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KEYPractice Problems for Chapter 12.5: Molarity and Dilution(Read pgs. 382– 387 in the chemistry textbook) Define the following term: Molaritya concentration that states the number of moles of solute in exactly 1 L of solution (solvent + solute) What is the equation to determine Molarity?Molarity (M) = moles of solute Liters of solution → M = molL3. (a) Compare the concentration of a 2.50 M and a 5.0 M solution of MgF2.the larger the numerical value the greater the concentration, therefore, a 5.0 M solution is 2X as concentrated as a 2.50 M solution(b) Compare the concentration of 1.0 liter of a 0.20 M solution and 2.00 L of a 0.10 M solution.the volume of solution makes no difference, it is only the molarity that determines concentration - therefore, the 0.20 M solution is 2X as concentrated as the 0.10 M solution, regardless of the fact that there is a greater volume of the 0.10 M solution4. Circle the MORE CONCENTRATED solution in each of the following:(a) 500 mL of a 1.0 M solutionor10.0 mL of a 1.5 M solution(b) 2.0 L of a 0.10 M solutionor0.500 L of a 0.010 M solution(a) 1 L of a 2.0 M solutionor50.0 mL of a 2.0 M solutionthese two solutions have the SAME concentration, the volume makes no difference5. Comment on the molarity of the remaining solution in each of the following situations:(a) A 2.0 M solution is spilled and about half is lost.molarity remains the same b/c an equal amount of solute and solvent are lost = less solution and that solution has the same concentration(b) The lid is left off of the container of a 2.0 M solution and about half the volume is lost due to evaporation. molarity increases b/c only solvent was lost....there is now the same amount of solute dissolved in a smaller amount of solvent = less solution and that solution is more concentrated(c) A student adds 100 mL of distilled water to 1 L of a 2.0 M solution.molarity decreases b/c additional solvent was added.....there is now the same amount of solute dissolved in a larger amount of solvent = more solution and that solution is less concentrated6. What is another way to read the term 2.0 M NaCl?2.0 moles NaCl 1 Liter solution → 2.0 molL7. Convert the following molarities to moles per liter(a) 6.0 M HCl6.0 moles HCl1 Liter of Solution → 6.0 molL(b) 0.20 M NaOH0.20 moles NaOH1 Liter of Solution → 0.20 molL (c) 2.50 M Sr(OH)22.50 moles Sr(OH)21 Liter of Solution → 2.50 molL C-LevelCalculating Molarity when you are given:1) moles solute, and2) volume of solution8. Calculate the molarity in each of the following:(a) 2.00 mol of glucose in 4.00 L of solution Molarity = 2.00 moles glucose4.00 Liter of solution = 0.500 M solution of glucose(b) 0.500 mol of sucrose in 0.200 L of solution Molarity = 0.500 moles sucrose2.00 Liter of solution = 0.250 M solution of sucrose(c) 0.0020 moles KOH in 0.100 L of solutionMolarity = 0.0020 moles KOH0.100 Liter of solution = 0.020 M solution of KOH(d) 5.0 × 10?2 moles of AlCl3 in 50.0 mL of solutioni) 50.0 mL = 0.0500 L of solutionii) Molarity = 5.0 × 10-2moles AlCl30.0500 Liter of solution = 1.00 M solution of AlCl3(e) Calculate the molarity of an aqueous solution containing 0.25 moles of NaNO3 dissolved in 500. mL of solution.i) 500. mL = 0.500 L of solutionii) Molarity = 0.25 moles NaNO30.500 Liter of solution = 0.500 M solution of NaNO3B-LevelCalculating Volume when you are given:1) moles solute, and2) Molarity 9. Calculate the volume of each of the following: (a) liters of 2.0 M solution which contain 0.50 moles of solute.i) 2.0 M is the same as 2.0 molesLii) use the Molarity equations 2.00 moles L × x liters = 0.50 moles → x = 0.50 moles2.00 molesL = 0.25 L Alternatively: set up an equality (Quicker!!) 2.0 molesL = 0.50 molesx → = 0.50 moles2.00 molesL = 0.25 L(b) liters of 0.10 M solution which contain 0.050 moles of solute.i) 0.10 M is the same as 0.10 molesLii) set up an equality0.10 molesL = 0.050 molesx → x = 0.50 L(c) milliliters of 0.10 M solution which contain 0.010 moles of solute.i) 0.10 M is the same as 0.10 molesLii) set up an equality0.10 molesL = 0.010 molesx → x = 0.100 Liii) convert Liters to milliliters………0.100 L = 100. mL(d) milliliters of 2.50 M solution which contain 0.10 moles of solute.i) 2.50 M is the same as 2.50 molesLii) set up an equality2.50 molesL = 0.10 molesx → x = 0.040 Liii) convert Liters to milliliters……… 0.040 L = 40. mL(e) Calculate the milliliters of 0.35 M KOH solution required in order to obtain 0.02 moles of KOH.0.35 M is the same as 0.35 molesLii) set up an equality0.35 molesL = 0.020 molesx → x = 0.057 Liii) convert Liters to milliliters……… 0.057 L = 57 mLB-LevelCalculating Moles when you are given:1) volume of solution, and2) Molarity 10. Calculate the number of moles in each of the following volumes: moles of solute in 0.500 L of a 2.0 M solution2.0 M is the same as 2.0 molesLii) 2.00 moles L × 0.500 L = 1.00 moles moles of solute in 0.050 L of a 0.10 M solution0.10 M is the same as 0.10 molesLii) 2.00 moles L × 0.050 L = 0.10 moles moles of solute in 250. mL of a 2.0 M solutionconvert milliliters to Liters ……… 250. m L = 0.250 L2.0 M is the same as 2.0 molesLii) 2.00 moles L × 0.250 L = 0.500 moles moles of solute in 25 mL of a 0.010 M solutionconvert milliliters to Liters ……… 25 m L = 0.025 L0.010 M is the same as 0.010 molesLii) 0.010 moles L × 0.025 L = 2.5 × 10─4 moles Calculate the moles of NaOH in 400. mL of a 0.50 M solution.convert milliliters to Liters ……… 400. m L = 0.400 L0.50 M is the same as 0.50 molesLii) 0.50 moles L × 0.400 L = 0.20 moles Calculate the moles of CH4 in 0.050 L of a 0.10 M solution0.10 M is the same as 0.10 molesLii) 0.10 moles L × 0.050 L = 5.0 × 10─3 moles A-LevelCalculating Molarity when you are given:grams of solute, andvolume of solution11. Calculate the molarity of each of the following:(a) 60.0 g of NaOH in 0.250 L of solution Molar mass of NaOH = 40.01 g/ molConvert grams to moles60.0 g ÷ 40.01 g/ mol = 1.50 moles of NaOHMoles ÷ volume (in liters!)Molarity = 1.50 moles NaOH0.250 Liter of solution = 6.00 M solution of NaOH (b) 75.0 g of KNO3 in 0.350 L of solutionMolar mass of KNO3 = 101.11 g/ molConvert grams to moles75.0 g ÷ 101.11 g/ mol = 0.742 moles of NaOHMoles ÷ volume (in liters!)Molarity = 0.742 moles KNO30.350 Liter of solution = 2.12 M solution of KNO311. Continued:(c) 73.0 g of HCl in 2.00 L of solutionMolar mass of HCl = 36.46 g/ molConvert grams to moles73.0 g ÷ 36.46 g/ mol = 2.00 moles of HClMoles ÷ volume (in liters!)Molarity = 2.00 moles HCl2.00 Liter of solution = 1.00 M solution of HCl(d) 5.85 g of NaCl in 40.0 mL of solutionMolar mass of NaCl= 58.44 g/ molConvert grams to moles5.85 g ÷ 58.44 g/ mol = 0.100 moles of NaClConvert milliliters to liters : 40.0 mL = 0.0400 LMoles ÷ volume (in liters!)Molarity = 0.100 moles NaCl0.0400 Liter of solution = 2.50 M solution of NaCl(e) 30.4 g of LiBr in 350. mL of solutionMolar mass of LiBr = 86.84 g/ molConvert grams to moles30.4 g ÷ 86.84 g/ mol = 0.350 moles of LiBrConvert milliliters to liters : 350. mL = 0.350 LMoles ÷ volume (in liters!)Molarity = 0.100 moles LiBr0.350 Liter of solution = 1.00 M solution of LiBr(f) Calculate the molarity of a solution that has 10.0 g of Li2O in 0.250 L of solution.Molar mass of Li2O = 29.88 g/ molConvert grams to moles10.0 g ÷ 29.88 g/ mol = 0.33467 moles of Li2OMoles ÷ volume (in liters!)Molarity = 0.33467 moles Li2O0.250 Liter of solution = 1.34 M solution of Li2OA-LevelCalculating volume of solution when you are given:grams of solute, andMolarity12. Calculate the volume of solution for each of the following: liters of a 2.00 M NaCl solution containing 67.3 g of NaCl:Molar mass of NaCl= 58.44 g/ molConvert grams to moles67.3 g ÷ 58.44 g/ mol = 1.15 moles of NaCl2.00 M is the same as 2.00 molesLiv)set up an equality2.00 molesL = 1.15 molesx → x = 0.575 L of solution Liters of 2.25 M HCl solution that will provide 4.12 g of HCl. Molar mass of HCl= 36.46 g/ molConvert grams to moles4.12 g ÷ 36.46 g/ mol = 0.113 moles of HCl2.25 M is the same as 2.25 molesLiv)set up an equality2.25 molesL = 0.113 molesx → x = 0.0502 L of solution(c) milliliters of a 0.120 M Na2CO3 solution that will yield 12.5 g of Na2CO3 Molar mass of Na2CO3 = 105.99 g/ molConvert grams to moles12.5 g ÷ 105.99 g/ mol = 0.118 moles of Na2CO30.120 M is the same as 0.120 molesLiv)set up an equality0.120 molesL = 0.118 molesx → x = 0.983 L of solutionConvert liters to milliliters : 0.983 L = 983 mL(d) Calculate the number of milliliters of 2.70 M LiOH solution that are needed to provide 30.0 g of LiOH .Molar mass of LiOH = 23.95 g/ molConvert grams to moles30.0 g ÷ 23.95 g/ mol = 1.25 moles of LiOH2.70 M is the same as 2.70 molesLiv)set up an equality2.70 molesL = 1.25 molesx → x = 0.463 L of solutionConvert liters to milliliters : 0.463 L = 463 mL13. What is a stock solution?a preexisting solution of known molarity14. (a) Define the term dilutionwhen more solvent is added to a a portion of stock solution to make it less concentrated (b) What is the dilution equation? Define each of its terms. M1V1 = M2V2C1 = Molarity of the MORE C2 = Molarity of the LESS Concentrated solustion concentrated solutionV1 = volume of the MOREV2 = volume of the LESSconcentrated solutionconcentrated solution15. List calculations for which the dilution equation may be used. Volume of stock solution required to make a dilution Volume of diluted solution madeMolarity of the dilutionMolarity of the original stock solutionB-LevelSolving for any variable using the dilution equation.16. Calculate the final concentration (Molarity) of the solution in each of the following:(a) Water is added to 0.150 L of a 6.00 M HCl stock solution to give a volume of 0.500 L.i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T knowM1 = 6.00 M M2 = xV1 = 0.150 LV2 = 0.500 Lii) Use the dilution equation to solve for the unknown:M1V1 = M2V2 *setup: (6.00 M) ( 0.150 L) = x (0.500L) → 6.00 M(0.150 L) = 0.500 L x→ 0.900 M-L = 0.500L x → 0.900 m-L0.500 L = x → x = 1.80 Molar HCl solution*NOTE: This is the only problem for which the algebra will be shown. For all other dilution problems, only the setup will be shown(b) A 10.0-mL sample of a 2.50 M KCl solution is diluted with water to 0.250 L.i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T knowM1 = 2.50 M M2 = xV1 = 10.0mL*V2 = 0.250 L = 250 mL*NOTE: volume can be in either mL or L, as long as both volumes are the sameii) Use the dilution equation to solve for the unknown:M1V1 = M2V2 setup: (2.50 M) ( 10.0m L) = x (250 mL) → x = 0.100 Molar KCl solution16. Continued:(c) Water is added to 0.250 L of a 12.0 M KBr solution to give a volume of 1.00 L.i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T knowM1 = 12.0 M M2 = xV1 = 0.250 LV2 = 1.00 L ii) Use the dilution equation to solve for the unknown:M1V1 = M2V2 setup: (12.0 M) ( 0.250 L) = x (1.00 L) → x = 3.00 M KBr sol’n17. Determine the final volume (mL) for each of the following:(a) diluting 50.0 mL of a 12.0 M NH4Cl solution to give a 2.00 M NH4Cl solution.i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T knowM1 = 12.0 M M2 = 2.00 MV1 = 50.0 mLV2 = x ii) Use the dilution equation to solve for the unknown:M1V1 = M2V2 setup: (12.0 M) ( 50.0 mL) = (2.00 M) x → x = 300. mL of 2.00 M NH4Cl solution(b) diluting 18.0 mL of a 15.0 M NaNO3 solution to give a 1.50 M NaNO3 solution.i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T knowM1 = 15.0 M M2 = 1.50 MV1 = 18.0 mLV2 = xii) Use the dilution equation to solve for the unknown:M1V1 = M2V2 setup: (15.0 M) ( 18.0 mL) = (1.50 M) x → x = 180. mL of 1.50 M NaNO3 solution17. Continued:(c) diluting 4.50 mL of a 18.0 M H2SO4 solution to give a 2.50 M H2SO4 solution.i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T knowM1 = 18.0 M M2 = 2.50 MV1 = 4.50 mLV2 = x ii) Use the dilution equation to solve for the unknown:M1V1 = M2V2 setup: (18.0 M) ( 4.50 mL) = (2.50 M) x → x = 32.4 mL of 2.50 M H2SO4 solution18. Determine the volume (mL) of stock solution required to prepare each of the following:(a) 255 mL of a 0.200 M HNO3 solution from a 4.00 M HNO3 solution:i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T knowM1 = 4.00 M M2 = 0.200 MV1 = xV2 = 255 mLii) Use the dilution equation to solve for the unknown:M1V1 = M2V2 setup: (4.00 M) x = (0.200 M) ( 255 mL) → x = 12.8 mL of 4.00 MHNO2 solution are required(b) 715 mL of a 0.100 M MgCl2 solution using a 6.00 M MgCl2 solution.i) first, assign values to C1V1 = C2V2 so that you can determine what you DON'T knowM1 = 6.00 M M2 = 0.100 MV1 = xV2 = 715 mLii) Use the dilution equation to solve for the unknown:M1V1 = M2V2 setup: (6.00 M) x = (0.100 M) ( 715 mL) → x = 11.9 mL of 6.00 M MgCl2 solution required ................
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