SQSM-02

# SQSM-02 - Chapter 2 Solutions 1. How many grams of CuSO4...

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Chapter 2 Solutions 2-1 1. How many grams of CuSO 4 are required to make 650. mL of a 0.115-M solution? 44 4 4 1 L solution 0.115 mol CuSO 159.608 g CuSO 650. mL solution 11.9 g CuSO 1000 mL solution L solution 1 mol CuSO ×× × = 3. How many grams of Na 2 SO 4 are required to make 90.0 mL of a solution that is 0.200 M in Na 1+ ? 1 24 1 1 L solution 0.200 mol Na 1 mol Na SO 142.042 g Na SO 90.0 mL solution 1.28 g Na SO 1000 mL solution L solution 1 mol Na SO 2 mol Na + + × × = 5. How many mmoles of chloride ion are in 55.0 mL of 0.0688 M BaCl 2 solution? 1- 1- 2 2 0.0688 mmol BaCl 2 mmol Cl 55.0 mL solution 7.57 mmol Cl 1 mL solution 1 mmol BaCl = 7. How many mL of 0.124 M Ba(OH) 2 are required to deliver 38.6 mmol of hydroxide ion ? 1- 2 1- 2 1 mmol Ba(OH) 1 mL solution 38.6 mmol OH 156 mL solution 0.124 mmol Ba(OH) 2 mmol OH = 9. How many grams of sodium should be added to 15.0 g Hg to make a mixture in which the mole fraction of sodium is 0.800? Na 1 mol Hg moles of mercury =15.0 g Hg 0.0748 mol Hg 200. 59 g Hg n Use moles of Hg and the mole fraction equation: X 0.800 , 0.0748 n 0.0598 rearrange to eliminate denominator: 0.0598 0.800n n n 0 ×= == + += = = 0.299 mol Na .200 22.99 g Na Calculate the grams of Na: 0.299 mol Na 6.88 g Na 1 mol Na 11. The density of a 1.140-M solution of NH 4 Cl at 20 o C is 1.0186 g/mL. What mass of water does 100. mL of this solution contain? The total mass of this solution is 1.0186 g/mL x 100. mL = 102 (three significant figures) The mass of NH 4 Cl in this solution is 4 4 1.140 mol NH Cl 53.49 g 0.100 L = 6.10g NH Cl 1 L NH Cl 1 mol NH Cl The mass of water is then the difference between the total mass of the solution and the mass of the NH 4 Cl: 102 g solution - 6 g NH 4 Cl = 96 g water. 13. What mass of Fe(ClO 4 ) 3 is required to make 275 mL of a solution that is 0.100 M in ClO 4 1- ? 1- 43 4 1 4 1 mol Fe(ClO ) 354 g Fe(ClO ) 0.100 mol ClO 0.0275 L 3.25 g Fe(ClO ) 1 L 1 mol 3 mol ClO × = 15. How many grams of CaCl 2 should be added to 50.0 g of water to make a solution in which the mole fraction of Cl 1- is 0.150? First get the number of moles of water: 2 22 2 1 mol H O 50.0 g H O = 2.78 mol H O 18.0 g H O × If n mol CaCl 2 are dissolved, n mol Ca 2+ and 2n mol Cl 1- form in solution, or 3n mol ions. -1 2 2 2 2 The mole fraction of chloride ion is 0.150, so 2n mol Cl 0.150 3n mol ions 2.78 mol of H O Rearrange and solve for n: 0.450n 0.417 2n n = 0.269 mol CaCl 111 g CaCl 0.269 mol CaCl 1 mol = + × 2 29.9 g CaCl =

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Solutions 2-2 17. A concentrated solution of phosphoric acid is 75% H 3 PO 4 by mass and has a density of 1.57 g/mL. a) What is the molarity of the concentrated phosphoric acid? Assume 100. g of concentrated phosphoric acid then 75% translates to 75 g of H 3 PO 4 . Calculate the moles of H 3 PO 4 . 34 1 mol H PO 75 g H PO =0.77 mol H PO 97.994 g × Use the density to calculate the volume of 100 g of concentrated phosphoric acid. 1 mL 100. g conc H PO =63.7 mL conc H PO 1.57 g × The molarity of the solution can then be calculated, 0.77 mol H PO =12. M H PO 0.0637 L solution b) How many mL of the concentrated acid would be required to prepare 1.5 L of a 0.20-M solution?
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## This note was uploaded on 01/27/2011 for the course CHEM 201 taught by Professor Wilson during the Spring '10 term at N.C. Central.

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SQSM-02 - Chapter 2 Solutions 1. How many grams of CuSO4...

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