Concentration questions
1. What is the molarity (molar concentration) of a solution made by dissolving 2.355 g of sulfuric acid (H2SO4) in water and diluting to a final volume of 50.0 mL?
2.355 g H2SO4 x 1 mol H2SO4 = 0.024 mol
98.08 g H2SO4
molar conc. = 0.024 mol = 0.48 mol/L
0.050 L
2. Hydrochloric acid is sold commercially as a 12.0 mol/L solution. How many moles of HCl are in 300.0 mL solution?
12.0 mol/L = # mol
0.300 L
# mol = 3.6 mol
3. The concentration of cholesterol in normal blood is (C27H46O) approximately 0.005 mol/L. How many grams of cholesterol are in 750 mL of blood?
0.005 mol/L = # mol 0.00375 mol x 386.73 C27H46O = 1.45 g
0.750 L 1 mol C27H46O
# mol = 0.00375 mol
4. What is the final concentration if 75.0 mL of a 3.50 mol/L glucose solution is diluted to a volume of 400.0 mL?
ci•Vi = cf•Vf
(3.5 mol/L) • (75.0 mL) = cf • (400.0 mL)
ci = 0.66 mol/L
5. Assume that you have a 5.75 % mass/mass solution of LiCl in water. What mass of solution (in grams) contains 1.60 g of LiCl?
5.75 % m/m = 1.60 g LiCl x 100%
mass sol’n
mass sol’n = 1.60 g LiCl x 100% = 27.83 g
5.75 % m/m
6. The legal limit for human exposure to carbon monoxide in the workplace is 35 ppm. Assuming that the density of air is 1.3 g/L, how many grams of carbon monoxide are in 1.0 L of air at the maximum allowable concentration?
35 ppm = mass solute x 106
1.3 g air
35 ppm x 1.3 g air = mass solute = 4.55 x 10-5 g
106
7. How many grams of NaOH would be required to prepare 800 grams of a 40% by mass NaOH solution? How many grams of water is required?
40 % m/m = x g NaOH x 100%
800 g sol’n
40 % m/m · 800 g sol’n = x g NaOH = 320 g NaOH
100%
8. Determine the molarity of a solution made by dissolving 20.0 g of NaOH in sufficient water to yield a 482 mL solution.
20.0 g NaOH x 1 mol NaOH = 0.50 mol
40.01 g NaOH
molar conc. = 0.50 mol = 1.04 mol/L
0.482 L
9. How would you prepare 500 ml of 3 mol/L HCl using 6 mol/L HCl from the stockroom. In other words how much water and how much 6 M HCl would you mix to accomplish this dilution?
ci•Vi = cf•Vf
(6 mol/L) • Vi = (3 mol/L) (500 mL)
Vi = 250 mL
VH2O = Vf – Vi
VH2O = 500 mL – 250 mL = 250 mL of water
10. A meteorologist indicates the level of a given pollutant in the air is 244.5 ppm. According to this value, what is the mass of pollutants in 234.56 kg of air? (3)
244.5 ppm = mass solute x 106
234.56 kg air
244.5 ppm x 234.56 kg air = mass solute = 0.057 kg
106
11. What is the molar concentration (mol/L) of single-single coffee from Tim’s Horton’s (what I would enjoy), if 5.0 g of sugar (C6H12O6) are dissolved in 630 mL (extra-large) of coffee? (4)
5.0 g C6H12O6 x 1 mol C6H12O6 = 0.028 mol
180.16 g C6H12O6
molar conc. = 0.028 mol = 0.044 mol/L
0.630 L
12. A saline solution (contact cleanser) contains 0.90 g of sodium chloride, dissolved to make a 100.0 mL solution. What is the molar concentration (mol/L) of this solution? (4)
0.90 g NaCl x 1 mol NaCl = 0.015 mol
58.44 g NaCl
molar conc. = 0.015 mol = 0.15 mol/L
0.100 L
13. At 20ºC, a solution of KClO3 will permit the dissolution of 10 g of solute and the solution obtained has a molar concentration of 0.150 mol/L. What must be the volume of this solution? (4)
10 g KClO3 x 1 mol KClO3 = 0.082 mol
122.55 g KClO3
0.150 mol/L = 0.082 mol ð x L = 0.082 mol = 0.54 L or 544 mL
x L 0.150 mol/L
14. What volume of ethanol (in wine) is necessary for preparing a 800 mL ethanol solution (wine) of 12% v/v? If more than 72 mL of alcohol passes the legal limit for consumption behind the wheel, is this person legally drunk? (4)
12.0 % v/v = vol. solute x 100%
800 mL sol’n
12.0% v/v x 800 mL sol’n = vol. solute = 96 mL
100%
Legally drunk, over the 72 mL limit.
15. Balsamic vinegar is sold as a solution of 6% v/v acetic acid in water. What quantity of balsamic vinegar contains exactly 20.0 mL of pure acetic acid? (3)
6 % v/v = 20.0 ml acetic acid x 100%
vol. sol’n
vol. sol’n = 20.0 mL acetic acid x 100% = 333 mL
6 % v/v
16. Hair dye can contain a solution that is 10.2% v/v hydrogen peroxide in water. What volume of the peroxide is to be found in 250.5 mL of the chemical hair treatment? (3)
10.2 % v/v = vol. peroxide x 100%
250.5 mL treatment
10.2 % v/v x 250.5 mL sol’n = vol. solute = 25.5 mL
100%
17. A boric acid solution is used in ophthalmic drops (for eyes). What mass of boric acid is present in 250.0 mL of a solution that is 2.25 % m/v of acid in water? (3)
2.25 % m/v = x g boric acid x 100%
250.0 mL sol’n
2.25 % m/v x 250.0 mL sol’n = x g boric acid = 5.62 mL
100%
18. Household chemical cleaners often contain ammonia. Industrial strength ammonia is 14.0 mol/L. If 3.0 L of an ammonia solution are needed to clean the house at a concentration of 0.10 mol/L, what would be the volume needed of the original solution that would be diluted? What volume of water needs to be added to dilute the ammonia? (4)
ci•Vi = cf•Vf
(14.0 mol/L) • Vi = (0.10 mol/L) (3.0 L)
Vi = 0.0214 L or 21.4 mL
VH2O = Vf – Vi
VH2O = 3.0 L – 0.0214 L = 2.9786 L of water