1. Define molarity of a solution. Be specific with solute, solvent or solution in the definition.

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1 Chem 115L Moles, Molarity, Dilution and Stoichiometry Name Two of the topics from Chem 110 that you must be able to use in Chem 115 are stoichiometry and solution concentrations/dilutions. The concentration unit most often used in lab is molarity. In all of the following calculations show work and express the final answer in the correct number of significant figures. 1. Define molarity of a solution. Be specific with solute, solvent or solution in the definition. 2. One can have a 6.0 M HCl solution a) only when one has 1 L of solution b) only when one has 100 ml of solution c) when one has any volume of solution 3. a) Formula of copper(ii) sulfate pentahydrate is : b) In what physical state (s, l, or g) will one find the above? c) Molar mass of the hydrated salt (with unit): Show work : 4. Calculate the moles mass of solute and moles of ions present in each solution: ml of 2.00 M sodium thiosulfate pentahydrate Moles of solute (mol) Mass of solute (g) Formula of cation Formula of anion Moles and M of cation Moles and M of anion Total moles and M of ions ml of.100 M ammonium sulfate 5. What volume of 6.00 M hydrochloric acid will contain moles of HCl? ml

2 6. Give the buret reading with units to the correct number of significant figures in the picture below. Refer to: 24 ml 25 ml buret reading : unit : 7) A sample of 98g of sulfuric acid is dissolved in water to prepare a.50m solution. What is the volume of the solution made, in liters? 8. If ml of water was added to the above solution, the new solution s molarity will be 9. To prepare 10. 0L of 0.15 M HCl, one would need ml of 15.0 M HCl and ml of water. 10) If 4.00L of water is added to 1.00 L of a 6.00M solution of calcium chloride, what is the new molarity of the solution made? Assume additive volumes. 11) 20.0 ml of M aq.solution of A is mixed with 60.0 ml of M aq. solution of B. If A and B do not react with each other, the final molarities of the solutions of A = B = Assume additive volumes.

3 12. The molarity of a solution prepared by mixing 300. ml of a M solution of sulfuric acid with 300. ml of a 4.00 M solution of sulfuric acid will be. 13) 50.0 ml of 3.00 M HCl is mixed with 50.0 ml of 1.00 M NaOH. What is the molarity of the HCl in the final solution. (Remember there is a reaction between HCl and NaOH) Setup the ICF Chart after writing the balanced equation: (Use the sample on the last page) Initial moles + + Which is the Limiting reactant? DO NOT USE THESE answers in the division in any subsequent calculations!!!!!!! They are only to be used to determine the LR. In all the following steps use initial moles of LR and mole ratios to find the desired moles Changed moles (used or produced) Final moles

4 ml of M FeCl 3 solution is reacted with ml of M NaOH solution. (a) Write the balanced molecular equation with physical states in paranthesis. (b) Setup the ICF Chart after writing the balanced equation: ( ) + ( ) ( ) + ( ) Initial moles Which is the Limiting reactant? DO NOT USE THESE answers in the division in any subsequent calculations!!!!!!! They are only to be used to determine the LR. Changed moles (used or produced) Final moles Final molarities of all solutions (c) From the above chart, mass of the precipitate formed: g (d) Calculate the total molarities of EACH ION in the resultant solution after reaction. 1. Solution : M ; cation : M; anion : M 2. Solution : M ; cation : M; anion : M 3. Solution : M ; cation : M; anion : M Total M for each ion: Ion formula Molarity

5 SAMPLE Limiting reactant calculation and ICF Chart 1. Consider the following reaction: NH4NO3 + Na3PO4 (NH4)3PO4 + NaNO3 Which reactant is limiting, assuming we started with 30.0 grams of ammonium nitrate and 50.0 grams of sodium phosphate. What is the mass of each product that can be formed? What mass of the excess reactant(s) is left over? 3 NH 4 NO 3 Na 3 PO 4 (NH 4 ) 3 PO 4 3 Na NO 3 Initial moles Zero Zero 30.0 g / (g/mole) 50.0 g/163.94(g/mole) Which is the Limiting reactant? /3 = Smaller than , Hence NH 4 NO 3 is LR /1 = Na 3 PO 4 is in excess DO NOT USE THESE answers in the division in any subsequent calculations!!!!!!! They are only to be used to determine the LR. In all the following steps use initial moles of LR and mole ratios to find the desired moles Changed moles (used or produced) (all of initial LR) (calculate from initial moles of LR and the mole ratio) Final moles Zero = Changed moles: moles (LR) NH 4 NO 3 used x { 1mol Na 3PO 4 used } = mol Na 3PO 4 used 3mole NH 4 NO 3 used moles (LR) NH 4 NO 3 used x { 1mol (NH 4 ) 3 PO 4 formed } = mol (NH 4 ) 3 PO 4 formed 3mole NH 4 NO 3 used moles (LR) NH 4 NO 3 used x { 3mol Na NO 3formed } = mol Na NO 3formed 3mole NH 4 NO 3 used

6 Products: (NH 4 ) 3 PO 4 : (149.09) = 18.6 g ; Na NO 3 : (84.99) = 31.8 g Left over excess reactant Na 3 PO 4 = (163.94) = 29.5 g

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