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4 Aqueous Reactions Solutions to Exercises 4.64 Calculate the mol of at the two concentrations; the difference is the mol NaCl required to increase the concentration to the desired level. 0.118 mol = 0.5428 = 0.54 mol L 0.138 mol = 0.6348 = 0.63 mol L (0.6348 - 0.5428) = 0.092 = 0.09 mol NaCl (2 decimal places and 1 sig fig) 0.092 mol NaCl 58.5 NaCl = 5.38 = 5 NaCl mol 4.65 Analyze. Given: g alcohol/100 mL blood; molecular formula of alcohol. Find: molarity (mol/L) of alcohol. Plan. Use the molar mass (MM) of alcohol to change (g/100) mL to (mol/100 mL) then mL to L. Solve. MM of alcohol = 2(12.01) + 6(1.1008) + 1(16.00) = 46.07 g alcohol/mol BAC = 100 0.08 mL g alcohol blood 46.07 1 mol g alcohol alcohol X 1000 1L mL = 0.0174 = 0.02 M alcohol 4.66 Analyze. Given: BAC (definition from Exercise 4.65), vol of blood. Find: mass alcohol in bloodstream. Plan. Change BAC (g/100 mL) to (g/L), then times vol of blood in L. Solve. BAC = 0.10 g/100 mL 100 mL blood 0.10 g alcohol X 1000 1L mL 5.0 L blood = 5.0 g alcohol 4.67 Plan. Proceed as in Sample Exercises 4.11 and 4.13. M = mol = MM g (MM is the symbol for molar mass in this manual.) Solve. (a) Check. (0.25 120) ≈ 30; 30 X 0.18 ≈ 5.4 g KBr (b) 14.75 Ca(NO₃)₂ 164.09 1 mol Ca(NO₃)₂ Ca(NO₃)₂ = 0.06537 M Ca(NO₃)₂ Check. (15/1.5) ≈ 10; 10/160 = 5/80 ≈ 0.06 M Ca(NO₃)₂ = 10.2 mL solution " 90