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Ag Chem
MOLARITY
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molarity_worksheet_.doc
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When completed check your work Molarity Worksheet #1 name _________________________ What does molarity mean? Number of moles of solute 1 liter solution What is the molarity of a solution that contains 4.53 moles of lithium nitrate in 2.85 liters of solution? 4.53 mol LiNO3 = 1.59 M LiN03 2.85 L soln What is the molarity of a solution that contains 0.00372 moles hydrochloric acid in 2.39 x 10-2 liters of solution? 0.00372 mol HCL = 0.156 M HCL 2.39x10-2 L soln A flask contains 85.5 g C12H22O11 (sucrose) in 1.00 liters of solution. What is the molarity? 85.5g sucrose x 1 mol sucrose = 0.250 M sucrose 1.00 L soln 342.34g sucrose A beaker contains 214.2 grams osmium (III) fluoride in 0.0673 liters of solution. What is the molarity? 214.2g OsF3 x 1 mol OsF3 = 12.9 M OsF3 0.0673 L soln 247.23g OsF3 Calculate the molarity if a flask contains 1.54 moles potassium sulfate in 125 ml of solution. 1.54 mol K2SO4 = 12.3 M K2SO4 0.125 L soln A chalice contains 36.45 grams ammonium chlorite in 2.36 liters of solution - calculate the molarity. 36.45g NH4ClO2 x 1 mol NH4ClO2 = 0.181 M NH4ClO2 2.36 L soln 85.50g NH4ClO2 What is the molarity of a solution that contains 14.92 grams magnesium oxalate in 3.65 ml of solution? 14.92g MgC2C4 x 1 mol MgC2C4 = 36.4 mol MgC2C4 0.00365 L soln 112.32g MgC2C4 What mass of lithium phosphate would you mass to make 2.5 liter of 1.06 M lithium phosphate solution? 2.5 L soln x 1.06 mol Li3PO4 x 115.79g Li3PO4 = 310g Li3PO4 1 L soln 1 mol Li3PO4 If you evaporated 250. mL of a 3.5 M solution of iron (II) nitrite, what mass of iron (II) nitrite would you recover? 0.250 L soln x 3.5 mol Fe(NO2)2 x 147.86g Fe(NO2)2 = 130g Fe(NO2)2 1 L soln 1 mol Fe(NO2)2 A chemist has 4.0 g of silver nitrate and needs to prepare 2.0 L of a 0.010 M solution. Will there be enough silver nitrate? If so, how much silver nitrate will be left over? 2.0 L soln x 0.010 mol AgNO3 x 169.88g AgNO3 = 3.4g AgNO3 Used/Needed 1 L soln 1 mol AgNO3 There is enough silver nitrate available. 4.0g AgNO3 – 3.4g AgNO3 = 0.6 g AgNO3 12. A rabbit pours 500.00 mL of a 3.0000 molar solution of sodium hydroxide into a 2.000 liter volumetric flask and fills the flask up with water. What is the new molarity of the solution? 0.50000 L soln x 3.0000 mol NaOH = 1.5000 mol NaOH 1 L soln 1.500 mol NaOH = 0.75000 M NaOH 2 L soln Solve the following solutions Stoichiometry problems: 1. How many grams of silver chromate will precipitate when 150. mL of 0.500 M silver nitrate are added to 100. mL of 0.400 M potassium chromate? 2 AgNO3(aq) + K2CrO4(aq) Ag2CrO4(s) + 2 KNO3(aq) 0.150 L AgNO3 0.500 moles AgNO3 1 moles Ag2CrO4 331.74 g Ag2CrO4 = 12.4 g Ag2CrO4 1 L 2 moles AgNO3 1 moles Ag2CrO4 0.100 L K2CrO4 0.400 moles K2CrO4 1 moles Ag2CrO4 331.74 g Ag2CrO4 = 13.3 g Ag2CrO4 1 L 1 moles K2CrO4 1 moles Ag2CrO4 2. How many mL of 0.280 M barium nitrate are required to precipitate as barium sulfate all the sulfate ions from 25.0 mL of 0.350 M aluminum sulfate? (93.8 mL barium nitrate) 3 Ba(NO3)2(aq) + Al2(SO4)3(aq) 3 BaSO4(s) + 2 Al(NO3)3(aq) 0.0250 L Al2(SO4)3 0.350 moles Al2(SO4)3 3 moles Ba(NO3)2 1 L = 0.0938 L Ba(NO3)2 1 L 1 moles Al2(SO4)3 0.280 moles Ba(NO3)2 3. 25.0 mL of 0.350 M NaOH are added to 45.0 mL of 0.125 M copper (II) sulfate. How many grams of copper (II) hydroxide will precipitate? 2 NaOH(aq) + CuSO4(aq) Cu(OH)2(s) + Na2SO4(aq) 0.0250 L NaOH 0.350 moles NaOH 1 moles Cu(OH)2 97.57 g Cu(OH)2 = 0.427 g Cu(OH)2 1 L NaOH 2 moles NaOH 1 mole Cu(OH)2 0.0450 L CuSO4 0.125 moles CuSO4 1 moles Cu(OH)2 97.57 g Cu(OH)2 = 0.549 g Cu(OH)2 1 L NaOH 1 moles CuSO4 1 mole Cu(OH)2 4. What volume of 0.415 M silver nitrate will be required to precipitate as silver bromide all the bromide ion in 35.0 mL of 0.128 M calcium bromide? 2 AgNO3(aq) + CaBr2(aq) Ca(NO3)2(aq) + 2 AgBr(s) 0.0350 L CaBr2 0.128 moles CaBr2 2 moles AgNO3 1 L AgNO3 = 0.0216 L AgNO3 1 L CaBr2 1 moles CaBr2 0.415 mole AgNO3 5. What volume of 0.496 M HCl is required to neutralize 20.0 mL of 0.809 M sodium hydroxide? HCl(aq) + NaOH(aq) NaCl(aq) + H(OH)(l) 0.0200 L NaOH 0.809 mole NaOH 1 mole HCl 1 L HCl = 0.0326 L HCl 1 L NaOH 1 mole NaOH 0.496 mole HCl 6. How many mL of 0.715 M HCl is required to neutralize 1.25 grams of sodium carbonate? (producing carbonic acid) 2 HCl(aq) + Na2CO3(s) 2 NaCl(aq) + H2CO3(aq) 1.25 g Na2CO3 1 mole Na2CO3 2 mole HCl 1 L HCl = 0.0330 L HCl 105.99 g Na2CO3 1 mole Na2CO3 0.715 mole HCl 7. How many grams of magnesium hydroxide will precipitate if 25.0 mL of 0.235 M magnesium nitrate are combined with 30.0 mL of 0.260 M potassium hydroxide? Mg(NO3)2(aq) + 2 KOH 2 KNO3(aq) + Mg(OH)2(s) 0.0250 L Mg(NO3)2 0.235 mole Mg(NO3)2 1 mole Mg(OH)2 58.33 g Mg(OH)2 = 0.343 Mg(OH)2 1 L Mg(NO3)2 1 mole Mg(NO3)2 1 mole Mg(OH)2 0.0300 L KOH 0.260 mole KOH 1 mole Mg(OH)2 58.33 g Mg(OH)2 = 0.227 g Mg(OH)2 1 L KOH 2 mole KOH 1 mole Mg(OH)2 8. 60.0 mL of 0.322 M potassium iodide are combined with 20.0 mL of 0.530 M lead (II) nitrate. How many grams of lead (II) iodide will precipitate? 2 KI(aq) + Pb(NO3)2(aq) 2 KNO3(aq) + PbI2(s) 0.0600 L KI 0.322 mole KI 1 mole PbI2 461.00 g PbI2 = 4.45 g PbI2 1 L KI 2 mole KI 1 mole PbI2 0.0200 L Pb(NO3)2 0.530 mole Pb(NO3)2 1 mole PbI2 461.00 g PbI2 = 4.89 g PbI2 1 L Pb(NO3)2 1 mole Pb(NO3)2 1 mole PbI2
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