Molality Problems
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:
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:
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:
Problem #4: Calculate the molality of 15.00 M HCl with a density of 1.0745 g/cm3
Solution:
1) Let us assume 1000. mL of solution are on hand. In that liter of 15-molar solution, there are:
Problem #5: 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:
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:
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:
Problem #8: What is the molality of NaCl in an aqueous solution in which the mole fraction of NaCl is 0.100?
Solution:
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:
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:
Solution:
1) Assume 1.00 L of the 4.20 M solution is present:
Problem #12: Calculate the molarity of a 3.58 m aqueous RbCl solution with a density of 1.12 g/mL.
Solution:
1) 3.58 m means this:
Problem #13: Calculate the molality of a solution containing 16.5 g of naphthalene (C10H8) in 54.3 g benzene (C6H6).
Solution:
Problem #14: What is the molality of a solution consisting of 1.34 mL of carbon tetrachloride (CCl4, density= 1.59 g/mL) in 65.0 mL of methylene chloride (CH2Cl2, density = 1.33 g/mL)?
Solution:
1) Moles CCl4:
Problem #15: Determine concentration of a solution that contains 825 mg of Na2HPO4 dissolved in 450.0 mL of water in (a) molarity, (b) molality, (c) mole fraction, (d) mass %, and (e) ppm. Assume the density of the solution is the same as water (1.00 g/mL). Assume no volume change upon the addition of the solute.
Solution:
1) Molarity:
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 solvent8.010 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 cm31.329 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 g2) Molarity:
moles of acetone: 2.6037 g / 58.0794 g/mol = 0.04483 mol <--- need to look up formula of acetone
mass of solution: (75.0 mL) (0.993 g/mL) = 74.475 g
mass of water in the solution: 74.475 g - 2.6037 g = 71.8713 g
moles of water: 71.8713 g / 18.015 g/mol = 3.9896 mol
0.04483 mol / 0.0750 L = 0.598 M3) Molality:
0.04483 mol / 0.0718713 kg = 0.624 m4) Mole fraction:
0.04483 mol / (0.04483 mol + 3.9896 mol) = 0.0111
Problem #4: Calculate the molality of 15.00 M HCl with a density of 1.0745 g/cm3
Solution:
1) Let us assume 1000. mL of solution are on hand. In that liter of 15-molar solution, there are:
15.00 mol/L times 1.000 L = 15.00 mole of HCl15.00 mol times 36.4609 g/mol = 546.9135 g of HCl2) Use the density to get mass of solution
1000. mL times 1.0745 g/cm3 = 1074.5 g of solution1074.5 g minus 546.9135 g = 527.5865 g of water = 0.5275865 kg3) Calculate molality:
15.00 mol / 0.5275865 kg = 28.43 m (to four sig figs)Note: the mole fractions of water and HCl can also be calculated with the above data. There are 29.286 moles of water and 15.00 moles of HCl. You may work out the mole fractions on your own.
Problem #5: 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:
0.7500 molal means 0.7500 mole of solute (the acetone) per 1000 g of watermass of acetone ---> 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 g33.6950 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 tetrahydrofuran1.42 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 kgmoles 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 g2) Determine mass of 3.75 mol of H2SO4:
3.75 mol x 98.0768 g/mol = 367.788 g3) Determine mass of solvent:
1230 - 367.788 = 862.212 g4) Determine molality:
3.75 mol / 0.862212 kg = 4.35 molal (to three sig figs)Problem #11: What is the molality of NaCl in an aqueous solution which is 4.20 molar? The density of the solution is 1.05 x 103 g/L.
Solution:
1) Assume 1.00 L of the 4.20 M solution is present:
1.00 L of this solution contains 4.20 mole of NaCl.1.00 L times 1050 g/L = 1050 g of solution.2) Determine mass of water in 1050 g of solution:
4.20 mol times 58.443 g/mol = 245.4606 g <--- mass of NaCl in solution1050 g - 245.4606 g = 804.5394 g3) Calculate the molality:
4.20 mol / 0.8045394 kg = 5.22 m (to three sig figs)
Problem #12: Calculate the molarity of a 3.58 m aqueous RbCl solution with a density of 1.12 g/mL.
Solution:
1) 3.58 m means this:
3.58 mole of RbCl in 1000 g of water.2) Determine total mass of solution:
3.58 mol times 120.921 g/mol = 432.89718 g1000 g + 432.89718 g = 1432.89718 g3) Determine volume of solution:
1432.89718 g / 1.12 g/mL = 1279.37 mL4) Determine molarity:
3.58 mol / 1.27937 L = 2.80 MHere's another problem of this type.
Problem #13: Calculate the molality of a solution containing 16.5 g of naphthalene (C10H8) in 54.3 g benzene (C6H6).
Solution:
molality = moles of naphthalene / kilograms of benzene(16.5 g / 128.1732 g/mol) / 0.0543 kg = 2.37 m
Problem #14: What is the molality of a solution consisting of 1.34 mL of carbon tetrachloride (CCl4, density= 1.59 g/mL) in 65.0 mL of methylene chloride (CH2Cl2, density = 1.33 g/mL)?
Solution:
1) Moles CCl4:
1.34 mL times 1.59 g/mL = 2.1306 g2.1306 g / 153.823 g/mol = 0.013851 mol2) Mass of the methylene chloride:
65.0 mL times 1.33 g/mL = 86.45 g = 0.08645 kg3) Molality:
0.013851 mol / 0.08645 kg = 0.160 m (to three sig figs)
Problem #15: Determine concentration of a solution that contains 825 mg of Na2HPO4 dissolved in 450.0 mL of water in (a) molarity, (b) molality, (c) mole fraction, (d) mass %, and (e) ppm. Assume the density of the solution is the same as water (1.00 g/mL). Assume no volume change upon the addition of the solute.
Solution:
1) Molarity:
MV = mass / molar mass(x) (0.4500 L) = 0.825 g / 141.9579 g/mol2) Molality:
x = 0.0129 M
0.825 g / 141.9579 g/mol = 0.00581158 mol0.00581158 mol / 0.4500 kg = 0.0129 m3) Mole fraction:
Na2HPO4 ---> 0.825 g / 141.9579 g/mol = 0.00581158 mol4) Mass percent:
H2O ---> 450.0 g / 18.015 g/mol = 24.97918401 molmole fraction of the water ---> 24.97918401 mol / (24.97918401 mol + 0.00581158 mol) = 0.9998
water ---> (450 g / 450.825 g) * 100 = 99.8%5) ppm:
ppm means the number of grams of solute per 1,000,000 grams of solution0.825 is to 450.825 as x is to 1,000,000x = 1830 ppm
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