L>ptcouncil.net: Calculations involving molality, molarity, density, mass percent, mole portion (Problems #11 - 25)Calculations entailing molality, molarity, density, mass percent, mole fractionProblems #11 - 25Fifteen ExamplesProblems 1 - 10Return to services MenuProblem #11: calculate the molarity and also mole fraction of acetone in a 2.28-molal solution of acetone (CH3COCH3) in ethanol (C2H5OH). (Density that acetone = 0.788 g/cm3; density of ethanol = 0.789 g/cm3.) Assume the the quantities of acetone and ethanol add.

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Solution because that molarity:Remember, 2.28-molal means 2.28 moles of acetone in 1.00 kilogram that ethanol.1) identify volumes that acetone and ethanol, then complete volume:acetone2.28 mol x 58.0794 g/mol = 132.421 g132.421 g separated by 0.788 g/cm3 = 168.047 cm3ethanol1000 g divided by 0.789 g/cm3 = 1267.427 cm3total volume168.047 + 1267.427 = 1435.474 cm32) determine molarity:2.28 mol / 1.435 l = 1.59 MSolution because that mole fraction:1) identify moles of ethanol:1000 g / 46.0684 g/mol = 21.71 mol2) determine mole fraction of acetone:2.28 / (2.28 + 21.71) = 0.0950Problem #12: calculation the normality the a 4.0 molal sulfuric acid equipment with a density of 1.2 g/mL.Reminders:N = #equivalents / l solution#equivalents = molecular weight / n (n = number of H+ or OH¯ released every dissociation.)molal = moles solute / kg solventSolution:1) identify grams the H2SO4 present:4.0 molal = 4.0 moles H2SO4 / 1000 g solution4.0 mol times 98.09 g/mol = 392.32 g2) determine equivalent load for H2SO4:98.09 g/mol / 2 dissociable hydrogen/mol = 49.05 g/equivalent3) determine # equivalents in 392.32 g:392.32 g time (1 tantamount / 49.05 g) = 8.0 equivalents4) recognize volume of solution:392.32 g + 1000 g = 1392.32 g (total massive of solution)1392.32 g / 1.2 g/mL = 1160.27 mL5) recognize normality:N = 8.0 equivalents / 1.16027 l = 6.9 NProblem #13: An auto antifreeze mixture is do by mix equal volumes of ethylene glycol (d = 1.114 g/mL, molar mass 62.07 g/mol) and also water (d = 1.000 g/mL) at 20.0 °C. The density of the systems is 1.070 g/mL.Express the concentration that ethylene glycol as:(a) volume percent(b) mass percent(c) molarity(d) molality(e) mole fractionSolution come (a):Since the volumes space equal, the volume percent the ethylene glycol is 50%Solution come (b):1) determine the masses that equal volumes (we"ll use 50.0 mL) the the 2 substances:ethylene glycol: (50.0 mL) (1.114 g/mL) = 55.7 gwater: (50.0 mL) (1.000 g/mL) = 50.0 g2) identify percent due to ethylene glycol:55.7 g / 105.7 g = 52.7%Solution to (c):1) recognize moles of ethylene glycol:55.7 g / 62.07 g/mol = 0.89737 mol2) recognize volume the solution:105.7 g / 1.070 g/mL = 98.785 mL3) identify molarity:0.89737 mol / 0.098785 l = 9.08 MSolution to (d):0.89737 mol / 0.050 kg = 17.9 mNote the large difference between the molarity and also the molality.Solution to (e):1) identify moles of water:50.0 g / 18.0 g/mol = 2.777782) identify mole portion for ethylene glycol:0.89737 mol / 3.67515 mol = 0.244Problem #14: What is the percent of CsCl by mass in a 0.0711 M CsCl systems that has a thickness of 1.09 g/mL?Solution:1) identify mass of dissolved CsCl:Let us assume 100.0 mL the solution.MV = grams / molar mass(0.0711 mol/L) (0.100 L) = x / 168.363 g/molx = 1.197 g2) recognize mass of solution:1.09 g/mL time 100.0 mL = 109 gDetermine massive percent the CsCl in solution:1.197 g / 109 g = 1.098%to three sig figs: 1.10%Problem #15: A 8.77 M equipment of one acid, HX, has a thickness of 0.853 g/mL.The acid, HX, has actually a molar mass of 31.00 g/mol. Recognize the molal concentration the this solution, ΧHX (mole portion of HX), and % w/w (percent by mass). The solvent in this solution is water, H2O.Comment: have the right to an 8.77 M solution of one acid have actually a density of 0.853 g/mL? who cares? We"ll just solve the problem.Solution:There is a trick to addressing this type of problem: let united state assume 1.00 l (or 1000 mL) of the solution is present. (Another location where a comparable trick is work is in determining empirical formulas, wherein you assume 100 g the the problem is present.)1) some preliminary calculations:moles acid: 1.00 together x (8.77 moles / L) = 8.77 moles HX (used in molality and mole fraction)mass acid: 8.77 mole x 31.00 g / mole = 272 g HX (percent through mass)mass the solution: 1000 mL x 0.853 g / mL = 853 g systems (percent by mass)mass of solvent: 853 g minus 272 g = 581 g = 0.581 kg (molality)moles solvent: 581 g separated by 18.015 g) = 32.25 (mole fraction)2) Calculations come answer the questions:molality = 8.77 moles / 0.581 kg = 15.1mmole portion = 8.77 / (8.77 + 32.25) = 0.214% w/w = (272 g / 853 g) x 100 = 31.9%Problem #16: A systems of hydrogen peroxide, H2O2, is 30.0% by mass and has a thickness of 1.11 g/cm3. Calculation the (a) molality, (b) molarity, and (c) mole fractionSolution:1) fixed of 1 liter of solution:1.11 g/cm3 time (1000 cm3 / L) = 1110 g/L2) fixed of the two contents of the solution:mass of H2O2 ---> 30.0% that 1110 g = 333 gmass that H2O ---> 70.0% that 1110 g = 777 g3) mole of hydrogen peroxide:333 g / 34.0138 g/mol = 9.79 mol4) Molality:9.79 mol / 0.777 kg = 12.6 m5) Molarity9.79 mol / 1.00 l = 9.79 M6) Mole fractionmole the H2O2 = 9.79mole of H2O = 43.13total moles ---> 9.79 + 43.13 = 52.92mole portion ---> 9.79 / 52.92 = 0.185Problem #17: family hydrogen peroxide is an aqueous systems containing 3.0% hydrogen peroxide through mass. What is the molarity that this solution? (Assume a density of 1.01 g/mL.)Solution:1) permit us have 1000 g of systems on hand. For the volume of solution:1000 g split by 1.01 g/mL = 990.1 mL2) because H2O2 is 3% by mass, we understand that there are 30 grams the H2O2 current in the 1000 g of solution.MV = mass / molar mass(x) (0.9901 L) = 30 g / 34.0138 g/molx = 0.890814 MTwo sig figs appears reasonable, so 0.89 M.Problem #18: A 6.90 M KOH systems in water has 30% by load KOH. Calculation the density of the KOH solution.Solution:Let us assume we have actually 1.00 liter of solution present. This means we have 6.90 mol that KOH. Let us determine the massive of KOH:6.90 mol times 56.1049 g/mol = 387.124 g387.124 g represents 30% of the full weight of the solution. (The water provides up the other 70%.) To acquire the massive of the solution, perform this:387.124 g is to 0.3 as x is to 1x = 1290 gdensity the the equipment ---> 1290 g / 1000 mL = 1.29 g/mL (to three sig figs)Problem #19: one aqueous NaCl solution is made using 138 g of NaCl diluted to a total solution volume that 1.30 L.(A) calculate the molarity of the solution.(B) calculation the molality of the solution. (Assume a thickness of 1.08 g/mL because that the solution.)(C) calculation the fixed percent that the solution. (Assume a thickness of 1.08 g/mL for the solution.)Solution:Part A:MV = mass / molar mass(x) (1.30 L) = 138 g / 58.443 g/molx = 1.82 MPart B:molality is mole solute per kg that solvent. Ns will usage 2.3613 mol (keeping a couple of guard digits).Let united state assume 1000 mL that the solution is present. This tells us that 2.3613 mol of the solute is current (that"s the 138 g that NaCl).1.08 g/mL times 1000 mL = 1080 g 1080 g minus 138 g = 942 g 942 g = 0.942 kg2.3613 mol / 0.942 kg = 2.51 mPart C:138 g that solute was dissolved in 1080 total grams the solution(138 / 1080) times 100 = 12.8% 100% minus 12.8% = 87.2% 2OAnother kind of inquiry is this area is to ask you to identify the mole portion for every substance. For the you will require to know the mole of water:942 g / 18.015 g/mol = 52.29 molThe mole portion of NaCl is this:2.3613 mol / (2.3613 mol + 52.29 mol) = 0.0432The mole fraction of the water is this:1 − 0.0432 = 0.9568Problem #20: A equipment is all set by dissolve 28.0 g that glucose (C6H12O6) in 350 g that water. The last volume the the solution is 384 mL . Because that this solution, calculate each the the following:1) molarity; 2) molality; 3) percent through mass; 4) mole fraction; 5) mole percentSolution:1) Molarity is the number of moles separated by volume that solvent in liters.28.0 g / 180 g/mol = 0.156 mol0.156 mol / 0.384 together = 0.405 M2) Molality is the number of moles separated by fixed of solvent in kilograms.0.156 mol / 0.350 kg = 0.444 m3) Percent through mass, together the name implies, is the fixed of solute divided by full mass time 100%.28.0 g / (350 g + 28 g) time 100 = 7.41% C6H12O6 by mass4) Mole fraction is the variety of moles of solute split by full moles.350 g / 18.015 g/mol = 19.428 mol of water0.156 mol / (19.428 + 0.156) = 0.0080 mol fraction C6H12O6The mole portion of water is:1 − 0.0080 = 0.9925) Mole percent is mole portion times 100%.0.0080 x 100 = 0.80 % C6H12O6 through molesProblem #21: How plenty of grams of glucose is necessary to dissolve in 3 litres the water to obtain 40% solution?Solution:Let x gms of glucose dissolve in 3 liters that water to type a 40% solution.glucose weight = xwater load = 3 liters = 3000 g (take water thickness as 1.00 g/mL)so complete solution load = x + 3000 gglucose percent ---> x / (x + 3000) = 0.40 x = 0.4 (x + 3000)x = 0.4x + 1200x − 0.4x = 12000.6x = 1200x = 2000 gProblem #22: The vinegar offered in the grocery stores is defined as 5% (v/v) acetic acid. What is the molarity the this solution (density the 100% acetic mountain is 1.05 g/mL)?Solution:1) 5% (v/v) method 5% that the volume is acetic acid. Therefore 1.00 together of vinegar includes 50 mL acetic acid:(0.05)(1000 mL) = 50 mL2) The fixed of this acetic mountain is:(1.05 g/mL)(50 mL) = 52.5 g3) because that the molarity, usage MV = massive / molar mass(x) (1.00 L) = 52.5 g / 60.0516 g/molx = 0.874 M (to three sig figs)Problem #23: by titration, the molarity of acetic acid in vinegar was established to be 0.870 M. Convert this come %(v/v). (The thickness of acetic mountain is 1.05 g/mL)Solution:1) i think 1.00 l of the systems to be present. Use MV = mass/molar fixed to identify mass the acetic acid present:(0.870 mol/L) (1.00 L) = x / 60.0516 g/molx = 52.245 g2) determine what volume the pure acetic acid this is:52.245 g divided by 1.05 g/mL = 49.757 mL3) identify %(v/v):(49.757 mL / 1000 mL) * 100 = 4.9757Rounded off, this is 4.98%(v/v), usually offered as 5%(v/v).Problem #24: In one aqueous systems of sulfuric acid, the acid concentration is 2.40 mole percent and the density of the equipment is 1.079 g/mL. Calculate (1) the molal concentration the the acid, (2) the weight percentage of the acid, and also (3) the molarity the the solutionSolution:2.40 mole percent that acid way 97.60 mole percent water.Let"s i think 100 moles of the solution is present. This means 2.40 mole the the systems is H2SO4 and also 97.60 mole is water.(1) for molality, we require to know kg of water ---> 97.60 mol time 18.015 g/mol = 1758.264 g = 1.758264 kgmolality ---> 2.40 mol / 1.758264 kg = 1.365 m (1.36 m to three sig figs)(2) for the load percent, we need the massive of H2SO4 ---> 2.40 mol times 98.0768 g/mol = 235.38432 gweight percent ---> 235.38432 g / (235.38432 + 1758.264 g) = 0.118067 = 11.8% (three sig figs)(3) because that molarity, we require to understand the volume the the systems ---> 1993.64832 g separated by 1.079 g/mL = 1847.68 mL = 1.84768 LNote: 1993.64832 g is the full mass that the solution.molarity ---> 2.40 mol / 1.84768 together = 1.2989 M (1.30 M to 3 sig figs) through the way, mole percent is mole fraction written as a percent. The mole fractions in the above problem room 0.0240 and also 0.9760.Problem #25: calculate the molality, molarity, and also mole fraction of FeCl3 in a 26.3% (w/w) equipment (density = 1.28 g/mL).Solution:Molarity:Assume 100. G of equipment present.26.3 g of FeCl3 is present.100. G divided by 1.28 g/mL = 78.125 mLUse MV = mass / molar mass(x) (0.078125 L) = 26.3 g / 162.204 g/molx = 2.08 M (to 3 sig figs)Molality100. G − 26.3 g = 73.7 g molality ---> (26.3 g / 162.204 g/mol) / 0.0737 kg = 2.20 m (to three sig figs)Mole portion of FeCl3:moles FeCl3 ---> 26.3 g / 162.204 g/mol = 0.1621415 molmoles water ---> 73.7 g / 18.015 g/mol = 4.091035 molmole fraction ---> <0.1621415 mol / (0.1621415 mol + 4.091035 mol)> = 0.0381 (to 3 sig figs)Problem #26: The mole portion in a solution of Na2S is 0.125. Calculate the fixed precent (w/w) of Na2S in this solution.Solution:The mole portion of Na2S is 0.125. Therefore, the mole fraction of water in the solution is 0.875.mass Na2S ---> 0.125 mol time 78.045 g/mol = 9.755625 gmass water ---> 0.875 mole 18.015 g/mol = 15.763125 g% (w/w) Na2S ---> <9.755625 / (9.755625 + 15.763125) * 100> = 38.2% (to 3 sig figs)Problem #27: In dilute nitric acid, the concentration the HNO3 is 6.00 M and the thickness of this equipment is 1.19 g/mL. Usage that information to calculate the mass percent and also mole portion of HNO3 in the solution.

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Solution #1:1) Let us assume 1000 mL of the systems is present. Some preliminary calculations:1000 mL x 1.19 g/mL = 1190 g (this is the mass of our 1000 mL)6.00 mol/L x 1.00 l = 6.00 mol (this is how countless moles of HNO3 room in ours 1000 mL)6.00 mol x 63.012 g/mol = 378.072 g (the fixed of HNO3 in our 1000 mL)1190 g minus 378.072 g = 811.928 g (the fixed of water in the 1000 mL that solution)2) calculate the massive percent the HNO3:378.072 g / 1190 g = 31.8% (to 3 sig figs)3) calculation the mole portion of HNO3:811.928 g / 18.015 g/mol = 45.07 mol of H2O6.00 mol / (6.00 mol + 45.07 mol) = 0.118 (to 3 sig figs)Solution #2:1) Let us assume 1000 g the the solution is present. Part preliminary calculations:1000 g / 1.19 g/mL = 840.336 mL (the volume of ours 1000 g that solution)6.00 mol/L x 0.840336 together = 5.042 mol (this is how countless moles the HNO3 space in our 840.336 mL)5.042 mol x 63.012 g/mol = 317.7065 g (the massive of HNO3 in our 840.336 mL)1000 g minus 317.7065 g = 682.2935 g (the massive of water in the 1000 g the solution)2) calculation the fixed percent of HNO3:317.7065 g / 1000 g = 31.8% (to 3 sig figs)3) calculate the mole fraction of HNO3:682.2935 g / 18.015 g/mol = 37.874 mol that H2O5.042 mol / (5.042 mol + 37.874 mol) = 0.117 (to 3 sig figs)Comment: In systems #2, I derived 0.117485 and that, technically, is 0.117 (not 0.118), as soon as rounded to three sig figs.Problem #28: A 0.100 M NaOH solution will be all set by dilution of a 50.0% (w/w) NaOH solution. This solution has a density of 1.53 g/mL. Compute the volume of this systems that is compelled to prepare 1.00 x 103 mL the 0.100 M NaOH.Solution:1) determine moles the NaOH in 1.0 x 103 mL the 0.10 M solution:MV = moles(0.100 mol/L) (1.00 L) = 0.100 mol2) recognize mass of 0.100 mol that NaOHmoles x molar massive = grams(0.100 mol) (40.00 g/mol) = 4.00 g3) fixed of 50.0% (w/w) equipment that includes 4.00 g of NaOH:use a ratio and proportion50 g is come 100 g as 4 g is come xx = 8.00 mL4) recognize volume of equipment that includes 8.00 g the NaOH:Luke, uuuuuuse the density!8.00 g / 1.53 g/mL = 5.23 mLProblem #29: What is the molarity that a NaOH solution with a density of 1.33 g/mL the was made with 70.0 ml of water if the molality is 10.7 molal? Solution:1) use the molality come determine just how much NaOH was supplied with the 70.0 g of water:10.7 molal = 10.7 mol solute every kg solvent10.7 mol/kg = x / 0.0700 kgx = 0.749 mol of NaOH2) identify the grams the NaOH:(0.749 mol) (40.0 g/mol) = 29.96 g 3) full mass of the solution:70.0 g + 30.0 g = 100.0 g4) Volume the the solution:100.0 g / 1.33 g/mL = 75.2 mL5) Molarity:0.749 mol / 0.0752 together = 9.96 MProblem #30a: How numerous mL of an 3.78% (w/w) solution can be ready from 18.00 g that sucrose?Solution:3.78 is come 18 together 100 is to xx = 476.19 g 476.19 − 18.00 = 458.19 g i think 1 g/mL for the density and also no volume readjust when combine 458.19 g of water and also 18.00 g that sucrose.Answer = 458 mL (to 3 sig figs)Problem #30b: How countless mL of an 3.78% (w/v) solution have the right to be prepared from 18.00 g of sucrose?Solution:3.78%(w/v) method 3.78 g of sucrose every 100 mL of solution3.78 is to 100 together 18 is to xx = 476 mL (to three sig figs)Fifteen ExamplesProblems 1 - 10Return to services Menu