KEY
Practice Problems for Chapter 12.5: Molarity and Dilution
(Read pgs. 382– 387 in the chemistry textbook)
1. Define the following term: Molarity
a concentration that states the number of moles of solute
in exactly 1 L of solution (solvent + solute)
2.
What is the equation to determine Molarity?
Molarity (M) =
3.
𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐞
𝐋𝐢𝐭𝐞𝐫𝐬 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
→
M=
𝒎𝒐𝒍
𝑳
(a) Compare the concentration of a 2.50 M and a 5.0 M solution of MgF2.
the larger the numerical value the greater the concentration, therefore,
a 5.0 M solution is 2X as concentrated as a 2.50 M solution
(b) Compare the concentration of 1.0 liter of a 0.20 M solution and 2.00 L of a 0.10 M solution.
the volume of solution makes no difference, it is only the molarity that
determines concentration - therefore, the 0.20 M solution is 2X as
concentrated as the 0.10 M solution, regardless of the fact that there is
a greater volume of the 0.10 M solution
4. Circle the MORE CONCENTRATED solution in each of the following:
(a)
500 mL of a 1.0 M solution
or
10.0 mL of a 1.5 M solution
(b)
2.0 L of a 0.10 M solution
or
0.500 L of a 0.010 M solution
(a)
1 L of a 2.0 M solution
or
50.0 mL of a 2.0 M solution
these two solutions have the SAME concentration, the volume
makes no difference
1
5. Comment on the molarity of the remaining solution in each of the following situations:
(a) A 2.0 M solution is spilled and about half is lost.
molarity remains the same b/c an equal amount of solute and solvent
are lost = less solution and that solution has the same concentration
(b) The lid is left off of the container of a 2.0 M solution and about half the volume is lost due to
evaporation.
molarity increases b/c only solvent was lost....there is now the same
amount of solute dissolved in a smaller amount of solvent = less
solution and that solution is more concentrated
(c) A student adds 100 mL of distilled water to 1 L of a 2.0 M solution.
molarity decreases b/c additional solvent was added.....there is now
the same amount of solute dissolved in a larger amount of solvent =
more solution and that solution is less concentrated
6. What is another way to read the term 2.0 M NaCl?
𝟐.𝟎 𝐦𝐨𝐥𝐞𝐬 𝐍𝐚𝐂𝐥
𝟏 𝐋𝐢𝐭𝐞𝐫 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
→
𝟐.𝟎 𝒎𝒐𝒍
𝑳
7. Convert the following molarities to moles per liter
(a) 6.0 M HCl
𝟔.𝟎 𝐦𝐨𝐥𝐞𝐬 𝐇𝐂𝐥
𝟏 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧
→
𝟔.𝟎 𝒎𝒐𝒍
𝑳
→
𝟎.𝟐𝟎 𝒎𝒐𝒍
𝑳
(b) 0.20 M NaOH
𝟎.𝟐𝟎 𝐦𝐨𝐥𝐞𝐬 𝐍𝐚𝐎𝐇
𝟏 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧
(c) 2.50 M Sr(OH)2
𝟐.𝟓𝟎 𝐦𝐨𝐥𝐞𝐬 𝐒𝐫(𝐎𝐇)𝟐
𝟏 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐒𝐨𝐥𝐮𝐭𝐢𝐨𝐧
→
𝟐.𝟓𝟎 𝒎𝒐𝒍
𝑳
2
C-Level
Calculating Molarity when you are given:
1) moles solute, and
2) volume of solution
8. Calculate the molarity in each of the following:
(a) 2.00 mol of glucose in 4.00 L of solution
Molarity =
𝟐.𝟎𝟎 𝐦𝐨𝐥𝐞𝐬 𝐠𝐥𝐮𝐜𝐨𝐬𝐞
𝟒.𝟎𝟎 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
= 0.500 M solution of glucose
(b) 0.500 mol of sucrose in 0.200 L of solution
Molarity =
𝟎.𝟓𝟎𝟎 𝐦𝐨𝐥𝐞𝐬 𝐬𝐮𝐜𝐫𝐨𝐬𝐞
𝟐.𝟎𝟎 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
= 0.250 M solution of sucrose
(c) 0.0020 moles KOH in 0.100 L of solution
Molarity =
𝟎.𝟎𝟎𝟐𝟎 𝐦𝐨𝐥𝐞𝐬 𝐊𝐎𝐇
𝟎.𝟏𝟎𝟎 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
= 0.020 M solution of KOH
(d) 5.0 × 10−2 moles of AlCl3 in 50.0 mL of solution
i) 50.0 mL = 0.0500 L of solution
ii) Molarity =
𝟓.𝟎 × 𝟏𝟎−𝟐 𝐦𝐨𝐥𝐞𝐬 𝐀𝐥𝐂𝐥𝟑
𝟎.𝟎𝟓𝟎𝟎 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
= 1.00 M solution of AlCl3
(e) Calculate the molarity of an aqueous solution containing 0.25 moles of NaNO3 dissolved in 500. mL of
solution.
i) 500. mL = 0.500 L of solution
ii) Molarity =
𝟎.𝟐𝟓 𝐦𝐨𝐥𝐞𝐬 𝐍𝐚𝐍𝐎𝟑
𝟎.𝟓𝟎𝟎 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
= 0.500 M solution of NaNO3
3
B-Level
Calculating Volume when you are given:
1) moles solute, and
2) Molarity
9. Calculate the volume of each of the following:
(a) liters of 2.0 M solution which contain 0.50 moles of solute.
i) 2.0 M is the same as
𝟐.𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
ii) use the Molarity equations
𝟐.𝟎𝟎 𝐦𝐨𝐥𝐞𝐬
×
𝐋
x liters = 0.50 moles → x =
𝟎.𝟓𝟎 𝐦𝐨𝐥𝐞𝐬
𝟐.𝟎𝟎
𝐦𝐨𝐥𝐞𝐬
𝐋
= 0.25 L
Alternatively: set up an equality (Quicker!!)
𝟐.𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
=
𝟎.𝟓𝟎 𝒎𝒐𝒍𝒆𝒔
𝒙
→ =
𝟎.𝟓𝟎 𝐦𝐨𝐥𝐞𝐬
𝟐.𝟎𝟎
𝐦𝐨𝐥𝐞𝐬
𝐋
= 0.25 L
(b) liters of 0.10 M solution which contain 0.050 moles of solute.
i) 0.10 M is the same as
𝟎.𝟏𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
ii) set up an equality
𝟎.𝟏𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
=
𝟎.𝟎𝟓𝟎 𝒎𝒐𝒍𝒆𝒔
𝒙
→ x = 0.50 L
4
(c) milliliters of 0.10 M solution which contain 0.010 moles of solute.
i) 0.10 M is the same as
𝟎.𝟏𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
ii) set up an equality
𝟎.𝟏𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
=
𝟎.𝟎𝟏𝟎 𝒎𝒐𝒍𝒆𝒔
𝒙
→ x = 0.100 L
iii) convert Liters to milliliters………0.100 L = 100. mL
(d) milliliters of 2.50 M solution which contain 0.10 moles of solute.
i) 2.50 M is the same as
𝟐.𝟓𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
ii) set up an equality
𝟐.𝟓𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
=
𝟎.𝟏𝟎 𝒎𝒐𝒍𝒆𝒔
𝒙
→ x = 0.040 L
iii) convert Liters to milliliters……… 0.040 L = 40. mL
(e) Calculate the milliliters of 0.35 M KOH solution required in order to obtain 0.02 moles of KOH.
i)
0.35 M is the same as
𝟎.𝟑𝟓 𝒎𝒐𝒍𝒆𝒔
𝑳
ii) set up an equality
𝟎.𝟑𝟓 𝒎𝒐𝒍𝒆𝒔
𝑳
=
𝟎.𝟎𝟐𝟎 𝒎𝒐𝒍𝒆𝒔
𝒙
→ x = 0.057 L
iii) convert Liters to milliliters……… 0.057 L = 57 mL
5
B-Level
Calculating Moles when you are given:
1) volume of solution, and
2) Molarity
10. Calculate the number of moles in each of the following volumes:
(a) moles of solute in 0.500 L of a 2.0 M solution
i)
ii)
2.0 M is the same as
𝟐.𝟎𝟎 𝐦𝐨𝐥𝐞𝐬
𝐋
×
𝟐.𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
0.500 L = 1.00 moles
(b) moles of solute in 0.050 L of a 0.10 M solution
i)
ii)
0.10 M is the same as
𝟐.𝟎𝟎 𝐦𝐨𝐥𝐞𝐬
𝐋
×
𝟎.𝟏𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
0.050 L = 0.10 moles
(c) moles of solute in 250. mL of a 2.0 M solution
i)
convert milliliters to Liters ……… 250. m L = 0.250 L
ii)
2.0 M is the same as
ii)
𝟐.𝟎𝟎 𝐦𝐨𝐥𝐞𝐬
𝐋
×
𝟐.𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
0.250 L = 0.500 moles
6
(d) moles of solute in 25 mL of a 0.010 M solution
i)
convert milliliters to Liters ……… 25 m L = 0.025 L
ii)
0.010 M is the same as
ii)
𝟎.𝟎𝟏𝟎 𝐦𝐨𝐥𝐞𝐬
𝐋
×
𝟎.𝟎𝟏𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
0.025 L = 2.5 × 10─4 moles
(e) Calculate the moles of NaOH in 400. mL of a 0.50 M solution.
i)
convert milliliters to Liters ……… 400. m L = 0.400 L
ii)
0.50 M is the same as
ii)
𝟎.𝟓𝟎 𝐦𝐨𝐥𝐞𝐬
𝐋
×
𝟎.𝟓𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
0.400 L = 0.20 moles
(f) Calculate the moles of CH4 in 0.050 L of a 0.10 M solution
i)
ii)
0.10 M is the same as
𝟎.𝟏𝟎 𝐦𝐨𝐥𝐞𝐬
𝐋
×
𝟎.𝟏𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
0.050 L = 5.0 × 10─3 moles
7
A-Level
Calculating Molarity when you are given:
1) grams of solute, and
2) volume of solution
11. Calculate the molarity of each of the following:
(a) 60.0 g of NaOH in 0.250 L of solution
i)
Molar mass of NaOH = 40.01 g/ mol
ii)
Convert grams to moles
60.0 g ÷ 40.01 g/ mol = 1.50 moles of NaOH
iii) Moles ÷ volume (in liters!)
Molarity =
𝟏.𝟓𝟎 𝐦𝐨𝐥𝐞𝐬 𝐍𝐚𝐎𝐇
𝟎.𝟐𝟓𝟎 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
= 6.00 M solution of NaOH
(b) 75.0 g of KNO3 in 0.350 L of solution
i)
Molar mass of KNO3 = 101.11 g/ mol
ii)
Convert grams to moles
75.0 g ÷ 101.11 g/ mol = 0.742 moles of NaOH
iii) Moles ÷ volume (in liters!)
Molarity =
𝟎.𝟕𝟒𝟐 𝐦𝐨𝐥𝐞𝐬 𝐊𝐍𝐎𝟑
𝟎.𝟑𝟓𝟎 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
= 2.12 M solution of KNO3
8
11. Continued:
(c) 73.0 g of HCl in 2.00 L of solution
i)
Molar mass of HCl = 36.46 g/ mol
ii)
Convert grams to moles
73.0 g ÷ 36.46 g/ mol = 2.00 moles of HCl
iii) Moles ÷ volume (in liters!)
Molarity =
𝟐.𝟎𝟎 𝐦𝐨𝐥𝐞𝐬 𝐇𝐂𝐥
𝟐.𝟎𝟎 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
= 1.00 M solution of HCl
(d) 5.85 g of NaCl in 40.0 mL of solution
i)
Molar mass of NaCl= 58.44 g/ mol
ii)
Convert grams to moles
5.85 g ÷ 58.44 g/ mol = 0.100 moles of NaCl
iii) Convert milliliters to liters : 40.0 mL = 0.0400 L
iv) Moles ÷ volume (in liters!)
Molarity =
𝟎.𝟏𝟎𝟎 𝐦𝐨𝐥𝐞𝐬 𝐍𝐚𝐂𝐥
𝟎.𝟎𝟒𝟎𝟎 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
= 2.50 M solution of NaCl
9
(e) 30.4 g of LiBr in 350. mL of solution
i)
Molar mass of LiBr = 86.84 g/ mol
ii)
Convert grams to moles
30.4 g ÷ 86.84 g/ mol = 0.350 moles of LiBr
iii) Convert milliliters to liters : 350. mL = 0.350 L
iv) Moles ÷ volume (in liters!)
Molarity =
𝟎.𝟏𝟎𝟎 𝐦𝐨𝐥𝐞𝐬 𝐋𝐢𝐁𝐫
𝟎.𝟑𝟓𝟎 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
= 1.00 M solution of LiBr
(f) Calculate the molarity of a solution that has 10.0 g of Li2O in 0.250 L of solution.
i)
Molar mass of Li2O = 29.88 g/ mol
ii)
Convert grams to moles
10.0 g ÷ 29.88 g/ mol = 0.33467 moles of Li2O
iii) Moles ÷ volume (in liters!)
Molarity =
𝟎.𝟑𝟑𝟒𝟔𝟕 𝐦𝐨𝐥𝐞𝐬 𝐋𝐢𝟐 𝐎
𝟎.𝟐𝟓𝟎 𝐋𝐢𝐭𝐞𝐫 𝐨𝐟 𝐬𝐨𝐥𝐮𝐭𝐢𝐨𝐧
= 1.34 M solution of Li2O
10
A-Level
Calculating volume of solution when you are given:
3) grams of solute, and
4) Molarity
12. Calculate the volume of solution for each of the following:
a)
liters of a 2.00 M NaCl solution containing 67.3 g of NaCl:
i)
Molar mass of NaCl= 58.44 g/ mol
ii)
Convert grams to moles
67.3 g ÷ 58.44 g/ mol = 1.15 moles of NaCl
iii)
2.00 M is the same as
iv)
set up an equality
𝟐.𝟎𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
b)
=
𝟐.𝟎𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
𝟏.𝟏𝟓 𝒎𝒐𝒍𝒆𝒔
→ x = 0.575 L of solution
𝒙
Liters of 2.25 M HCl solution that will provide 4.12 g of HCl.
i)
Molar mass of HCl= 36.46 g/ mol
ii)
Convert grams to moles
4.12 g ÷ 36.46 g/ mol = 0.113 moles of HCl
iii)
2.25 M is the same as
iv)
set up an equality
𝟐.𝟐𝟓 𝒎𝒐𝒍𝒆𝒔
𝑳
=
𝟐.𝟐𝟓 𝒎𝒐𝒍𝒆𝒔
𝑳
𝟎.𝟏𝟏𝟑 𝒎𝒐𝒍𝒆𝒔
𝒙
→ x = 0.0502 L of solution
11
(c) milliliters of a 0.120 M Na2CO3 solution that will yield 12.5 g of Na2CO3
i)
Molar mass of Na2CO3 = 105.99 g/ mol
ii)
Convert grams to moles
12.5 g ÷ 105.99 g/ mol = 0.118 moles of Na2CO3
iii)
0.120 M is the same as
iv)
set up an equality
𝟎.𝟏𝟐𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
v)
=
𝟎.𝟏𝟐𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
𝟎.𝟏𝟏𝟖 𝒎𝒐𝒍𝒆𝒔
𝒙
→ x = 0.983 L of solution
Convert liters to milliliters : 0.983 L = 983 mL
(d) Calculate the number of milliliters of 2.70 M LiOH solution that are needed to provide 30.0 g of
LiOH .
iii) Molar mass of LiOH = 23.95 g/ mol
iv) Convert grams to moles
30.0 g ÷ 23.95 g/ mol = 1.25 moles of LiOH
iv)
2.70 M is the same as
iv)
set up an equality
𝟐.𝟕𝟎 𝒎𝒐𝒍𝒆𝒔
𝑳
=
𝟐.𝟕𝟎 𝒎𝒐𝒍𝒆𝒔
𝟏.𝟐𝟓 𝒎𝒐𝒍𝒆𝒔
𝒙
𝑳
→ x = 0.463 L of solution
vi) Convert liters to milliliters : 0.463 L = 463 mL
12
13. What is a stock solution?
a preexisting solution of known molarity
14.
(a) Define the term dilution
when more solvent is added to a a portion of stock solution to
make it less concentrated
(b) What is the dilution equation? Define each of its terms.
M1V1 = M2V2
C1 = Molarity of the MORE
Concentrated solustion
C2 = Molarity of the LESS
concentrated solution
V1 = volume of the MORE
concentrated solution
V2 = volume of the LESS
concentrated solution
15. List calculations for which the dilution equation may be used.
i) Volume of stock solution required to make a dilution
ii) Volume of diluted solution made
iii) Molarity of the dilution
iv) Molarity of the original stock solution
13
B-Level
Solving for any variable using the dilution equation.
16. Calculate the final concentration (Molarity) of the solution in each of the following:
(a) Water is added to 0.150 L of a 6.00 M HCl stock solution to give a volume of 0.500 L.
i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T know
M1 = 6.00 M
M2 = x
V1 = 0.150 L
V2 = 0.500 L
ii) Use the dilution equation to solve for the unknown:
M1V1 = M2V2
*setup: (6.00 M) ( 0.150 L) = x (0.500L) → 6.00 M(0.150 L) = 0.500 L x
→ 0.900 M-L = 0.500L x →
𝟎.𝟗𝟎𝟎 𝒎−𝑳
𝟎.𝟓𝟎𝟎 𝑳
= x → x = 1.80 Molar HCl
solution
*NOTE: This is the only problem for which the algebra will be shown. For all other dilution problems,
only the setup will be shown
(b) A 10.0-mL sample of a 2.50 M KCl solution is diluted with water to 0.250 L.
i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T know
M1 = 2.50 M
M2 = x
V1 = 10.0mL*
V2 = 0.250 L = 250 mL
*NOTE: volume can be in either mL or L, as long as both volumes are the same
ii) Use the dilution equation to solve for the unknown:
M1V1 = M2V2
setup:
(2.50 M) ( 10.0m L) = x (250 mL) → x = 0.100 Molar KCl
solution
14
16. Continued:
(c) Water is added to 0.250 L of a 12.0 M KBr solution to give a volume of 1.00 L.
i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T know
M1 = 12.0 M
M2 = x
V1 = 0.250 L
V2 = 1.00 L
ii) Use the dilution equation to solve for the unknown:
M1V1 = M2V2
setup:
(12.0 M) ( 0.250 L) = x (1.00 L) → x = 3.00 M KBr sol’n
17. Determine the final volume (mL) for each of the following:
(a) diluting 50.0 mL of a 12.0 M NH4Cl solution to give a 2.00 M NH4Cl solution.
i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T know
M1 = 12.0 M
M2 = 2.00 M
V1 = 50.0 mL
V2 = x
ii) Use the dilution equation to solve for the unknown:
M1V1 = M2V2
setup:
(12.0 M) ( 50.0 mL) = (2.00 M) x → x = 300. mL of 2.00 M
NH4Cl solution
(b) diluting 18.0 mL of a 15.0 M NaNO3 solution to give a 1.50 M NaNO3 solution.
i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T know
M1 = 15.0 M
M2 = 1.50 M
V1 = 18.0 mL
V2 = x
ii) Use the dilution equation to solve for the unknown:
M1V1 = M2V2
setup:
(15.0 M) ( 18.0 mL) = (1.50 M) x → x = 180. mL of 1.50 M
NaNO3 solution
15
17. Continued:
(c) diluting 4.50 mL of a 18.0 M H2SO4 solution to give a 2.50 M H2SO4 solution.
i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T know
M1 = 18.0 M
V1 = 4.50 mL
M2 = 2.50 M
V2 = x
ii) Use the dilution equation to solve for the unknown:
M1V1 = M2V2
setup:
(18.0 M) ( 4.50 mL) = (2.50 M) x → x = 32.4 mL of 2.50 M
H2SO4 solution
18. Determine the volume (mL) of stock solution required to prepare each of the following:
(a) 255 mL of a 0.200 M HNO3 solution from a 4.00 M HNO3 solution:
i) first, assign values to M1V1 = M2V2 so that you can determine what you DON'T know
M1 = 4.00 M
V1 = x
M2 = 0.200 M
V2 = 255 mL
ii) Use the dilution equation to solve for the unknown:
M1V1 = M2V2
setup:
(4.00 M) x = (0.200 M) ( 255 mL) → x = 12.8 mL of 4.00 M
HNO2 solution are
required
(b) 715 mL of a 0.100 M MgCl2 solution using a 6.00 M MgCl2 solution.
i) first, assign values to C1V1 = C2V2 so that you can determine what you DON'T know
M1 = 6.00 M
V1 = x
M2 = 0.100 M
V2 = 715 mL
ii) Use the dilution equation to solve for the unknown:
M1V1 = M2V2
setup:
(6.00 M) x = (0.100 M) ( 715 mL) → x = 11.9 mL of 6.00 M
MgCl2 solution
required
16