# Solution determine some fundamental quantities mass g

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Solution: Determine some fundamental quantities: Mass (g) of NH 3 = 3 8.00% NH 100 g solution 100% solution = 8.00 g NH 3 Mass (g) H 2 O = mass of solution – mass NH 3 = (100.00 – 8.00) g = 92.00 g H 2 O Mass (kg) of H 2 O = 2 3 1 kg 92.00 g H O 10 g = 0.09200 kg H 2 O Moles of NH 3 = 3 3 3 1 mol NH 8.00 g NH 17.03 g NH = 0.469759 mol NH 3 Moles of H 2 O = 2 2 2 1 mol H O 92.00 g H O 18.02 g H O = 5.1054 mol H 2 O 13-13
Volume (L) of solution = 3 1 mL solution 10 L 100.00 g solution 0.9651 g solution 1 mL = 0.103616 L Using the above fundamental quantities and the definitions of the various units: Molality = m = moles of solute kg of solvent = 3 2 0.469759 mol NH 0.09200 kg H O = 5.106076 = 5.11 m NH 3 Molarity = M = moles of solute L of solution = 3 0.469759 mol NH 0.103616 L = 4.53365 = 4.53 M NH 3 Mole fraction = X = 3 moles of NH total moles = 3 0.469759 mol NH 0.469759 5.1054 mol = 0.084259 = 0.0843 13.75 The information given is 28.8 mass % FeCl 3 solution with a density of 1.280 g/mL. For convenience, choose exactly 100.00 g of solution. Determine some fundamental quantities: Mass (g) of FeCl 3 = (100.00 g solution)(28.8% FeCl 3 /100%) = 28.8 g FeCl 3 Mass (g) of H 2 O = mass of solution – mass FeCl 3 = (100.00 – 28.8) g = 71.20 g H 2 O Moles of FeCl 3 = (28.80 g FeCl 3 )(1 mol FeCl 3 /162.20 g FeCl 3 ) = 0.1775586 mol FeCl 3 Moles of H 2 O = (71.20 g H 2 O)(1 mol H 2 O/18.02 g H 2 O) = 3.951165 mol H 2 O Volume of solution = (100.00 g solution)(1 mL/1.280 g)(10 –3 L/1 mL) = 0.078125 L Using the above fundamental quantities and the definitions of the various units: Molality = M = moles solute/kg solvent = 3 3 2 0.1775586 mol FeCl 10 g 71.20 g H O 1 kg  = 2.49380 = 2.49 m FeCl 3 Molarity = m = moles solute/L solution = 3 0.1775586 mol FeCl 0.078125 L = 2.272750 = 2.27 M FeCl 3 Mole fraction = X = moles substance/total moles = 3 0.1775586 mol FeCl 0.1775586 3.951165 mol = 0.043005688 = 0.0430 13.76 Plan: Use the equation for parts per million, ppm. Use the given density of solution to find the mass of solution; divide the mass of each ion by the mass of solution and multiply by 1x10 6 . Solution: Mass (g) of solution is 3 1 mL 1.001 g 100.0 L solution 1 mL 10 L    = 1.001x10 5 g ppm = 6 mass solute x 10 mass solution ppm Ca 2+ = 2+ 6 5 0.25 g Ca x 10 1.001x10 g solution = 2.49750 = 2.5 ppm Ca 2+ ppm Mg 2+ = 2+ 6 5 0.056 g Mg x 10 1.001x10 g solution = 0.5594406 = 0.56 ppm Mg 2+ 13.77 The information given is that ethylene glycol has a density of 1.114 g/mL and a molar mass of 62.07 g/mol. Water has a density of 1.00 g/mL. The solution has a density of 1.070 g/mL. For convenience, choose exactly 1.0000 L as the equal volumes mixed. Ethylene glycol will be designated EG. Determine some fundamental quantities: Mass (g) of EG = (1.0000 L EG)(1mL/10 –3 L)(1.114 g EG/mL) = 1114 g EG Mass (g) of H 2 O = (1.0000 L H 2 O)(1mL/10 –3 L)(1.00 g H 2 O/mL) = 1.00x10 3 g H 2 O Moles of EG = (1114 g EG)(1 mol EG/62.07 g EG) = 17.94747865 mol EG Moles of H 2 O = (1.00x10 3 g H 2 O)(1 mol H 2 O/18.02 g H 2 O) = 55.49389567 mol H 2 O 13-14