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
0.050 L
2.
= 0.48 mol/L
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
0.750 L
0.00375 mol x
386.73 C27H46O = 1.45 g
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
mass sol’n
mass sol’n = 1.60 g LiCl
5.75 % m/m
6.
x 100%
x 100%
= 27.83 g
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
1.3 g air
35 ppm
x 106
x 1.3 g air = mass solute
106
= 4.55 x 10-5 g
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
800 g sol’n
x 100%
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
0.482 L
9.
= 1.04 mol/L
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
106
= 0.057 kg
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
0.630 L
= 0.044 mol/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
0.100 L
= 0.15 mol/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
x L
 x L =
0.082 mol = 0.54 L or 544 mL
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 =
12.0% v/v
vol. solute
800 mL sol’n
x 100%
x 800 mL sol’n = vol. solute
100%
= 96 mL
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
vol. sol’n
vol. sol’n = 20.0 mL acetic acid
6 % v/v
x 100%
x 100%
= 333 mL
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 =
10.2 % v/v
vol. peroxide
x 100%
250.5 mL treatment
x 250.5 mL sol’n = vol. solute
100%
= 25.5 mL
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
250.0 mL sol’n
2.25 % m/v
x 100%
x 250.0 mL sol’n = x g boric acid
100%
= 5.62 mL
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