Chem quiz - In the laboratory you dissolve 16.4 g of zinc...

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In the laboratory you dissolve 16.4 g of zinc chloride in a volumetric flask and add water to a total volume of 250 mL. What is the molarity of the solution ? 0.481 M. What is the concentration of the zinc cation ? 0.481 M. What is the concentration of the chloride anion ? 0.963 M. Feedback: zinc chloride = ZnCl 2 = 136 g/mol Molarity = moles solute liters soln 1. Determine the number of moles of solute present: mol ZnCl 2 = 16.4 g ZnCl 2 1 mol = 0.120 mol ZnCl 2 136 g ZnCl 2 2. Divide by the volume to get the molarity of the solution: L soln = 250 mL soln 1 L = 0.250 L soln 1000 mL M ZnCl 2 soln = 0.120 mol ZnCl 2 = 0.481 M ZnCl 2 0.250 L soln 3. Determine the molarity for each ion from the number of ions formed when ZnCl 2 dissolves in water: ZnCl 2 (s) Zn 2+ (aq) + 2 Cl - (aq) M Zn 2+ (aq) = 1 mol Zn 2+ 0.481 M = 0.481 M Zn 2+ (aq) mol ZnCl 2 M Cl - (aq) = 2 mol Cl - 0.481 M = 0.963 M Cl - (aq) mol ZnCl 2 You need to make an aqueous solution of 0.241 M aluminum acetate for an experiment in lab, using a 500 mL volumetric flask. How much solid aluminum acetate should you add ? ... 24.6 grams Feedback: aluminum acetate = Al(CH 3 COO) 3 = 204 g/mol Molarity = moles solute liters soln 0.241 M Al(CH 3 COO) 3 = 0.241 mol Al(CH 3 COO) 3 L soln 1. Determine the number of moles of solute present in 500 mL, or 0.500 L, of the solution:
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mol Al(CH 3 COO) 3 = 0.500 L soln 0.241 mol Al(CH 3 COO) 3 = 0.121 mol Al(CH 3 COO) 3 L soln 2. Convert to grams of solute: g Al(CH 3 COO) 3 = 0.121 mol Al(CH 3 COO) 3 204 g Al(CH 3 COO) 3 = 24.6 g Al(CH 3 COO) 3 mol Al(CH 3 COO) 3 Put 24.6 grams of solid aluminum acetate into the flask and fill to the line with distilled water. How many milliliters of an aqueous solution of 0.248 M chromium(III) sulfate is needed to obtain 15.8 grams of the salt ? .... 162 mL Feedback: chromium(III) sulfate = Cr 2 (SO 4 ) 3 = 392.2 g/mol Molarity = moles solute liters soln 0.248 M Cr 2 (SO 4 ) 3 = 0.248 mol Cr 2 (SO 4 ) 3 L soln 1. Determine the number of moles of solute: mol Cr 2 (SO 4 ) 3 = 15.8 g Cr 2 (SO 4 ) 3 1 mol = 4.029E-2 mol Cr 2 (SO 4 ) 3 392.2 g Cr 2 (SO 4 ) 3 2. Determine the volume of solution required: mL Cr 2 (SO 4 ) 3 soln = 4.029E-2 mol Cr 2 (SO 4 ) 3 1 L soln 1000 mL = 162 mL Cr 2 (SO 4 ) 3 soln 0.248 mol Cr 2 (SO 4 ) 3 L You need to make an aqueous solution of 0.195 M chromium(II) acetate for an experiment in lab, using a 125 mL volumetric flask. How much solid chromium(II) acetate should you add ? ... 4.15 grams Feedback: chromium(II) acetate = Cr(CH 3 COO) 2 = 170 g/mol Molarity = moles solute liters soln 0.195 M Cr(CH 3 COO) 2 = 0.195 mol Cr(CH 3 COO) 2 L soln 1. Determine the number of moles of solute present in 125 mL, or 0.125 L, of the solution: mol Cr(CH 3 COO) 2 = 0.125 L soln 0.195 mol Cr(CH 3 COO) 2 = 2.44E-2 mol Cr(CH 3 COO) 2 L soln
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2. Convert to grams of solute: g Cr(CH 3 COO) 2 = 2.44E-2 mol Cr(CH 3 COO) 2 170 g Cr(CH 3 COO) 2 = 4.15 g Cr(CH
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Chem quiz - In the laboratory you dissolve 16.4 g of zinc...

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