Chapter 10 problems

Chapter 10 problems - 229/229 Solutions to Problems from...

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Unformatted text preview: 229/229 Solutions to Problems from Chapter 10 1. Complete the following table for aqueous solutions of sodium sulfate. [Show work for partial credit.] Mass of Solute Volume of Solution Molarity (a) 18.8 g 352 mL 0.375 M (b) 102 g 2.92 L 0.246 M (c) 4.13 g 0.291 L 0.100 M (a) MM of Na 2 SO 4 = [2(22.99) + 32.07 + 4(16.00)] g/mol = 142.05 g/mol 32/32 2 4 2 4 2 4 2 4 1 mol Na SO 18.8 g Na SO = 0.132 mol Na SO 142.05 g Na SO . +16 1 L 352 mL = 0.352 L 1000 mL . 2 4 2 4 2 4 0.132 moles Na SO molarity Na SO solution = = 0.375 M Na SO 0.352 liters solution (b) 2 4 2 4 2 4 0.246 mol Na SO 2.92 L = 0.718 mol Na SO 1 L Na SO solution . +8 2 4 2 4 2 4 2 4 142.05 g Na SO 0.718 mol Na SO = 102 g Na SO 1 mol Na SO . (c) 2 4 2 4 2 4 2 4 1 mol Na SO 4.13 g Na SO = 0.0291 mol Na SO 142.05 g Na SO . +8 2 4 2 4 1 L solution 0.0291 mol Na SO = 0.291 L solution = 291 mL solution 0.100 mol Na SO . 10 - 1 2. Complete the following table for aqueous solutions of iron(III) chloride. [Show work for par- tial credit.] Density (g/mL) Molarity Molality Mass Percent of Solute (a) 1.080 0.6203 0.634 9.315 (b) 1.230 1.845 1.982 24.33 (c) 1.350 2.869 3.243 34.47 (a) Find the molality and mass percent of 0.6203 M FeCl 3 solution. 79/79 First find molality. (a) = 25 1. [ ] = . M = . mol 1 L solution mol kg FeCl FeCl FeCl FeCl H O 3 3 3 3 2 0 6203 0 6203 2. We take 1 L of solution as a convenient quantity. This is an exact number. 3. 0.6203 mol 0.6203 mol We must make the conversions 1 L solution mass solution mass solvent = mass of H 2 O using the conversion factors 1 L = 1000 mL 1.080 g solution/1 mL solution (density) MM = 162.20 g FeCl 3 /mol mass mass mass mass mass mass solution FeCl H O H O solution FeCl = + or = 3 2 2 3- 1 kg = 1000 g +4 1000 solution 1.080 g solution solution = 1.080 g solution mL mL 10 3 10 - 2 +6 0 6203 162 20 100 6 3 3 3 3 . mol . g 1 mol = . g FeCl FeCl FeCl FeCl +3 1.080 10 3 g solution 100.6 g FeCl 3 = 979 g H 2 O +4 979 2 2 2 2 g 1 kg 1000 g = 0.979 kg H O H O H O H O +4 4. molality = . mol 0.979 kg = . / 0 6203 0 634 3 2 3 FeCl H O mol FeCl kg water We have already calculated the quantities that we need to find the mass % FeCl 3 . mass percent = mass mass solution 100% FeCl FeCl 3 3 +4 mass percent = . g 1.080 g solution 100% = . FeCl FeCl FeCl 3 3 3 3 100 6 10 9 315% (b) Find the molarity and mass percent of a 1.982 molal FeCl 3 solution....
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This note was uploaded on 04/07/2008 for the course CHEM 112 taught by Professor Dow during the Spring '08 term at Suffolk.

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Chapter 10 problems - 229/229 Solutions to Problems from...

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