five numericals of molality
Answers
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 <--- 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
2) Molarity:
0.04483 mol / 0.0750 L = 0.598 M
3) Molality:
0.04483 mol / 0.0718713 kg = 0.624 m
4) 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 HCl
15.00 mol times 36.4609 g/mol = 546.9135 g of HCl
2) Use the density to get mass of solution
1000. mL times 1.0745 g/cm3 = 1074.5 g of solution
1074.5 g minus 546.9135 g = 527.5865 g of water = 0.5275865 kg
3) 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 water
mass 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
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