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Calculating Concentrations WKST – Vanden Bout – (51165)

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Calculating Concentrations WKST – Vanden Bout – (51165) Version 001 – Calculating Concentrations WKST – vanden bout –(51165) 1 This print-out should have 22 questions. 003 10.0 points Multiple-choice questions may continue on A chemist studying the properties of photo- the next column or page – find all choices graphic emulsions needed to prepare 500 mL before answering. of 0.178 M AgNO3(aq). What mass of silver nitrate must be placed into a 500 mL volumet- ChemPrin3e G 13 ric flask, dissolved, and diluted to the mark 001 (part 1 of 2) 10.0 points with water? Determine the mass of anhydrous copper(II) sulfate that must be used to prepare 250 mL Correct answer: 15.1187 g. of 1.9 M CuSO4(aq). Explanation: v =500mL=0.5 L M =0.178 M Correct answer: 75.8148 g. FW = 107.8682 g/mol + 14.0067 g/mol Explanation: AgNO3 V =250mL=0.25 L M =1.9 M +3(15.9994 g/mol) = 169.873 g/mol FWCuSO4 = 63.546 g/mol mAgNO3 = (0.178 M)(0.5 L) + 32.066 g/mol × (169.873 g/mol) + 4(15.9994 g/mol) = 15.1187 g = 159.61 g/mol Brodbelt 013 015 004 10.0 points m = (1.9 M)(0.25 L)(159.61 g/mol) CuSO4 How much NaNO3 is needed to prepare 225 = 75.8148 g mL of a 1.55 M solution of NaNO3? 1. 4.10 g 002 (part 2 of 2) 10.0 points · Determine the mass of CuSO4 5H2O that 2. 29.6 g correct must be used to prepare 250 mL of 1.9 M CuSO4(aq). 3. 12.3 g Correct answer: 118.603 g. 4. 0.132 g Explanation: 5. 0.244 g Explanation: FWCuSO4·5H2O = FWCuSO4 + 5(FWH2O) V = 225 mL M =1.55 M = 159.61 g/mol +10(1.0079 g/mol) 1 L soln ? g NaNO = 225 mL × +5(15.9994 g/mol) 3 1000 mL = 249.69 g/mol × 1.55 mol NaNO3 1 L soln × 85 g NaNO3 mCuSO4·5H2O = (1.9 M)(0.25 L)(249.69 g/mol) 1 mol NaNO3 = 118.603 g = 29.6 g NaNO3 ChemPrin3e G 05 Brodbelt 03 13 Version 001 – Calculating Concentrations WKST – vanden bout –(51165) 2 005 10.0 points in the two individual solutions: How many moles of HCl are present in 40.0 mL of a 0.035 M solution? ? mol Ca(OH)2 (soln 1) =0.255 L soln 1. 0.012 mol × 0.250 mol Ca(OH)2 1 L soln 2. 0.0060 mol =0.06375 mol Ca(OH)2 3. 0.25 mol ? mol Ca(OH)2 (soln 2) =0.055 L soln 4. 0.0012 mol × 0.65 mol Ca(OH)2 5. 0.0014 mol correct 1 L soln =0.03575 mol Ca(OH)2 Explanation: Total mol Ca(OH)2 =0.06375 mol V = 40.0 mL M =0.035 M +0.03575 mol =0.0995 mol Ca(OH)2 The total volume of the final solution is the ?molHCl = 40.0mLsoln sum of the volumes of the individual solutions. × 1 L × 0.035 mol HCl Total L = 0.255 L + 0.055 L = 0.31 L soln 1000 mL 1 L soln Molarity is moles solute per L of solution. =0.0014 mol HCl 0.0995 mol Ca(OH) ? M Ca(OH) = 2 2 0.31 L soln =0.321 M Ca(OH)2 Brodbelt 013 515 006 10.0 points Msci 14 0813 What is the final concentration of Ca(OH)2 007 10.0 points when 255 mL of 0.250 M Ca(OH)2 is mixed What is the effective molality of a solution with 55.0 mL of 0.65 M Ca(OH)2? containing 12.0 g of KI in 550 g water? As- 1. 0.390 M sume 100 percent ionic dissociation. 2. 0.642 M 1. 0.072 molal 3. 0.780 M 2. 0.26 molal correct 4. 2.90 M 3. 0.42 molal 5. 0.900 M 4. 0.066 molal 6. 0.321 M correct 5. 0.13 molal 6. 0.59 molal Explanation: V1 Ca(OH)2 = 255 mL [Ca(OH)2]1 = 0.250 M Explanation: mKI =12.0g mH2O = 550 g V2 Ca(OH)2 =55mL [Ca(OH)2]2 = 0.65 M The total moles of Ca(OH)2 in the final Because KI is a 1:1 salt, you get one cation solution will be the sum of the moles present and one anion for every single formula unit Version 001 – Calculating Concentrations WKST – vanden bout –(51165) 3 that dissolves. Therefore, you’ll get a DOU- BLING of the stated molality for the effective Holt da 13 1 sample 1 molality. The formula weight of KI is 166 010 10.0 points g/mol, so the number of moles of KI is What is the molarity of a 3.047 L solution 1 mol KI that is made from 11.29 g of NaCl? 12.0g KI =0.0723 mol KI , 166 g KI Correct answer: 0.0634032 M. and the (stated) molality of the solution Explanation: would then be Vsolution =3.047L mNaCl = 11.29 g 0.0723 mol KI =0.131 m. M = ? 0.550 kg H2O mol solute But recall that the effective molality would be M = twice the stated molality here, so the effective L soln molality is 0.262 m. ? mol NaCl = 11.29 g NaCl × 1 mol NaCl Nlib 11 0020 58.44 g NaCl 008 10.0 points =0.19319 mol NaCl If you mix 3 moles of ethylene glycol (an- tifreeze) in 4165 grams of water, what is the mol NaCl 0.19319 mol molality of the solution? ? M = = L soln 3.047 L =0.0634032 M Correct answer: 0.720288 m. Explanation: nethylene glycol =3mol mwater = 4165 g Molality 08 44a Molality (m) is moles solute per kilogram 011 10.0 points of solvent. The solute is ethylene glycol. The Calculate the molality of sucrose in a solution solvent is water and 4165 g = 4.165 kg. composed of 11.31 g of sucrose (C12H22O11) dissolved in 606 mL of water. 3 mol ethylene glycol ? m = =0.720288 m 4.165 kg H2O Correct answer: 0.054524 m. Nsci 14 0808 Explanation: 009 10.0 points mC12H22O11 = 11.31 g 1.9g of NaCl and 6.1 g of KBr were dissolved VH2O =606mL=0.606 L in 48 g of water. What is the molality of NaCl in the solution? MWC12H22O11 = 342.296 g/mol Correct answer: 0.6773 m. 1 kg m = (0.606 L) =0.606 kg Explanation: H2O 1 L mNaCl =1.9g mKBr =6.1g Thus the molality is mwater =48g mol NaCl moles solute m = mC12H22O11 = NaCl kg water kg solvent 1.9g 11.31 g sucrose 58.4g · mol−1 NaCl 342.296 g/mol sucrose = = 0.048 kg water 0.606 kg H2O =0.6773 m =0.054524 m Version 001 – Calculating Concentrations WKST – vanden bout –(51165) 4 d =0.9344 g/cm3 Msci 14 0809 Assume 1 L of 10.5 M NH3(aq); it will 012 10.0 points contain 10.5 mol NH3 with a mass of What additional information, if any, would enable you to calculate the molality of a 7.35 (10.5 mol)(17.0305 g/mol) = 178.82 g . molar solution of a nonelectrolyte solid dis- solved in water? The density of the 1 L of solution is 1000 cm3 1. Both the density of the solution and 0.9344 g/cm3 · = 934.4g/L , 1 L the molecular weight of the solute would be needed. correct so the total mass of the solution is 934.4 g, which leaves 934.4g − 178.82 g = 755.58 g of 2. Only the density of the solution would be water. needed. Therefore, the molality is 10.5 mol NH 3. Only the molecular weight of the solute m = 3 = 13.8966 m . would be needed. 0.75558 kg solvent 4. Only the density of water would be Nsci 14 0803 needed. 014 10.0 points Formalin is a solution of 40.0% formaldehyde 5. None is needed. (H2CO), 10.0% methyl alcohol (CH3OH), and 50.0% water by mass. Calculate the mole Explanation: fraction of methyl alcohol in formalin. mol solute molarity = L solution Correct answer: 0.0706436. mol solute Explanation: molality = kg solvent In a 100 g formalin, solution the masses are formaldehyde 40.0 g, methyl alcohol 10.0 g, The density of the solution can be used to and water 50.0 g. convert volume (1 L) of solution into mass of ntotal = nCH3OH + nH2CO + nwater solution. Then the molecular weight of the 10 g 40 g solute (given or calculated from the formula) = + can be used to convert the number of moles 32 g/mol 30 g/mol 50 g solute in 1 L solution into mass of solute + in grams. The mass of the solvent is the 18 g/mol difference between the mass of the solution =4.42361 mol and the mass of the solute (both of which The mole fraction of methyl alcohol is have been calculated). Substitute the values nCH3OH into the molality formula and calculate. X = ntotal 10.0g Molality 08 46c 32.0g/mol 013 10.0 points = Calculate the molality of 10.5 M NH3(aq) 4.42361 mol with a density of 0.9344 g/cm3. =0.0706436 Correct answer: 13.8966 m. Nsci 14 0801 Explanation: 015 10.0 points MW = 17.0305 g/mol M = 10.5 M Toluene (C6H5CH3) is a liquid compound Version 001 – Calculating Concentrations WKST – vanden bout –(51165) 5 similar to benzene (C6H6).
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