Chem Team Molality Problems #1-10 PDF

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Molality Problems #1-10 Go to molality problems #11-25 Return to the molality discussion Return to Solutions Menu Problem #1: A solution of H2SO4 with a molal concentration of 8.010 m has a density of 1.354 g/mL. What is the molar concentration of this solution? Solution: 8.010 m means 8.010 mol / 1 kg of solvent 8.010 mol times 98.0768 g/mol = 785.6 g of solute 785.6 g + 1000 g = 1785.6 g total for solute and solvent in the 8.010 m solution. 1785.6 g divided by 1.354 g/mL = 1318.76 mL 8.01 moles / 1.31876 L = 6.0739 M 6.074 M (to four sig figs) Problem #2: A sulfuric acid solution containing 571.4 g of H2SO4 per liter of solution has a density of 1.329 g/cm3. Calculate the molality of H2SO4 in this solution Solution: 1 L of solution = 1000 mL = 1000 cm3 1.329 g/cm3 times 1000 cm3 = 1329 g (the mass of the entire solution) 1329 g minus 571.4 g = 757.6 g = 0.7576 kg (the mass of water in the solution) 571.4 g / 98.0768 g/mol = 5.826 mol of H2SO4 5.826 mol / 0.7576 kg = 7.690 m Problem #3: An aqueous solution is prepared by diluting 3.30 mL acetone (d = 0.789 g/mL) with water to a final volume of 75.0 mL. The density of the solution is 0.993 g/mL. What is the molarity, molality and mole fraction of acetone in this solution?

Solution: 1) Preliminary calculations: mass of acetone: (3.30 mL) (0.789 g/mL) = 2.6037 g moles of acetone: 2.6037 g / 58.0794 g/mol = 0.04483 mol 58.0794 g/mol times 0.7500 mol = 43.56 g mass of solution ---> 1000 g + 43.56 g = 1043.56 g 43.56 is to 1043.56 as x is to 450 x = 18.78 g Problem #6: A 0.391 m solution of the solute hexane dissolved in the solvent benzene is available. Calculate the mass (g) of the solution that must be taken to obtain 247 g of hexane (C6H14). Solution: 0.391 mol times 86.1766 g/mol = 33.6950 g 33.6950 g + 1000 g = 1033.6950 g In other words, every 1033.6950 g of 0.391 m solution delivers 33.6950 g of hexane 33.6950 is to 1033.6950 as 247 is to x x = 7577.46446 g to three sig figs, 7.58 kg of solution Problem #7: Calculate the mass of the solute C6H6 and the mass of the solvent tetrahydrofuran that should be added to prepare 1.63 kg of a solution that is 1.42 m. Solution: 1.42 m means 1.42 mole of C6H6 in 1 kg of tetrahydrofuran 1.42 mol times 78.1134 g/mol = 110.921 g 110.921 g + 1000 g = 1110.921 g 110.921 is to 1110.921 as x is to 1630

x = 162.75 g To check, do this: 162.75 g / 78.1134 g/mol = 2.08351 mol 1630 g - 162.75 g = 1467.25 g 2.08351 mol / 1.46725 kg = 1.42 m Problem #8: What is the molality of NaCl in an aqueous solution in which the mole fraction of NaCl is 0.100? Solution: A mole fraction of 0.100 for NaCl means the mole fraction of water is 0.900. Let us assume a solution is present made up of 0.100 mole of NaCl and 0.900 mole of water. mass of water present ---> 0.900 mol times 18.015 g/mol = 16.2135 g molality of solution ---> 0.100 mol / 0.0162135 kg = 6.1677 m to three sig figs, 6.17 m Problem #9: Calculate the molality (m) of a 7.55 kg sample of a solution of the solute CH2Cl2 (molar mass = 84.93 g/mol) dissolved in the solvent acetone (CH3COH3C) if the sample contains 929 g of methylene chloride Solution: mass solvent ---> 7550 g - 929 g = 6621 g = 6.621 kg moles solute ---> 929 g/ 84.93 g/mol = 10.9384 mol molality = 10.9384 mol / 6.621 kg = 1.65 m Problem #10: What is the molality of a 3.75 M H2SO4 solution with a density of 1.230 g/mL? Solution: 1) Determine mass of 1.00 L of solution: 1000 mL x 1.230 g/mL = 1230 g

2) Determine mass of 3.75 mol of H2SO4: 3.75 mol x 98.0768 g/mol = 367.788 g 3) Determine mass of solvent: 1230 - 367.788 = 862.212 g 4) Determine molality: 3.75 mol / 0.862212 kg = 4.35 molal (to three sig figs) Bonus Problem: You are given 450.0 g of a 0.7500 molal solution of acetone dissolved in water. How many grams of acetone are in this amount of solution? Solution: 1) Molality is moles solute dissolved per kilogram of solvent. mol solute m = ––––––––– kg solvent 2) Let moles of solute be represented by 'n.' mass solute n = ––––––––––––––––– molar mass of solute 3) The formula for acetone is C3H6O and its molar mass is 58.0794 g/mol, which equals 0.0580794 kg/mol. I'm going to use the kg/mol amount and the reason will show up in a moment. mass solute n = –––––––––––– 0.0580794 kg/mol 4) Given the following: mass of solution = mass of solvent + mass of solute = 450.0 g = 0.45 kg I'll ignore those two trailing zeros for the moment. We write this:

mass of solvent = 0.45 − mass of solute Here's the reason why I must use the kg/mol unit: In the subtraction just above, the 0.45 is in kilograms. That means the mass of solute must also be in kilograms. You can't subtract two numbers using different units. Also, the bottom unit must be in kilograms because the 0.75 molal value is determined with kg in the denominator. Using grams in the denominator is not done with molality. 5) We are now ready to solve the problem. Let the mass of solute (in kilograms) be represented by 'x.' x / 0.0580794 0.75 = ––––––––––– 0.45 − x x 0.3375 − 0.75x = ––––––––– 0.0580794 x = 0.019602 − 0.04356x 1.04356x = 0.019602 x = 0.01878 kg = 18.78 g Go to molality problems #11-25 Return to the molality discussion Return to Solutions Menu...