 0.9973 kg Malality= 10:022 m Imole fraction of maks of glyceral Glycerad in total mols obrelutron 10-022 m How we have makes of Glycerol = 0.022.onal man of water = 997.2.9 vs. the so mals of water mass o molar mass of the = 997 39 4 188/mole i -155.4 mole So teeltal moles of selufoon - 1022 + 55.4 550422 mars]

somel fraction of Glyceral = 2.422 mel - 0.022 mal 55.422 enel T&grigenal> 3-96410-4 1 ② Now we have man of glyceral z males x melarman of I glyard -0.022 mol x 92 81onale = 2.024 al We have many of water used =997-39! so tatal man of solutsomai 997.3+2.034 1999-3249] man % concentration of solitoon. sni m an of soluta 1 t efal man sopor xloo d i v in 3:024. X 100 9996324% . 0:203% (019)

Ippon concentration: 1mgoman of salute 1 kg man of solution a man of solutie x 106 mass of sol? So for our glyceral solution, concentration = 2.024 9 x 106 999.324 g 2025 ppm .

Answer 2

Part A:

Concentration of glycerol = 2.200 x 10^-2 M

Volume of solution = 1.000 L

Moles of glycerol = (Concentration of glycerol) x (Volume of solution)

Moles of glycerol = (2.200 x 10^-2 M) x (1.000 L)

Moles of glycerol = 2.200 x 10^-2 mole

Volume of water = 999.1 mL

Mass of water = (Volume of water) x (density of water)

Mass of water = (999.1 mL) x (0.9982 g/mL)

Mass of water = 997.3 g

Mass of water = 0.9973 kg

Molality of glycerol = (Moles of glycerol) / (Mass of water in kg)

Molality of glycerol = (2.200 x 10^-2 mole) / (0.9973 kg)

Molality of glycerol = 2.194 x 10^-2 m

Part B:

Moles of water = mass/MW = 997.3 g/18.015 g/mole = 55.359 mole

Mole fraction of glycerol = Moles of glycerol/total moles = 2.200 x 10^-2 mole/(2.200 x 10^-2 mole + 55.359 mole) = 3.972 x 10^-4

Part C:

Mass of glycerol = moles x MW = 2.200 x 10^-2 mole x 92 g/mole = 2.024 g

Mass of solution = 2.024 g + 997.3 g = 999.324 g

Mass % =  (2.024 g/999.324 g) x 100 = 0.2025 %

Part D:

ppm = (Mass of solute/mass of solution) x 10⁶

ppm = (2.024 g/999.324 g) x 10⁶

ppm = 2.025 x 10³