CHEM 105 - ELEMENTS OF
GENERAL CHEMISTRY - Molarity
and Concentrations
Question Bank - Set 2
Liberty University
Question 1
Question
A chemist wants to prepare 500 mL of a 0.2 M copper(II) sulfate solution. The
molar mass of copper(II) sulfate is 159.61 g/mol. How many grams of copper(II)
sulfate should be dissolved in the solution?
Solution
Step 1: Calculate the number of moles of copper(II) sulfate needed. Given:
Volume of solution = 500 mL = 0.5 L Molarity of solution = 0.2 M Molar mass
of copper(II) sulfate = 159.61 g/mol
We can use the formula for molarity:
Molarity =Number of moles
Volume in liters
Since we want to find the number of moles, we rearrange the formula to:
Number of moles = Molarity ×Volume in liters
Substitute the values:
Number of moles = 0.2 mol/L ×0.5 L
Number of moles = 0.1 mol
Therefore, 0.1 moles of copper(II) sulfate are needed.
Step 2: Calculate the mass of copper(II) sulfate needed. The mass of cop-
per(II) sulfate can be calculated using the formula:
Mass = Number of moles ×Molar mass
Substitute the values:
Mass = 0.1 mol ×159.61 g/mol
Mass = 15.961 g
Therefore, 15.961 grams of copper(II) sulfate should be dissolved in the
solution.
Question 2
Question
A student is given a 500 mL solution of hydrochloric acid with a concentration
of 4.5 M. The student needs to make a 250 mL solution of hydrochloric acid
with a concentration of 2.5 M. How much of the original solution should the
student use, and how much water should be added?
Solution
Let xbe the volume of the original 4.5 M solution that the student needs to
use. Let ybe the volume of water that needs to be added.
Step 1: Write the equation based on the conservation of moles: The num-
ber of moles of hydrochloric acid initially is equal to the number of moles of
hydrochloric acid in the final solution.
(xmL) ×(4.5 M) = (250 mL) ×(2.5 M)
Step 2: Solve the equation:
4.5x= 2.5(250)
4.5x= 625
x=625
4.5
x≈138.89 mL
So, the student needs to use approximately 138.89 mL of the original 4.5 M
solution.
Step 3: Calculate the volume of water needed:
Total volume of final solution = 250 mL
Volume of original solution used = 138.89 mL
2
Volume of water added = 250 −138.89 = 111.11 mL
Therefore, the student should use approximately 138.89 mL of the original
4.5 M solution and add 111.11 mL of water to make a 250 mL solution with a
concentration of 2.5 M.
Question 3
Question
A chemist needs to prepare 500 mL of a 0.4 M copper(II) sulfate solution. The
chemist has a stock solution containing 1 M copper(II) sulfate. How many
milliliters of the stock solution should the chemist use to prepare the desired
solution?
Solution
Step 1: Determine the number of moles of copper(II) sulfate needed to prepare
500 mL of a 0.4 M solution. This can be calculated using the formula:
moles = Molarity ×Volume (L)
Given that the desired molarity is 0.4 M and the volume is 500 mL (which
is 0.5 L), we have:
moles = 0.4 mol/L ×0.5 L = 0.2 mol
Step 2: Calculate the volume of the 1 M stock solution needed to supply 0.2
moles of copper(II) sulfate. This can be done using the formula:
Volume of stock solution (L) = moles
Molarity
Substitute the values given:
Volume of stock solution (L) = 0.2 mol
1 mol/L = 0.2 L
Step 3: Convert the volume needed from liters to milliliters:
0.2 L ×1000 = 200 mL
Therefore, the chemist needs to use 200 mL of the 1 M stock solution to
prepare 500 mL of a 0.4 M copper(II) sulfate solution.
3
Question 4
Question
A chemist has a solution of hydrochloric acid that has a concentration of 12.0
M. The chemist needs to dilute this solution to make 500.0 mL of a 3.00 M HCl
solution. How much water (in mL) should be added to the original solution to
achieve the desired concentration?
Solution
Step 1: Let the volume of the original 12.0 M HCl solution be V1in mL. The
molarity-volume equation, M1V1=M2V2, can be used to solve this problem.
Step 2: We are given: - M1= 12.0 M (initial concentration of HCl solution)
-M2= 3.00 M (desired concentration of HCl solution) - V2= 500.0 mL (final
diluted volume of HCl solution)
Step 3: Using the molarity-volume equation, we can find V1:
M1V1=M2V2
12.0×V1= 3.00 ×500.0
Step 4: Solving for V1:
V1=3.00 ×500.0
12.0= 125.0 mL
Step 5: To find the volume of water that needs to be added, we subtract the
volume of the original solution from the total desired volume:
Volume of water to be added = V2−V1= 500.0−125.0 = 375.0 mL
Therefore, the chemist should add 375.0 mL of water to the original 12.0 M
HCl solution to achieve a final concentration of 3.00 M.
Question 5
Question
A solution is prepared by dissolving 4.5 grams of potassium permanganate
(KMnO4) in enough water to make 250 mL of solution. Calculate the molarity
of the solution.
Solution
Step 1: Calculate the molar mass of KMnO4. The molar mass of K = 39.10
g/mol, Mn = 54.94 g/mol, and O4= 4 ×16.00 g/mol. Adding them up, we get:
39.10 + 54.94 + 4 ×16.00 = 39.10 + 54.94 + 64.00 = 158.04 g/mol
4
Step 2: Calculate the number of moles of potassium permanganate in the
solution. The number of moles is given by:
moles = mass
molar mass =4.5 g
158.04 g/mol = 0.0285 mol
Step 3: Calculate the volume of the solution in liters.
250 mL = 250 ×10−3L=0.250 L
Step 4: Calculate the molarity of the solution. The molarity is given by:
Molarity =moles
volume in liters =0.0285 mol
0.250 L = 0.114 mol/L
Therefore, the molarity of the solution is 0.114 mol/L.
Question 6
Question
A chemist mixes 250 mL of a 0.5 M solution of NaCl with 500 mL of a 1 M
solution of NaCl. What is the final molarity of NaCl in the resulting solution?
Solution
Step 1: Calculate the moles of NaCl in the first solution.
Moles of NaCl = Molarity ×Volume (L)
Moles of NaCl = 0.5 M ×0.250 L = 0.125 moles
Step 2: Calculate the moles of NaCl in the second solution.
Moles of NaCl = Molarity ×Volume (L)
Moles of NaCl = 1 M ×0.500 L = 0.500 moles
Step 3: Add the moles of NaCl from both solutions to find the total moles
of NaCl in the final solution.
Total moles of NaCl = 0.125 moles + 0.500 moles = 0.625 moles
Step 4: Calculate the total volume of the resulting solution.
Total volume (L) = 0.250 L + 0.500 L = 0.750 L
Step 5: Calculate the final molarity of NaCl in the resulting solution using
the total moles and total volume.
Molarity = Total moles of NaCl
Total volume (L)
Molarity = 0.625 moles
0.750 L = 0.833 M
Answer: The final molarity of NaCl in the resulting solution is 0.833 M.
5
Question 7
Question
A chemist mixes 250 mL of a 0.5 M solution of hydrochloric acid with 500 mL of
a 1.2 M solution of hydrochloric acid. What is the final molarity of the resulting
solution?
Solution
Step 1: Calculate the moles of hydrochloric acid in the first solution. Given:
Volume of solution 1, V1= 250 mL = 0.25 L Molarity of solution 1, M1= 0.5
M
Using the formula for moles,
molessolution 1 =M1×V1= 0.5 mol/L ×0.25 L = 0.125 mol
Step 2: Calculate the moles of hydrochloric acid in the second solution.
Given: Volume of solution 2, V2= 500 mL = 0.5 L Molarity of solution 2,
M2= 1.2 M
Using the formula for moles,
molessolution 2 =M2×V2= 1.2 mol/L ×0.5 L = 0.6 mol
Step 3: Calculate the total moles of hydrochloric acid in the final solution.
Total moles of HCl = molessolution 1+molessolution 2 = 0.125 mol+0.6 mol = 0.725 mol
Step 4: Calculate the total volume of the final solution.
Total volume of solution = V1+V2= 0.25 L + 0.5 L = 0.75 L
Step 5: Calculate the final molarity of the resulting solution.
Final molarity = Total moles of HCl
Total volume of solution =0.725 mol
0.75 L = 0.9667 M
Therefore, the final molarity of the resulting solution is 0.9667 M.
Question 8
Question
Calculate the molarity of a solution obtained by dissolving 5.8 g of potassium
permanganate (KMnO4) in enough water to make 350 mL of solution.
6
Solution
Step 1: Calculate the molar mass of potassium permanganate (KMnO4).
Molar mass of KMnO4= Atomic mass of K + Atomic mass of Mn + 4(Atomic mass of O)
= 39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol)
= 158.04 g/mol
Step 2: Calculate the number of moles of potassium permanganate.
Number of moles = Mass
Molar mass
=5.8 g
158.04 g/mol
≈0.0367 mol
Step 3: Calculate the volume of the solution in liters.
Volume = 350 mL = 0.350 L
Step 4: Calculate the molarity of the solution.
Molarity = Number of moles
Volume in liters
=0.0367 mol
0.350 L
≈0.105 M
Therefore, the molarity of the solution is approximately 0.105 M.
Question 9
Question
A chemist prepares a solution by dissolving 5.00 g of sodium hydroxide in enough
water to make 250.0 mL of solution. Calculate the molarity of the resulting
solution. The molar mass of NaOH is 40.00 g/mol.
Solution
Step 1: Calculate the number of moles of sodium hydroxide.
Moles of NaOH = Mass
Molar mass =5.00 g
40.00 g/mol = 0.125 mol
Step 2: Convert the volume of the solution to liters.
Volume of solution = 250.0 mL = 250.0×10−3L=0.2500 L
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Step 3: Calculate the molarity of the solution.
Molarity = Moles of solute
Volume of solution (in L) =0.125 mol
0.2500 L = 0.500 M
Therefore, the molarity of the resulting solution is 0.500 M.
Question 10
Question
A chemist has a solution of acetic acid that has a molarity of 1.50 M. She wants
to dilute this solution to create 500.0 mL of a new solution with a molarity of
0.500 M. How much of the original solution should she use and how much water
should she add to achieve this dilution?
Solution
Step 1: Let Voriginal be the volume of the original 1.50 M acetic acid solution
that the chemist will use. The final volume of the diluted solution will be 500.0
mL or 0.500 L. Let Vwater be the volume of water that will be added to dilute
the solution. Then the total volume Vtotal is given by:
Vtotal =Voriginal +Vwater = 0.500 L
Step 2: Recall that the relationship between the molarity, volume, and num-
ber of moles of a solution is given by the formula:
M1V1=M2V2
Where M1and V1are the initial molarity and volume, and M2and V2are the
final molarity and volume.
Step 3: We can set up two equations based on the given information: First,
based on the amount of acetic acid in the original solution and the diluted
solution, we get:
1.50 M ×Voriginal = 0.500 M ×0.500 L
Step 4: Solve the equation for Voriginal:
Voriginal =0.500 M ×0.500 L
1.50 M = 0.167 L = 167 mL
So, the chemist should use 167 mL of the original 1.50 M acetic acid solution.
Step 5: Calculate the volume of water needed to dilute the solution:
Vtotal =Voriginal +Vwater
0.500 L = 0.167 L + Vwater
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Vwater = 0.500 L −0.167 L = 0.333 L = 333 mL
The chemist should add 333 mL of water to the 167 mL of the original 1.50
M acetic acid solution to create 500.0 mL of a new solution with a molarity of
0.500 M.
Question 11
Question
A solution is prepared by dissolving 15.0 g of glucose (C6H12O6) in enough water
to make 250.0 mL of solution. What is the molarity of the glucose solution?
(Glucose molar mass: 180.16 g/mol)
Solution
Step 1: Calculate the number of moles of glucose in the solution.
Moles of glucose = Mass of glucose (g)
Molar mass of glucose (g/mol)
Moles of glucose = 15.0 g
180.16 g/mol = 0.0833 mol
Step 2: Convert the volume of solution to liters.
Volume of solution (L) = 250.0 mL
1000 mL/L = 0.2500 L
Step 3: Calculate the molarity of the glucose solution.
Molarity (M) = Moles of solute
Volume of solution (L)
Molarity (M) = 0.0833 mol
0.2500 L = 0.333M
Therefore, the molarity of the glucose solution is 0.333 M.
Question 12
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of potassium perman-
ganate (KMnO4) from a stock solution that has a molarity of 1.0 M. What
volume of the stock solution should be used and how much water should be
added to dilute it?
9
Solution
Let V1be the volume of the stock solution to be used and V2be the volume of
water to be added to dilute it.
Step 1: Determine the moles of KMnO4needed for the final solution.
Moles of KMnO4= Molarity ×Volume (in L)
Moles of KMnO4= 0.2 M ×0.5 L = 0.1 mol
Step 2: Calculate the moles of KMnO4present in the stock solution.
Moles of KMnO4= Molarity ×Volume (in L)
0.1 mol = 1.0 M ×V1L
V1= 0.1 L = 100 mL
Step 3: Calculate the volume of water needed to dilute the stock solution.
V2= Volume of final solution −V1
V2= 500 mL −100 mL = 400 mL
Step 4: Write the final answer. The chemist should use 100 mL of the
stock solution and add 400 mL of water to prepare 500 mL of a 0.2 M KMnO4
solution.
Question 13
Question
A chemist prepares a solution by mixing 50.0 mL of a 0.200 M potassium per-
manganate solution with 150.0 mL of water. What is the molarity of the final
solution?
Solution
Step 1: Calculate the moles of potassium permanganate in the original solution.
Step 2: Determine the volume of the final solution. Step 3: Calculate the
molarity of the final solution.
Step 1: The moles of potassium permanganate in the original solution can
be calculated using the formula:
moles = Molarity ×Volume (L)
Given: Molarity = 0.200 M Volume = 50.0 mL = 0.050 L
Substitute the values into the formula:
moles = 0.200 M ×0.050 L = 0.010 moles
10
Step 2: The total volume of the final solution can be calculated by adding
the volume of the potassium permanganate solution and the volume of water:
Total volume = 50.0 mL + 150.0 mL = 200.0 mL = 0.200 L
Step 3: The molarity of the final solution can be calculated using the
formula:
Molarity = moles
Total volume (L)
Substitute the values into the formula:
Molarity = 0.010 moles
0.200 L = 0.050 M
Therefore, the molarity of the final solution is 0.050 M.
Question 14
Question
Calculate the volume of a 5.0 M sulfuric acid solution needed to neutralize 250.0
mL of a 2.0 M sodium hydroxide solution.
Solution
Step 1: Write the balanced chemical equation for the reaction between sulfuric
acid (H2SO4) and sodium hydroxide (NaOH):
H2SO4+ 2NaOH →Na2SO4+ 2H2O
Step 2: Determine the moles of sodium hydroxide (NaOH) using the given
molarity and volume:
Moles of NaOH = Molarity ×Volume = 2.0 M ×0.250 L = 0.500 moles NaOH
Step 3: From the balanced equation, we see that 1 mole of sulfuric acid
(H2SO4) reacts with 2 moles of sodium hydroxide (NaOH). Therefore, the moles
of sulfuric acid required will be half of the moles of sodium hydroxide used in
the reaction.
Moles of H2SO4=1
2×Moles of NaOH = 1
2×0.500 = 0.250 moles H2SO4
Step 4: Calculate the volume of 5.0 M sulfuric acid solution needed to contain
0.250 moles of H2SO4:
Volume = Moles
Molarity =0.250 moles
5.0 M = 0.0500 L = 50.0 mL
Therefore, 50.0 mL of 5.0 M sulfuric acid solution is needed to neutralize
250.0 mL of 2.0 M sodium hydroxide solution.
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Question 15
Question
A chemist prepared a solution by dissolving 15.0 g of glucose (C6H12O6) in
enough water to make 250.0 mL of solution.
1. Calculate the molarity of the glucose solution.
2. If 20.0 mL of the solution is diluted to 100.0 mL, what is the new molarity
of the solution?
Solution
1. To calculate the molarity of the glucose solution, we first need to find the
number of moles of glucose in the solution.
Step 1: Calculate moles of glucose
Given: Mass of glucose = 15.0 g Molar mass of glucose (C6H12O6) = 180.16
g/mol
Number of moles of glucose:
moles = mass
molar mass =15.0 g
180.16 g/mol
moles = 0.0833 mol
Step 2: Calculate molarity
Molarity is defined as moles of solute per liter of solution. Given: Volume of
solution = 250.0 mL = 0.2500 L
Molarity of the glucose solution:
Molarity = moles of solute
volume of solution in liters =0.0833 mol
0.2500 L
Molarity = 0.3332 M
2. When 20.0 mL of the solution is diluted to 100.0 mL, the moles of solute
remain the same. Therefore, the molarity of the solution after dilution can be
calculated as follows:
Step 3: Calculate new molarity
New volume of solution after dilution = 100.0 mL = 0.1000 L
New molarity:
New molarity = moles of solute
volume of solution in liters =0.0833 mol
0.1000 L
New molarity = 0.8330 M
Therefore, the new molarity of the solution after dilution is 0.8330 M.
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Question 16
Question
A chemist wants to prepare 500 mL of a solution with a molarity of 0.25 M. The
chemist has a stock solution of 1 M available to make the dilution. What volume
of the stock solution should be used, and how much solvent (water) needs to be
added to make the desired solution?
Solution
Let V1be the volume of the stock solution to be used and V2be the volume of
water (solvent) to be added.
Step 1: Use the formula for dilution of solutions:
M1V1=M2V2
where M1= initial molarity of stock solution = 1 M, M2= final molarity of
desired solution = 0.25 M, V1= volume of stock solution to be used (unknown),
V2= volume of water to be added (unknown).
Step 2: Substitute the given values into the formula:
1 M ·V1= 0.25 M ·500 mL
Step 3: Solve for V1:
V1=0.25 M ·500 mL
1 M
V1= 125 mL
Therefore, 125 mL of the stock solution should be used.
Step 4: Calculate the volume of water to be added:
V2= 500 mL −125 mL = 375 mL
So, 375 mL of water needs to be added to make the desired 500 mL solution
with a molarity of 0.25 M.
Question 17
Question
A solution is prepared by dissolving 25.0 g of potassium hydroxide (KOH) in
enough water to make 250.0 mL of solution. Calculate the molarity of the
solution.
13
Solution
Step 1: Calculate the molar mass of KOH.
Molar mass of K = 39.10 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of KOH = 39.10 + 16.00 + 1.01 = 56.11 g/mol
Step 2: Calculate the number of moles of KOH in 25.0 g.
Number of moles = Mass
Molar mass
Number of moles = 25.0 g
56.11 g/mol = 0.445 mol
Step 3: Calculate the molarity of the solution.
Molarity = Number of moles
Volume of solution in L
Volume of solution = 250.0 mL = 0.250 L
Molarity = 0.445 mol
0.250 L = 1.78 M
Therefore, the molarity of the solution is 1.78 M.
Question 18
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of potassium chloride
(KCl). However, the chemist only has a 0.5 M stock solution of potassium
chloride. How can the chemist prepare the desired solution using the stock
solution and distilled water?
Solution
Let V1be the volume of the 0.5 M stock solution of potassium chloride and V2
be the volume of distilled water that needs to be added to prepare 500 mL of a
0.2 M solution.
Step 1: Write the equation for dilution of solutions, which states that the
amount of solute remains constant when a solution is diluted:
V1×C1= (V1+V2)×C2
where: - C1is the initial concentration of the stock solution (0.5 M), - C2is
the final concentration of the diluted solution (0.2 M), and - V1and V2are the
volumes of the stock solution and distilled water, respectively.
14
Step 2: Plug in the known values into the dilution equation:
V1×0.5=(V1+ 500) ×0.2
Step 3: Solve for V1:
0.5V1= 0.2(V1+ 500)
0.5V1= 0.2V1+ 100
0.3V1= 100
V1=100
0.3= 333.3 mL
Step 4: Calculate the volume of distilled water required:
V2= 500 −V1= 500 −333.3 = 166.6 mL
Therefore, the chemist needs to mix 333.3mLofthe0.5Mstocksolutionwith166.6mLofdistilledwatertoprepare500mLofa0.2Msolutionofpotassiumchloride.
Question 19
Question
A chemist wants to prepare 500 mL of a 0.2 M hydrochloric acid solution. The
stock solution available is 6 M. What volume of the stock solution should be
used and how much water should be added to prepare the desired solution?
Solution
Step 1: Calculate the volume of the stock solution needed.
Given: Desired concentration of hydrochloric acid solution, Cfinal = 0.2 M
Concentration of the stock solution, Cstock = 6 M Volume of the final solution,
Vfinal = 500 mL
Using the formula for dilution:
Cstock ·Vstock =Cfinal ·Vfinal
Let Vstock be the volume of stock solution needed.
Substitute the given values:
6 M ·Vstock = 0.2 M ·500 mL
Vstock =0.2 M ·500 mL
6 M
Vstock =100
6= 16.67 mL
15
Therefore, 16.67 mL of the stock solution should be used.
Step 2: Calculate the volume of water needed.
The total volume of the final solution is given as 500 mL.
Volume of water required, Vwater =Vfinal −Vstock
Substitute the values:
Vwater = 500 mL −16.67 mL = 483.33 mL
Therefore, 483.33 mL of water should be added to the stock solution to
prepare a 500 mL of 0.2 M hydrochloric acid solution.
Question 20
Question
A chemist prepares a solution by dissolving 5.00 g of sodium hydroxide in enough
water to make 200.0 mL of solution. Calculate the molarity of the resulting
solution.
Solution
Step 1: Calculate the molar mass of NaOH Given that: - The molar mass of Na
is 22.99 g/mol - The molar mass of O is 16.00 g/mol - The molar mass of H is
1.01 g/mol
The molar mass of NaOH is:
22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
Step 2: Calculate the number of moles of NaOH Given that the mass of
NaOH is 5.00 g, and using the molar mass calculated in Step 1:
Number of moles = 5.00 g
40.00 g/mol = 0.125 mol
Step 3: Calculate the molarity of the solution The volume of solution is 200.0
mL, which is equivalent to 0.2000 L. Using the number of moles calculated in
Step 2:
Molarity = 0.125 mol
0.2000 L = 0.625 M
Therefore, the molarity of the resulting solution is 0.625 M.
Question 21
Question
Calculate the molarity of a solution prepared by dissolving 25.0 g of sodium
chloride (NaCl) in enough water to make 500.0 mL of solution.
16
Solution
Step 1: Calculate the molar mass of sodium chloride (NaCl).
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Step 2: Convert the mass of sodium chloride to moles.
Moles of NaCl = 25.0 g
58.44 g/mol = 0.428 mol
Step 3: Convert the volume of the solution from milliliters to liters.
Volume of solution = 500.0 mL = 0.500 L
Step 4: Calculate the molarity of the solution.
Molarity = Moles of solute
Volume of solution in liters =0.428 mol
0.500 L = 0.856 M
Therefore, the molarity of the solution prepared by dissolving 25.0 g of
sodium chloride in enough water to make 500.0 mL of solution is 0.856 M.
Question 22
Question
A chemist prepares a solution by dissolving 10.0 g of potassium iodide (KI)
in enough water to make 250.0 mL of solution. Calculate the molarity of the
potassium iodide solution.
Solution
Step 1: Calculate the molar mass of potassium iodide (KI). The molar mass of
KI can be calculated as follows:
Molar mass of KI = Atomic mass of K + Atomic mass of I
Molar mass of KI = 39.10 g/mol + 126.90 g/mol = 166.00 g/mol
Step 2: Determine the number of moles of potassium iodide (KI) in 10.0 g.
The number of moles can be calculated using the formula:
moles = mass
molar mass
moles of KI = 10.0 g
166.00 g/mol = 0.0602 mol
Step 3: Calculate the molarity of the potassium iodide solution. The formula
for molarity is:
Molarity = moles of solute
volume of solution in liters
17
Volume of solution = 250.0 mL
1000 mL/L = 0.2500 L
Molarity = 0.0602 mol
0.2500 L = 0.2408 M
Therefore, the molarity of the potassium iodide solution is 0.2408 M.
Question 23
Question
A solution is prepared by dissolving 7.50 g of potassium permanganate (KMnO4)
in enough water to make 500.0 mL of solution. What is the molarity of the
solution?
Solution
Step 1: Calculate the molar mass of KMnO4. The molar mass of K is 39.10
g/mol, Mn is 54.94 g/mol, and O is 16.00 g/mol. So, the molar mass of KMnO4
is
molar mass of KMnO4= 39.10 + 54.94 + 4(16.00) = 158.04 g/mol
Step 2: Calculate the number of moles of KMnO4in 7.50 g. Using the
formula n=m
M, where nis the number of moles, mis the mass, and Mis the
molar mass, we have
n=7.50 g
158.04 g/mol = 0.0474 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as the
number of moles of solute per liter of solution. Given that the number of moles
of KMnO4is 0.0474 mol and the volume of the solution is 500.0 mL (or 0.500
L), we have
Molarity (M) = 0.0474 mol
0.500 L= 0.0948 M
Therefore, the molarity of the solution is 0.0948 M.
Question 24
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of a compound, but
only has a stock solution that is 1 M. What volume of the stock solution should
be used and how should the solution be prepared?
18
Solution
Step 1: Use the formula for Molarity to find the volume of the stock solution
needed. The formula for Molarity (M) is:
M=moles of solute
liters of solution
Given that the chemist wants to prepare 500 mL of a 0.2 M solution, we can
rearrange the formula to solve for the moles of solute:
moles of solute = M×liters of solution
moles of solute = 0.2 M ×0.5 L = 0.1 moles
Step 2: Use the moles of solute to find the volume of the stock solution
needed. Since the stock solution is 1 M, the moles of solute present in 1 L of
the solution is equal to the Molarity of the solution. So, the volume of the stock
solution needed can be calculated using the formula:
moles of solute = M×volume of solution =⇒0.1 moles = 1 M×volume of solution
Volume of solution = 0.1 moles
1 M = 0.1 liters = 100 mL
Step 3: Prepare the solution. To prepare the 0.2 M solution, the chemist
should measure out 100 mL of the 1 M stock solution using a volumetric flask and
then dilute it with water to reach a total volume of 500 mL. Additional water
should be added up to the 500 mL mark on the flask and then the solution
should be thoroughly mixed to ensure homogeneity.
Question 25
Question
A chemist wants to prepare a solution by diluting 50.0 mL of a 3.00 M sulfuric
acid (H2SO4) solution to a final volume of 250.0 mL. What will be the molarity
of the final solution?
Solution
Step 1: Calculate the number of moles of sulfuric acid in the initial solution.
mol = M ×L = 3.00 mol/L ×0.0500 L = 0.150 mol
Step 2: Use the moles of sulfuric acid and final volume to determine the
molarity of the final solution.
Mfinal =mol
Lfinal
=0.150 mol
0.250 L = 0.60 M
Answer: The molarity of the final solution will be 0.60 M.
19
Question 26
Question
A chemist has a solution containing 25.0 g of NaCl dissolved in 500.0 mL of
water. What is the molarity of the solution? (Molar mass of NaCl is 58.44
g/mol)
Solution
Step 1: Calculate the number of moles of NaCl in the solution.
Moles of NaCl = Mass of NaCl
Molar mass of NaCl =25.0 g
58.44 g/mol
Moles of NaCl = 0.428 mol
Step 2: Calculate the volume of the solution in liters.
Volume of solution = 500.0 mL = 0.500 L
Step 3: Calculate the molarity of the solution.
Molarity = Moles of solute
Volume of solution in liters =0.428 mol
0.500 L
Molarity = 0.856 M
Therefore, the molarity of the solution is 0.856 M.
Question 27
Question
A solution is prepared by dissolving 25.0 g of potassium chloride (KCl) in enough
water to make 500.0 mL of solution. What is the molarity of the solution?
Solution
Step 1: Calculate the molar mass of potassium chloride (KCl). The molar mass
of KCl is the sum of the atomic masses of potassium (K) and chlorine (Cl).
Molar mass of KCl = Atomic mass of K + Atomic mass of Cl
The atomic mass of potassium (K) is 39.10 g/mol, and the atomic mass of
chlorine (Cl) is 35.45 g/mol.
Molar mass of KCl = 39.10 g/mol + 35.45 g/mol = 74.55 g/mol
20
Step 2: Calculate the number of moles of potassium chloride dissolved. Given
that 25.0 g of KCl is dissolved, we can calculate the number of moles using the
formula:
Number of moles = Mass
Molar mass
Number of moles = 25.0 g
74.55 g/mol = 0.335 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as the
number of moles of solute per liter of solution. To calculate molarity, we use
the formula:
Molarity = Number of moles of solute
Volume of solution in liters
Given that the volume of the solution is 500.0 mL (or 0.500 L), we can
calculate the molarity:
Molarity = 0.335 mol
0.500 L = 0.670 M
Therefore, the molarity of the solution prepared is 0.670 M.
Question 28
Question
A solution is prepared by dissolving 5.00 grams of sodium chloride (NaCl) in
enough water to make 250.0 mL of solution. Calculate the molarity (M) of the
resulting solution.
Solution
Step 1: Calculate the molar mass of NaCl. The molar mass of NaCl is the sum
of the atomic masses of sodium (Na) and chlorine (Cl). Molar mass of NaCl =
Molar mass of Na + Molar mass of Cl = 22.99 g/mol+35.45 g/mol = 58.44 g/mol
Step 2: Calculate the number of moles of NaCl. Given mass of NaCl =
5.00 grams and molar mass of NaCl = 58.44 g/mol Number of moles of NaCl =
5.00 g
58.44 g/mol = 0.0856 mol
Step 3: Calculate the molarity of the solution. Molarity (M) is defined as
the number of moles of solute per liter of solution. Volume of solution = 250.0
mL = 0.250 L Molarity = 0.0856 mol
0.250 L = 0.3424 M
Therefore, the molarity of the resulting solution is 0.3424 M.
21
Question 29
Question
A chemist wants to prepare 2.00 L of a 0.100 M solution of sodium chloride
(NaCl). The only stock solution available is a 2.00 M solution of NaCl. How
many milliliters of the stock solution should the chemist use to make the desired
solution?
Solution
Let’s denote the volume of the 2.00 M stock solution that the chemist needs to
use as xmL.
Step 1: Calculate the amount of solute in the desired solution. The
amount of solute in the final solution can be calculated using the formula:
Amount of solute = Molarity ×Volume
For the final solution:
Amount of solute in final solution = 0.100 mol/L ×2.00 L = 0.200 mol
Step 2: Calculate the amount of solute in the stock solution. Using
the same formula as Step 1:
Amount of solute in stock solution = 2.00 mol/L ×(xmL/1000) = 2x/1000 mol
Step 3: Set up the equation and solve for x.Since the amount of solute
in the final solution should equal the amount of solute in the stock solution, we
have:
0.200 mol = 2x/1000 mol
x= (0.200 ×1000)/2 = 100 mL
Therefore, the chemist should use 100 mL of the 2.00 M stock solution to
prepare the desired 0.100 M solution of NaCl.
Question 30
Question
A solution is prepared by dissolving 15.0 g of glucose (C6H12O6) in enough water
to make 250.0 mL of solution. What is the molarity of the glucose solution?
(Molar mass of glucose = 180.16 g/mol)
22
Question 4
Question
A chemist has a solution of hydrochloric acid that has a concentration of 12.0
M. The chemist needs to dilute this solution to make 500.0 mL of a 3.00 M HCl
solution. How much water (in mL) should be added to the original solution to
achieve the desired concentration?
Solution
Step 1: Let the volume of the original 12.0 M HCl solution be V1in mL. The
molarity-volume equation, M1V1=M2V2, can be used to solve this problem.
Step 2: We are given: - M1= 12.0 M (initial concentration of HCl solution)
-M2= 3.00 M (desired concentration of HCl solution) - V2= 500.0 mL (final
diluted volume of HCl solution)
Step 3: Using the molarity-volume equation, we can find V1:
M1V1=M2V2
12.0×V1= 3.00 ×500.0
Step 4: Solving for V1:
V1=3.00 ×500.0
12.0= 125.0 mL
Step 5: To find the volume of water that needs to be added, we subtract the
volume of the original solution from the total desired volume:
Volume of water to be added = V2−V1= 500.0−125.0 = 375.0 mL
Therefore, the chemist should add 375.0 mL of water to the original 12.0 M
HCl solution to achieve a final concentration of 3.00 M.
Question 5
Question
A solution is prepared by dissolving 4.5 grams of potassium permanganate
(KMnO4) in enough water to make 250 mL of solution. Calculate the molarity
of the solution.
Solution
Step 1: Calculate the molar mass of KMnO4. The molar mass of K = 39.10
g/mol, Mn = 54.94 g/mol, and O4= 4 ×16.00 g/mol. Adding them up, we get:
39.10 + 54.94 + 4 ×16.00 = 39.10 + 54.94 + 64.00 = 158.04 g/mol
4
Step 2: Calculate the number of moles of potassium permanganate in the
solution. The number of moles is given by:
moles = mass
molar mass =4.5 g
158.04 g/mol = 0.0285 mol
Step 3: Calculate the volume of the solution in liters.
250 mL = 250 ×10−3L=0.250 L
Step 4: Calculate the molarity of the solution. The molarity is given by:
Molarity =moles
volume in liters =0.0285 mol
0.250 L = 0.114 mol/L
Therefore, the molarity of the solution is 0.114 mol/L.
Question 6
Question
A chemist mixes 250 mL of a 0.5 M solution of NaCl with 500 mL of a 1 M
solution of NaCl. What is the final molarity of NaCl in the resulting solution?
Solution
Step 1: Calculate the moles of NaCl in the first solution.
Moles of NaCl = Molarity ×Volume (L)
Moles of NaCl = 0.5 M ×0.250 L = 0.125 moles
Step 2: Calculate the moles of NaCl in the second solution.
Moles of NaCl = Molarity ×Volume (L)
Moles of NaCl = 1 M ×0.500 L = 0.500 moles
Step 3: Add the moles of NaCl from both solutions to find the total moles
of NaCl in the final solution.
Total moles of NaCl = 0.125 moles + 0.500 moles = 0.625 moles
Step 4: Calculate the total volume of the resulting solution.
Total volume (L) = 0.250 L + 0.500 L = 0.750 L
Step 5: Calculate the final molarity of NaCl in the resulting solution using
the total moles and total volume.
Molarity = Total moles of NaCl
Total volume (L)
Molarity = 0.625 moles
0.750 L = 0.833 M
Answer: The final molarity of NaCl in the resulting solution is 0.833 M.
5
Question 7
Question
A chemist mixes 250 mL of a 0.5 M solution of hydrochloric acid with 500 mL of
a 1.2 M solution of hydrochloric acid. What is the final molarity of the resulting
solution?
Solution
Step 1: Calculate the moles of hydrochloric acid in the first solution. Given:
Volume of solution 1, V1= 250 mL = 0.25 L Molarity of solution 1, M1= 0.5
M
Using the formula for moles,
molessolution 1 =M1×V1= 0.5 mol/L ×0.25 L = 0.125 mol
Step 2: Calculate the moles of hydrochloric acid in the second solution.
Given: Volume of solution 2, V2= 500 mL = 0.5 L Molarity of solution 2,
M2= 1.2 M
Using the formula for moles,
molessolution 2 =M2×V2= 1.2 mol/L ×0.5 L = 0.6 mol
Step 3: Calculate the total moles of hydrochloric acid in the final solution.
Total moles of HCl = molessolution 1+molessolution 2 = 0.125 mol+0.6 mol = 0.725 mol
Step 4: Calculate the total volume of the final solution.
Total volume of solution = V1+V2= 0.25 L + 0.5 L = 0.75 L
Step 5: Calculate the final molarity of the resulting solution.
Final molarity = Total moles of HCl
Total volume of solution =0.725 mol
0.75 L = 0.9667 M
Therefore, the final molarity of the resulting solution is 0.9667 M.
Question 8
Question
Calculate the molarity of a solution obtained by dissolving 5.8 g of potassium
permanganate (KMnO4) in enough water to make 350 mL of solution.
6
Solution
Step 1: Calculate the molar mass of potassium permanganate (KMnO4).
Molar mass of KMnO4= Atomic mass of K + Atomic mass of Mn + 4(Atomic mass of O)
= 39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol)
= 158.04 g/mol
Step 2: Calculate the number of moles of potassium permanganate.
Number of moles = Mass
Molar mass
=5.8 g
158.04 g/mol
≈0.0367 mol
Step 3: Calculate the volume of the solution in liters.
Volume = 350 mL = 0.350 L
Step 4: Calculate the molarity of the solution.
Molarity = Number of moles
Volume in liters
=0.0367 mol
0.350 L
≈0.105 M
Therefore, the molarity of the solution is approximately 0.105 M.
Question 9
Question
A chemist prepares a solution by dissolving 5.00 g of sodium hydroxide in enough
water to make 250.0 mL of solution. Calculate the molarity of the resulting
solution. The molar mass of NaOH is 40.00 g/mol.
Solution
Step 1: Calculate the number of moles of sodium hydroxide.
Moles of NaOH = Mass
Molar mass =5.00 g
40.00 g/mol = 0.125 mol
Step 2: Convert the volume of the solution to liters.
Volume of solution = 250.0 mL = 250.0×10−3L=0.2500 L
7
Step 3: Calculate the molarity of the solution.
Molarity = Moles of solute
Volume of solution (in L) =0.125 mol
0.2500 L = 0.500 M
Therefore, the molarity of the resulting solution is 0.500 M.
Question 10
Question
A chemist has a solution of acetic acid that has a molarity of 1.50 M. She wants
to dilute this solution to create 500.0 mL of a new solution with a molarity of
0.500 M. How much of the original solution should she use and how much water
should she add to achieve this dilution?
Solution
Step 1: Let Voriginal be the volume of the original 1.50 M acetic acid solution
that the chemist will use. The final volume of the diluted solution will be 500.0
mL or 0.500 L. Let Vwater be the volume of water that will be added to dilute
the solution. Then the total volume Vtotal is given by:
Vtotal =Voriginal +Vwater = 0.500 L
Step 2: Recall that the relationship between the molarity, volume, and num-
ber of moles of a solution is given by the formula:
M1V1=M2V2
Where M1and V1are the initial molarity and volume, and M2and V2are the
final molarity and volume.
Step 3: We can set up two equations based on the given information: First,
based on the amount of acetic acid in the original solution and the diluted
solution, we get:
1.50 M ×Voriginal = 0.500 M ×0.500 L
Step 4: Solve the equation for Voriginal:
Voriginal =0.500 M ×0.500 L
1.50 M = 0.167 L = 167 mL
So, the chemist should use 167 mL of the original 1.50 M acetic acid solution.
Step 5: Calculate the volume of water needed to dilute the solution:
Vtotal =Voriginal +Vwater
0.500 L = 0.167 L + Vwater
8
Vwater = 0.500 L −0.167 L = 0.333 L = 333 mL
The chemist should add 333 mL of water to the 167 mL of the original 1.50
M acetic acid solution to create 500.0 mL of a new solution with a molarity of
0.500 M.
Question 11
Question
A solution is prepared by dissolving 15.0 g of glucose (C6H12O6) in enough water
to make 250.0 mL of solution. What is the molarity of the glucose solution?
(Glucose molar mass: 180.16 g/mol)
Solution
Step 1: Calculate the number of moles of glucose in the solution.
Moles of glucose = Mass of glucose (g)
Molar mass of glucose (g/mol)
Moles of glucose = 15.0 g
180.16 g/mol = 0.0833 mol
Step 2: Convert the volume of solution to liters.
Volume of solution (L) = 250.0 mL
1000 mL/L = 0.2500 L
Step 3: Calculate the molarity of the glucose solution.
Molarity (M) = Moles of solute
Volume of solution (L)
Molarity (M) = 0.0833 mol
0.2500 L = 0.333M
Therefore, the molarity of the glucose solution is 0.333 M.
Question 12
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of potassium perman-
ganate (KMnO4) from a stock solution that has a molarity of 1.0 M. What
volume of the stock solution should be used and how much water should be
added to dilute it?
9
Solution
Let V1be the volume of the stock solution to be used and V2be the volume of
water to be added to dilute it.
Step 1: Determine the moles of KMnO4needed for the final solution.
Moles of KMnO4= Molarity ×Volume (in L)
Moles of KMnO4= 0.2 M ×0.5 L = 0.1 mol
Step 2: Calculate the moles of KMnO4present in the stock solution.
Moles of KMnO4= Molarity ×Volume (in L)
0.1 mol = 1.0 M ×V1L
V1= 0.1 L = 100 mL
Step 3: Calculate the volume of water needed to dilute the stock solution.
V2= Volume of final solution −V1
V2= 500 mL −100 mL = 400 mL
Step 4: Write the final answer. The chemist should use 100 mL of the
stock solution and add 400 mL of water to prepare 500 mL of a 0.2 M KMnO4
solution.
Question 13
Question
A chemist prepares a solution by mixing 50.0 mL of a 0.200 M potassium per-
manganate solution with 150.0 mL of water. What is the molarity of the final
solution?
Solution
Step 1: Calculate the moles of potassium permanganate in the original solution.
Step 2: Determine the volume of the final solution. Step 3: Calculate the
molarity of the final solution.
Step 1: The moles of potassium permanganate in the original solution can
be calculated using the formula:
moles = Molarity ×Volume (L)
Given: Molarity = 0.200 M Volume = 50.0 mL = 0.050 L
Substitute the values into the formula:
moles = 0.200 M ×0.050 L = 0.010 moles
10
Step 2: The total volume of the final solution can be calculated by adding
the volume of the potassium permanganate solution and the volume of water:
Total volume = 50.0 mL + 150.0 mL = 200.0 mL = 0.200 L
Step 3: The molarity of the final solution can be calculated using the
formula:
Molarity = moles
Total volume (L)
Substitute the values into the formula:
Molarity = 0.010 moles
0.200 L = 0.050 M
Therefore, the molarity of the final solution is 0.050 M.
Question 14
Question
Calculate the volume of a 5.0 M sulfuric acid solution needed to neutralize 250.0
mL of a 2.0 M sodium hydroxide solution.
Solution
Step 1: Write the balanced chemical equation for the reaction between sulfuric
acid (H2SO4) and sodium hydroxide (NaOH):
H2SO4+ 2NaOH →Na2SO4+ 2H2O
Step 2: Determine the moles of sodium hydroxide (NaOH) using the given
molarity and volume:
Moles of NaOH = Molarity ×Volume = 2.0 M ×0.250 L = 0.500 moles NaOH
Step 3: From the balanced equation, we see that 1 mole of sulfuric acid
(H2SO4) reacts with 2 moles of sodium hydroxide (NaOH). Therefore, the moles
of sulfuric acid required will be half of the moles of sodium hydroxide used in
the reaction.
Moles of H2SO4=1
2×Moles of NaOH = 1
2×0.500 = 0.250 moles H2SO4
Step 4: Calculate the volume of 5.0 M sulfuric acid solution needed to contain
0.250 moles of H2SO4:
Volume = Moles
Molarity =0.250 moles
5.0 M = 0.0500 L = 50.0 mL
Therefore, 50.0 mL of 5.0 M sulfuric acid solution is needed to neutralize
250.0 mL of 2.0 M sodium hydroxide solution.
11
Question 15
Question
A chemist prepared a solution by dissolving 15.0 g of glucose (C6H12O6) in
enough water to make 250.0 mL of solution.
1. Calculate the molarity of the glucose solution.
2. If 20.0 mL of the solution is diluted to 100.0 mL, what is the new molarity
of the solution?
Solution
1. To calculate the molarity of the glucose solution, we first need to find the
number of moles of glucose in the solution.
Step 1: Calculate moles of glucose
Given: Mass of glucose = 15.0 g Molar mass of glucose (C6H12O6) = 180.16
g/mol
Number of moles of glucose:
moles = mass
molar mass =15.0 g
180.16 g/mol
moles = 0.0833 mol
Step 2: Calculate molarity
Molarity is defined as moles of solute per liter of solution. Given: Volume of
solution = 250.0 mL = 0.2500 L
Molarity of the glucose solution:
Molarity = moles of solute
volume of solution in liters =0.0833 mol
0.2500 L
Molarity = 0.3332 M
2. When 20.0 mL of the solution is diluted to 100.0 mL, the moles of solute
remain the same. Therefore, the molarity of the solution after dilution can be
calculated as follows:
Step 3: Calculate new molarity
New volume of solution after dilution = 100.0 mL = 0.1000 L
New molarity:
New molarity = moles of solute
volume of solution in liters =0.0833 mol
0.1000 L
New molarity = 0.8330 M
Therefore, the new molarity of the solution after dilution is 0.8330 M.
12
Question 16
Question
A chemist wants to prepare 500 mL of a solution with a molarity of 0.25 M. The
chemist has a stock solution of 1 M available to make the dilution. What volume
of the stock solution should be used, and how much solvent (water) needs to be
added to make the desired solution?
Solution
Let V1be the volume of the stock solution to be used and V2be the volume of
water (solvent) to be added.
Step 1: Use the formula for dilution of solutions:
M1V1=M2V2
where M1= initial molarity of stock solution = 1 M, M2= final molarity of
desired solution = 0.25 M, V1= volume of stock solution to be used (unknown),
V2= volume of water to be added (unknown).
Step 2: Substitute the given values into the formula:
1 M ·V1= 0.25 M ·500 mL
Step 3: Solve for V1:
V1=0.25 M ·500 mL
1 M
V1= 125 mL
Therefore, 125 mL of the stock solution should be used.
Step 4: Calculate the volume of water to be added:
V2= 500 mL −125 mL = 375 mL
So, 375 mL of water needs to be added to make the desired 500 mL solution
with a molarity of 0.25 M.
Question 17
Question
A solution is prepared by dissolving 25.0 g of potassium hydroxide (KOH) in
enough water to make 250.0 mL of solution. Calculate the molarity of the
solution.
13
Solution
Step 1: Calculate the molar mass of KOH.
Molar mass of K = 39.10 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of KOH = 39.10 + 16.00 + 1.01 = 56.11 g/mol
Step 2: Calculate the number of moles of KOH in 25.0 g.
Number of moles = Mass
Molar mass
Number of moles = 25.0 g
56.11 g/mol = 0.445 mol
Step 3: Calculate the molarity of the solution.
Molarity = Number of moles
Volume of solution in L
Volume of solution = 250.0 mL = 0.250 L
Molarity = 0.445 mol
0.250 L = 1.78 M
Therefore, the molarity of the solution is 1.78 M.
Question 18
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of potassium chloride
(KCl). However, the chemist only has a 0.5 M stock solution of potassium
chloride. How can the chemist prepare the desired solution using the stock
solution and distilled water?
Solution
Let V1be the volume of the 0.5 M stock solution of potassium chloride and V2
be the volume of distilled water that needs to be added to prepare 500 mL of a
0.2 M solution.
Step 1: Write the equation for dilution of solutions, which states that the
amount of solute remains constant when a solution is diluted:
V1×C1= (V1+V2)×C2
where: - C1is the initial concentration of the stock solution (0.5 M), - C2is
the final concentration of the diluted solution (0.2 M), and - V1and V2are the
volumes of the stock solution and distilled water, respectively.
14
Step 2: Plug in the known values into the dilution equation:
V1×0.5=(V1+ 500) ×0.2
Step 3: Solve for V1:
0.5V1= 0.2(V1+ 500)
0.5V1= 0.2V1+ 100
0.3V1= 100
V1=100
0.3= 333.3 mL
Step 4: Calculate the volume of distilled water required:
V2= 500 −V1= 500 −333.3 = 166.6 mL
Therefore, the chemist needs to mix 333.3mLofthe0.5Mstocksolutionwith166.6mLofdistilledwatertoprepare500mLofa0.2Msolutionofpotassiumchloride.
Question 19
Question
A chemist wants to prepare 500 mL of a 0.2 M hydrochloric acid solution. The
stock solution available is 6 M. What volume of the stock solution should be
used and how much water should be added to prepare the desired solution?
Solution
Step 1: Calculate the volume of the stock solution needed.
Given: Desired concentration of hydrochloric acid solution, Cfinal = 0.2 M
Concentration of the stock solution, Cstock = 6 M Volume of the final solution,
Vfinal = 500 mL
Using the formula for dilution:
Cstock ·Vstock =Cfinal ·Vfinal
Let Vstock be the volume of stock solution needed.
Substitute the given values:
6 M ·Vstock = 0.2 M ·500 mL
Vstock =0.2 M ·500 mL
6 M
Vstock =100
6= 16.67 mL
15
Therefore, 16.67 mL of the stock solution should be used.
Step 2: Calculate the volume of water needed.
The total volume of the final solution is given as 500 mL.
Volume of water required, Vwater =Vfinal −Vstock
Substitute the values:
Vwater = 500 mL −16.67 mL = 483.33 mL
Therefore, 483.33 mL of water should be added to the stock solution to
prepare a 500 mL of 0.2 M hydrochloric acid solution.
Question 20
Question
A chemist prepares a solution by dissolving 5.00 g of sodium hydroxide in enough
water to make 200.0 mL of solution. Calculate the molarity of the resulting
solution.
Solution
Step 1: Calculate the molar mass of NaOH Given that: - The molar mass of Na
is 22.99 g/mol - The molar mass of O is 16.00 g/mol - The molar mass of H is
1.01 g/mol
The molar mass of NaOH is:
22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
Step 2: Calculate the number of moles of NaOH Given that the mass of
NaOH is 5.00 g, and using the molar mass calculated in Step 1:
Number of moles = 5.00 g
40.00 g/mol = 0.125 mol
Step 3: Calculate the molarity of the solution The volume of solution is 200.0
mL, which is equivalent to 0.2000 L. Using the number of moles calculated in
Step 2:
Molarity = 0.125 mol
0.2000 L = 0.625 M
Therefore, the molarity of the resulting solution is 0.625 M.
Question 21
Question
Calculate the molarity of a solution prepared by dissolving 25.0 g of sodium
chloride (NaCl) in enough water to make 500.0 mL of solution.
16
Solution
Step 1: Calculate the molar mass of sodium chloride (NaCl).
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Step 2: Convert the mass of sodium chloride to moles.
Moles of NaCl = 25.0 g
58.44 g/mol = 0.428 mol
Step 3: Convert the volume of the solution from milliliters to liters.
Volume of solution = 500.0 mL = 0.500 L
Step 4: Calculate the molarity of the solution.
Molarity = Moles of solute
Volume of solution in liters =0.428 mol
0.500 L = 0.856 M
Therefore, the molarity of the solution prepared by dissolving 25.0 g of
sodium chloride in enough water to make 500.0 mL of solution is 0.856 M.
Question 22
Question
A chemist prepares a solution by dissolving 10.0 g of potassium iodide (KI)
in enough water to make 250.0 mL of solution. Calculate the molarity of the
potassium iodide solution.
Solution
Step 1: Calculate the molar mass of potassium iodide (KI). The molar mass of
KI can be calculated as follows:
Molar mass of KI = Atomic mass of K + Atomic mass of I
Molar mass of KI = 39.10 g/mol + 126.90 g/mol = 166.00 g/mol
Step 2: Determine the number of moles of potassium iodide (KI) in 10.0 g.
The number of moles can be calculated using the formula:
moles = mass
molar mass
moles of KI = 10.0 g
166.00 g/mol = 0.0602 mol
Step 3: Calculate the molarity of the potassium iodide solution. The formula
for molarity is:
Molarity = moles of solute
volume of solution in liters
17
Volume of solution = 250.0 mL
1000 mL/L = 0.2500 L
Molarity = 0.0602 mol
0.2500 L = 0.2408 M
Therefore, the molarity of the potassium iodide solution is 0.2408 M.
Question 23
Question
A solution is prepared by dissolving 7.50 g of potassium permanganate (KMnO4)
in enough water to make 500.0 mL of solution. What is the molarity of the
solution?
Solution
Step 1: Calculate the molar mass of KMnO4. The molar mass of K is 39.10
g/mol, Mn is 54.94 g/mol, and O is 16.00 g/mol. So, the molar mass of KMnO4
is
molar mass of KMnO4= 39.10 + 54.94 + 4(16.00) = 158.04 g/mol
Step 2: Calculate the number of moles of KMnO4in 7.50 g. Using the
formula n=m
M, where nis the number of moles, mis the mass, and Mis the
molar mass, we have
n=7.50 g
158.04 g/mol = 0.0474 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as the
number of moles of solute per liter of solution. Given that the number of moles
of KMnO4is 0.0474 mol and the volume of the solution is 500.0 mL (or 0.500
L), we have
Molarity (M) = 0.0474 mol
0.500 L= 0.0948 M
Therefore, the molarity of the solution is 0.0948 M.
Question 24
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of a compound, but
only has a stock solution that is 1 M. What volume of the stock solution should
be used and how should the solution be prepared?
18
Solution
Step 1: Use the formula for Molarity to find the volume of the stock solution
needed. The formula for Molarity (M) is:
M=moles of solute
liters of solution
Given that the chemist wants to prepare 500 mL of a 0.2 M solution, we can
rearrange the formula to solve for the moles of solute:
moles of solute = M×liters of solution
moles of solute = 0.2 M ×0.5 L = 0.1 moles
Step 2: Use the moles of solute to find the volume of the stock solution
needed. Since the stock solution is 1 M, the moles of solute present in 1 L of
the solution is equal to the Molarity of the solution. So, the volume of the stock
solution needed can be calculated using the formula:
moles of solute = M×volume of solution =⇒0.1 moles = 1 M×volume of solution
Volume of solution = 0.1 moles
1 M = 0.1 liters = 100 mL
Step 3: Prepare the solution. To prepare the 0.2 M solution, the chemist
should measure out 100 mL of the 1 M stock solution using a volumetric flask and
then dilute it with water to reach a total volume of 500 mL. Additional water
should be added up to the 500 mL mark on the flask and then the solution
should be thoroughly mixed to ensure homogeneity.
Question 25
Question
A chemist wants to prepare a solution by diluting 50.0 mL of a 3.00 M sulfuric
acid (H2SO4) solution to a final volume of 250.0 mL. What will be the molarity
of the final solution?
Solution
Step 1: Calculate the number of moles of sulfuric acid in the initial solution.
mol = M ×L = 3.00 mol/L ×0.0500 L = 0.150 mol
Step 2: Use the moles of sulfuric acid and final volume to determine the
molarity of the final solution.
Mfinal =mol
Lfinal
=0.150 mol
0.250 L = 0.60 M
Answer: The molarity of the final solution will be 0.60 M.
19
Question 26
Question
A chemist has a solution containing 25.0 g of NaCl dissolved in 500.0 mL of
water. What is the molarity of the solution? (Molar mass of NaCl is 58.44
g/mol)
Solution
Step 1: Calculate the number of moles of NaCl in the solution.
Moles of NaCl = Mass of NaCl
Molar mass of NaCl =25.0 g
58.44 g/mol
Moles of NaCl = 0.428 mol
Step 2: Calculate the volume of the solution in liters.
Volume of solution = 500.0 mL = 0.500 L
Step 3: Calculate the molarity of the solution.
Molarity = Moles of solute
Volume of solution in liters =0.428 mol
0.500 L
Molarity = 0.856 M
Therefore, the molarity of the solution is 0.856 M.
Question 27
Question
A solution is prepared by dissolving 25.0 g of potassium chloride (KCl) in enough
water to make 500.0 mL of solution. What is the molarity of the solution?
Solution
Step 1: Calculate the molar mass of potassium chloride (KCl). The molar mass
of KCl is the sum of the atomic masses of potassium (K) and chlorine (Cl).
Molar mass of KCl = Atomic mass of K + Atomic mass of Cl
The atomic mass of potassium (K) is 39.10 g/mol, and the atomic mass of
chlorine (Cl) is 35.45 g/mol.
Molar mass of KCl = 39.10 g/mol + 35.45 g/mol = 74.55 g/mol
20
Step 2: Calculate the number of moles of potassium chloride dissolved. Given
that 25.0 g of KCl is dissolved, we can calculate the number of moles using the
formula:
Number of moles = Mass
Molar mass
Number of moles = 25.0 g
74.55 g/mol = 0.335 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as the
number of moles of solute per liter of solution. To calculate molarity, we use
the formula:
Molarity = Number of moles of solute
Volume of solution in liters
Given that the volume of the solution is 500.0 mL (or 0.500 L), we can
calculate the molarity:
Molarity = 0.335 mol
0.500 L = 0.670 M
Therefore, the molarity of the solution prepared is 0.670 M.
Question 28
Question
A solution is prepared by dissolving 5.00 grams of sodium chloride (NaCl) in
enough water to make 250.0 mL of solution. Calculate the molarity (M) of the
resulting solution.
Solution
Step 1: Calculate the molar mass of NaCl. The molar mass of NaCl is the sum
of the atomic masses of sodium (Na) and chlorine (Cl). Molar mass of NaCl =
Molar mass of Na + Molar mass of Cl = 22.99 g/mol+35.45 g/mol = 58.44 g/mol
Step 2: Calculate the number of moles of NaCl. Given mass of NaCl =
5.00 grams and molar mass of NaCl = 58.44 g/mol Number of moles of NaCl =
5.00 g
58.44 g/mol = 0.0856 mol
Step 3: Calculate the molarity of the solution. Molarity (M) is defined as
the number of moles of solute per liter of solution. Volume of solution = 250.0
mL = 0.250 L Molarity = 0.0856 mol
0.250 L = 0.3424 M
Therefore, the molarity of the resulting solution is 0.3424 M.
21
Question 29
Question
A chemist wants to prepare 2.00 L of a 0.100 M solution of sodium chloride
(NaCl). The only stock solution available is a 2.00 M solution of NaCl. How
many milliliters of the stock solution should the chemist use to make the desired
solution?
Solution
Let’s denote the volume of the 2.00 M stock solution that the chemist needs to
use as xmL.
Step 1: Calculate the amount of solute in the desired solution. The
amount of solute in the final solution can be calculated using the formula:
Amount of solute = Molarity ×Volume
For the final solution:
Amount of solute in final solution = 0.100 mol/L ×2.00 L = 0.200 mol
Step 2: Calculate the amount of solute in the stock solution. Using
the same formula as Step 1:
Amount of solute in stock solution = 2.00 mol/L ×(xmL/1000) = 2x/1000 mol
Step 3: Set up the equation and solve for x.Since the amount of solute
in the final solution should equal the amount of solute in the stock solution, we
have:
0.200 mol = 2x/1000 mol
x= (0.200 ×1000)/2 = 100 mL
Therefore, the chemist should use 100 mL of the 2.00 M stock solution to
prepare the desired 0.100 M solution of NaCl.
Question 30
Question
A solution is prepared by dissolving 15.0 g of glucose (C6H12O6) in enough water
to make 250.0 mL of solution. What is the molarity of the glucose solution?
(Molar mass of glucose = 180.16 g/mol)
22
Question 4
Question
A chemist has a solution of hydrochloric acid that has a concentration of 12.0
M. The chemist needs to dilute this solution to make 500.0 mL of a 3.00 M HCl
solution. How much water (in mL) should be added to the original solution to
achieve the desired concentration?
Solution
Step 1: Let the volume of the original 12.0 M HCl solution be V1in mL. The
molarity-volume equation, M1V1=M2V2, can be used to solve this problem.
Step 2: We are given: - M1= 12.0 M (initial concentration of HCl solution)
-M2= 3.00 M (desired concentration of HCl solution) - V2= 500.0 mL (final
diluted volume of HCl solution)
Step 3: Using the molarity-volume equation, we can find V1:
M1V1=M2V2
12.0×V1= 3.00 ×500.0
Step 4: Solving for V1:
V1=3.00 ×500.0
12.0= 125.0 mL
Step 5: To find the volume of water that needs to be added, we subtract the
volume of the original solution from the total desired volume:
Volume of water to be added = V2−V1= 500.0−125.0 = 375.0 mL
Therefore, the chemist should add 375.0 mL of water to the original 12.0 M
HCl solution to achieve a final concentration of 3.00 M.
Question 5
Question
A solution is prepared by dissolving 4.5 grams of potassium permanganate
(KMnO4) in enough water to make 250 mL of solution. Calculate the molarity
of the solution.
Solution
Step 1: Calculate the molar mass of KMnO4. The molar mass of K = 39.10
g/mol, Mn = 54.94 g/mol, and O4= 4 ×16.00 g/mol. Adding them up, we get:
39.10 + 54.94 + 4 ×16.00 = 39.10 + 54.94 + 64.00 = 158.04 g/mol
4
Step 2: Calculate the number of moles of potassium permanganate in the
solution. The number of moles is given by:
moles = mass
molar mass =4.5 g
158.04 g/mol = 0.0285 mol
Step 3: Calculate the volume of the solution in liters.
250 mL = 250 ×10−3L=0.250 L
Step 4: Calculate the molarity of the solution. The molarity is given by:
Molarity =moles
volume in liters =0.0285 mol
0.250 L = 0.114 mol/L
Therefore, the molarity of the solution is 0.114 mol/L.
Question 6
Question
A chemist mixes 250 mL of a 0.5 M solution of NaCl with 500 mL of a 1 M
solution of NaCl. What is the final molarity of NaCl in the resulting solution?
Solution
Step 1: Calculate the moles of NaCl in the first solution.
Moles of NaCl = Molarity ×Volume (L)
Moles of NaCl = 0.5 M ×0.250 L = 0.125 moles
Step 2: Calculate the moles of NaCl in the second solution.
Moles of NaCl = Molarity ×Volume (L)
Moles of NaCl = 1 M ×0.500 L = 0.500 moles
Step 3: Add the moles of NaCl from both solutions to find the total moles
of NaCl in the final solution.
Total moles of NaCl = 0.125 moles + 0.500 moles = 0.625 moles
Step 4: Calculate the total volume of the resulting solution.
Total volume (L) = 0.250 L + 0.500 L = 0.750 L
Step 5: Calculate the final molarity of NaCl in the resulting solution using
the total moles and total volume.
Molarity = Total moles of NaCl
Total volume (L)
Molarity = 0.625 moles
0.750 L = 0.833 M
Answer: The final molarity of NaCl in the resulting solution is 0.833 M.
5
Question 7
Question
A chemist mixes 250 mL of a 0.5 M solution of hydrochloric acid with 500 mL of
a 1.2 M solution of hydrochloric acid. What is the final molarity of the resulting
solution?
Solution
Step 1: Calculate the moles of hydrochloric acid in the first solution. Given:
Volume of solution 1, V1= 250 mL = 0.25 L Molarity of solution 1, M1= 0.5
M
Using the formula for moles,
molessolution 1 =M1×V1= 0.5 mol/L ×0.25 L = 0.125 mol
Step 2: Calculate the moles of hydrochloric acid in the second solution.
Given: Volume of solution 2, V2= 500 mL = 0.5 L Molarity of solution 2,
M2= 1.2 M
Using the formula for moles,
molessolution 2 =M2×V2= 1.2 mol/L ×0.5 L = 0.6 mol
Step 3: Calculate the total moles of hydrochloric acid in the final solution.
Total moles of HCl = molessolution 1+molessolution 2 = 0.125 mol+0.6 mol = 0.725 mol
Step 4: Calculate the total volume of the final solution.
Total volume of solution = V1+V2= 0.25 L + 0.5 L = 0.75 L
Step 5: Calculate the final molarity of the resulting solution.
Final molarity = Total moles of HCl
Total volume of solution =0.725 mol
0.75 L = 0.9667 M
Therefore, the final molarity of the resulting solution is 0.9667 M.
Question 8
Question
Calculate the molarity of a solution obtained by dissolving 5.8 g of potassium
permanganate (KMnO4) in enough water to make 350 mL of solution.
6
Solution
Step 1: Calculate the molar mass of potassium permanganate (KMnO4).
Molar mass of KMnO4= Atomic mass of K + Atomic mass of Mn + 4(Atomic mass of O)
= 39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol)
= 158.04 g/mol
Step 2: Calculate the number of moles of potassium permanganate.
Number of moles = Mass
Molar mass
=5.8 g
158.04 g/mol
≈0.0367 mol
Step 3: Calculate the volume of the solution in liters.
Volume = 350 mL = 0.350 L
Step 4: Calculate the molarity of the solution.
Molarity = Number of moles
Volume in liters
=0.0367 mol
0.350 L
≈0.105 M
Therefore, the molarity of the solution is approximately 0.105 M.
Question 9
Question
A chemist prepares a solution by dissolving 5.00 g of sodium hydroxide in enough
water to make 250.0 mL of solution. Calculate the molarity of the resulting
solution. The molar mass of NaOH is 40.00 g/mol.
Solution
Step 1: Calculate the number of moles of sodium hydroxide.
Moles of NaOH = Mass
Molar mass =5.00 g
40.00 g/mol = 0.125 mol
Step 2: Convert the volume of the solution to liters.
Volume of solution = 250.0 mL = 250.0×10−3L=0.2500 L
7
Step 3: Calculate the molarity of the solution.
Molarity = Moles of solute
Volume of solution (in L) =0.125 mol
0.2500 L = 0.500 M
Therefore, the molarity of the resulting solution is 0.500 M.
Question 10
Question
A chemist has a solution of acetic acid that has a molarity of 1.50 M. She wants
to dilute this solution to create 500.0 mL of a new solution with a molarity of
0.500 M. How much of the original solution should she use and how much water
should she add to achieve this dilution?
Solution
Step 1: Let Voriginal be the volume of the original 1.50 M acetic acid solution
that the chemist will use. The final volume of the diluted solution will be 500.0
mL or 0.500 L. Let Vwater be the volume of water that will be added to dilute
the solution. Then the total volume Vtotal is given by:
Vtotal =Voriginal +Vwater = 0.500 L
Step 2: Recall that the relationship between the molarity, volume, and num-
ber of moles of a solution is given by the formula:
M1V1=M2V2
Where M1and V1are the initial molarity and volume, and M2and V2are the
final molarity and volume.
Step 3: We can set up two equations based on the given information: First,
based on the amount of acetic acid in the original solution and the diluted
solution, we get:
1.50 M ×Voriginal = 0.500 M ×0.500 L
Step 4: Solve the equation for Voriginal:
Voriginal =0.500 M ×0.500 L
1.50 M = 0.167 L = 167 mL
So, the chemist should use 167 mL of the original 1.50 M acetic acid solution.
Step 5: Calculate the volume of water needed to dilute the solution:
Vtotal =Voriginal +Vwater
0.500 L = 0.167 L + Vwater
8
Vwater = 0.500 L −0.167 L = 0.333 L = 333 mL
The chemist should add 333 mL of water to the 167 mL of the original 1.50
M acetic acid solution to create 500.0 mL of a new solution with a molarity of
0.500 M.
Question 11
Question
A solution is prepared by dissolving 15.0 g of glucose (C6H12O6) in enough water
to make 250.0 mL of solution. What is the molarity of the glucose solution?
(Glucose molar mass: 180.16 g/mol)
Solution
Step 1: Calculate the number of moles of glucose in the solution.
Moles of glucose = Mass of glucose (g)
Molar mass of glucose (g/mol)
Moles of glucose = 15.0 g
180.16 g/mol = 0.0833 mol
Step 2: Convert the volume of solution to liters.
Volume of solution (L) = 250.0 mL
1000 mL/L = 0.2500 L
Step 3: Calculate the molarity of the glucose solution.
Molarity (M) = Moles of solute
Volume of solution (L)
Molarity (M) = 0.0833 mol
0.2500 L = 0.333M
Therefore, the molarity of the glucose solution is 0.333 M.
Question 12
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of potassium perman-
ganate (KMnO4) from a stock solution that has a molarity of 1.0 M. What
volume of the stock solution should be used and how much water should be
added to dilute it?
9
Solution
Let V1be the volume of the stock solution to be used and V2be the volume of
water to be added to dilute it.
Step 1: Determine the moles of KMnO4needed for the final solution.
Moles of KMnO4= Molarity ×Volume (in L)
Moles of KMnO4= 0.2 M ×0.5 L = 0.1 mol
Step 2: Calculate the moles of KMnO4present in the stock solution.
Moles of KMnO4= Molarity ×Volume (in L)
0.1 mol = 1.0 M ×V1L
V1= 0.1 L = 100 mL
Step 3: Calculate the volume of water needed to dilute the stock solution.
V2= Volume of final solution −V1
V2= 500 mL −100 mL = 400 mL
Step 4: Write the final answer. The chemist should use 100 mL of the
stock solution and add 400 mL of water to prepare 500 mL of a 0.2 M KMnO4
solution.
Question 13
Question
A chemist prepares a solution by mixing 50.0 mL of a 0.200 M potassium per-
manganate solution with 150.0 mL of water. What is the molarity of the final
solution?
Solution
Step 1: Calculate the moles of potassium permanganate in the original solution.
Step 2: Determine the volume of the final solution. Step 3: Calculate the
molarity of the final solution.
Step 1: The moles of potassium permanganate in the original solution can
be calculated using the formula:
moles = Molarity ×Volume (L)
Given: Molarity = 0.200 M Volume = 50.0 mL = 0.050 L
Substitute the values into the formula:
moles = 0.200 M ×0.050 L = 0.010 moles
10
Step 2: The total volume of the final solution can be calculated by adding
the volume of the potassium permanganate solution and the volume of water:
Total volume = 50.0 mL + 150.0 mL = 200.0 mL = 0.200 L
Step 3: The molarity of the final solution can be calculated using the
formula:
Molarity = moles
Total volume (L)
Substitute the values into the formula:
Molarity = 0.010 moles
0.200 L = 0.050 M
Therefore, the molarity of the final solution is 0.050 M.
Question 14
Question
Calculate the volume of a 5.0 M sulfuric acid solution needed to neutralize 250.0
mL of a 2.0 M sodium hydroxide solution.
Solution
Step 1: Write the balanced chemical equation for the reaction between sulfuric
acid (H2SO4) and sodium hydroxide (NaOH):
H2SO4+ 2NaOH →Na2SO4+ 2H2O
Step 2: Determine the moles of sodium hydroxide (NaOH) using the given
molarity and volume:
Moles of NaOH = Molarity ×Volume = 2.0 M ×0.250 L = 0.500 moles NaOH
Step 3: From the balanced equation, we see that 1 mole of sulfuric acid
(H2SO4) reacts with 2 moles of sodium hydroxide (NaOH). Therefore, the moles
of sulfuric acid required will be half of the moles of sodium hydroxide used in
the reaction.
Moles of H2SO4=1
2×Moles of NaOH = 1
2×0.500 = 0.250 moles H2SO4
Step 4: Calculate the volume of 5.0 M sulfuric acid solution needed to contain
0.250 moles of H2SO4:
Volume = Moles
Molarity =0.250 moles
5.0 M = 0.0500 L = 50.0 mL
Therefore, 50.0 mL of 5.0 M sulfuric acid solution is needed to neutralize
250.0 mL of 2.0 M sodium hydroxide solution.
11
Question 15
Question
A chemist prepared a solution by dissolving 15.0 g of glucose (C6H12O6) in
enough water to make 250.0 mL of solution.
1. Calculate the molarity of the glucose solution.
2. If 20.0 mL of the solution is diluted to 100.0 mL, what is the new molarity
of the solution?
Solution
1. To calculate the molarity of the glucose solution, we first need to find the
number of moles of glucose in the solution.
Step 1: Calculate moles of glucose
Given: Mass of glucose = 15.0 g Molar mass of glucose (C6H12O6) = 180.16
g/mol
Number of moles of glucose:
moles = mass
molar mass =15.0 g
180.16 g/mol
moles = 0.0833 mol
Step 2: Calculate molarity
Molarity is defined as moles of solute per liter of solution. Given: Volume of
solution = 250.0 mL = 0.2500 L
Molarity of the glucose solution:
Molarity = moles of solute
volume of solution in liters =0.0833 mol
0.2500 L
Molarity = 0.3332 M
2. When 20.0 mL of the solution is diluted to 100.0 mL, the moles of solute
remain the same. Therefore, the molarity of the solution after dilution can be
calculated as follows:
Step 3: Calculate new molarity
New volume of solution after dilution = 100.0 mL = 0.1000 L
New molarity:
New molarity = moles of solute
volume of solution in liters =0.0833 mol
0.1000 L
New molarity = 0.8330 M
Therefore, the new molarity of the solution after dilution is 0.8330 M.
12
Question 16
Question
A chemist wants to prepare 500 mL of a solution with a molarity of 0.25 M. The
chemist has a stock solution of 1 M available to make the dilution. What volume
of the stock solution should be used, and how much solvent (water) needs to be
added to make the desired solution?
Solution
Let V1be the volume of the stock solution to be used and V2be the volume of
water (solvent) to be added.
Step 1: Use the formula for dilution of solutions:
M1V1=M2V2
where M1= initial molarity of stock solution = 1 M, M2= final molarity of
desired solution = 0.25 M, V1= volume of stock solution to be used (unknown),
V2= volume of water to be added (unknown).
Step 2: Substitute the given values into the formula:
1 M ·V1= 0.25 M ·500 mL
Step 3: Solve for V1:
V1=0.25 M ·500 mL
1 M
V1= 125 mL
Therefore, 125 mL of the stock solution should be used.
Step 4: Calculate the volume of water to be added:
V2= 500 mL −125 mL = 375 mL
So, 375 mL of water needs to be added to make the desired 500 mL solution
with a molarity of 0.25 M.
Question 17
Question
A solution is prepared by dissolving 25.0 g of potassium hydroxide (KOH) in
enough water to make 250.0 mL of solution. Calculate the molarity of the
solution.
13
Solution
Step 1: Calculate the molar mass of KOH.
Molar mass of K = 39.10 g/mol
Molar mass of O = 16.00 g/mol
Molar mass of H = 1.01 g/mol
Molar mass of KOH = 39.10 + 16.00 + 1.01 = 56.11 g/mol
Step 2: Calculate the number of moles of KOH in 25.0 g.
Number of moles = Mass
Molar mass
Number of moles = 25.0 g
56.11 g/mol = 0.445 mol
Step 3: Calculate the molarity of the solution.
Molarity = Number of moles
Volume of solution in L
Volume of solution = 250.0 mL = 0.250 L
Molarity = 0.445 mol
0.250 L = 1.78 M
Therefore, the molarity of the solution is 1.78 M.
Question 18
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of potassium chloride
(KCl). However, the chemist only has a 0.5 M stock solution of potassium
chloride. How can the chemist prepare the desired solution using the stock
solution and distilled water?
Solution
Let V1be the volume of the 0.5 M stock solution of potassium chloride and V2
be the volume of distilled water that needs to be added to prepare 500 mL of a
0.2 M solution.
Step 1: Write the equation for dilution of solutions, which states that the
amount of solute remains constant when a solution is diluted:
V1×C1= (V1+V2)×C2
where: - C1is the initial concentration of the stock solution (0.5 M), - C2is
the final concentration of the diluted solution (0.2 M), and - V1and V2are the
volumes of the stock solution and distilled water, respectively.
14
Step 2: Plug in the known values into the dilution equation:
V1×0.5=(V1+ 500) ×0.2
Step 3: Solve for V1:
0.5V1= 0.2(V1+ 500)
0.5V1= 0.2V1+ 100
0.3V1= 100
V1=100
0.3= 333.3 mL
Step 4: Calculate the volume of distilled water required:
V2= 500 −V1= 500 −333.3 = 166.6 mL
Therefore, the chemist needs to mix 333.3mLofthe0.5Mstocksolutionwith166.6mLofdistilledwatertoprepare500mLofa0.2Msolutionofpotassiumchloride.
Question 19
Question
A chemist wants to prepare 500 mL of a 0.2 M hydrochloric acid solution. The
stock solution available is 6 M. What volume of the stock solution should be
used and how much water should be added to prepare the desired solution?
Solution
Step 1: Calculate the volume of the stock solution needed.
Given: Desired concentration of hydrochloric acid solution, Cfinal = 0.2 M
Concentration of the stock solution, Cstock = 6 M Volume of the final solution,
Vfinal = 500 mL
Using the formula for dilution:
Cstock ·Vstock =Cfinal ·Vfinal
Let Vstock be the volume of stock solution needed.
Substitute the given values:
6 M ·Vstock = 0.2 M ·500 mL
Vstock =0.2 M ·500 mL
6 M
Vstock =100
6= 16.67 mL
15
Therefore, 16.67 mL of the stock solution should be used.
Step 2: Calculate the volume of water needed.
The total volume of the final solution is given as 500 mL.
Volume of water required, Vwater =Vfinal −Vstock
Substitute the values:
Vwater = 500 mL −16.67 mL = 483.33 mL
Therefore, 483.33 mL of water should be added to the stock solution to
prepare a 500 mL of 0.2 M hydrochloric acid solution.
Question 20
Question
A chemist prepares a solution by dissolving 5.00 g of sodium hydroxide in enough
water to make 200.0 mL of solution. Calculate the molarity of the resulting
solution.
Solution
Step 1: Calculate the molar mass of NaOH Given that: - The molar mass of Na
is 22.99 g/mol - The molar mass of O is 16.00 g/mol - The molar mass of H is
1.01 g/mol
The molar mass of NaOH is:
22.99 g/mol + 16.00 g/mol + 1.01 g/mol = 40.00 g/mol
Step 2: Calculate the number of moles of NaOH Given that the mass of
NaOH is 5.00 g, and using the molar mass calculated in Step 1:
Number of moles = 5.00 g
40.00 g/mol = 0.125 mol
Step 3: Calculate the molarity of the solution The volume of solution is 200.0
mL, which is equivalent to 0.2000 L. Using the number of moles calculated in
Step 2:
Molarity = 0.125 mol
0.2000 L = 0.625 M
Therefore, the molarity of the resulting solution is 0.625 M.
Question 21
Question
Calculate the molarity of a solution prepared by dissolving 25.0 g of sodium
chloride (NaCl) in enough water to make 500.0 mL of solution.
16
Solution
Step 1: Calculate the molar mass of sodium chloride (NaCl).
Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Step 2: Convert the mass of sodium chloride to moles.
Moles of NaCl = 25.0 g
58.44 g/mol = 0.428 mol
Step 3: Convert the volume of the solution from milliliters to liters.
Volume of solution = 500.0 mL = 0.500 L
Step 4: Calculate the molarity of the solution.
Molarity = Moles of solute
Volume of solution in liters =0.428 mol
0.500 L = 0.856 M
Therefore, the molarity of the solution prepared by dissolving 25.0 g of
sodium chloride in enough water to make 500.0 mL of solution is 0.856 M.
Question 22
Question
A chemist prepares a solution by dissolving 10.0 g of potassium iodide (KI)
in enough water to make 250.0 mL of solution. Calculate the molarity of the
potassium iodide solution.
Solution
Step 1: Calculate the molar mass of potassium iodide (KI). The molar mass of
KI can be calculated as follows:
Molar mass of KI = Atomic mass of K + Atomic mass of I
Molar mass of KI = 39.10 g/mol + 126.90 g/mol = 166.00 g/mol
Step 2: Determine the number of moles of potassium iodide (KI) in 10.0 g.
The number of moles can be calculated using the formula:
moles = mass
molar mass
moles of KI = 10.0 g
166.00 g/mol = 0.0602 mol
Step 3: Calculate the molarity of the potassium iodide solution. The formula
for molarity is:
Molarity = moles of solute
volume of solution in liters
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Volume of solution = 250.0 mL
1000 mL/L = 0.2500 L
Molarity = 0.0602 mol
0.2500 L = 0.2408 M
Therefore, the molarity of the potassium iodide solution is 0.2408 M.
Question 23
Question
A solution is prepared by dissolving 7.50 g of potassium permanganate (KMnO4)
in enough water to make 500.0 mL of solution. What is the molarity of the
solution?
Solution
Step 1: Calculate the molar mass of KMnO4. The molar mass of K is 39.10
g/mol, Mn is 54.94 g/mol, and O is 16.00 g/mol. So, the molar mass of KMnO4
is
molar mass of KMnO4= 39.10 + 54.94 + 4(16.00) = 158.04 g/mol
Step 2: Calculate the number of moles of KMnO4in 7.50 g. Using the
formula n=m
M, where nis the number of moles, mis the mass, and Mis the
molar mass, we have
n=7.50 g
158.04 g/mol = 0.0474 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as the
number of moles of solute per liter of solution. Given that the number of moles
of KMnO4is 0.0474 mol and the volume of the solution is 500.0 mL (or 0.500
L), we have
Molarity (M) = 0.0474 mol
0.500 L= 0.0948 M
Therefore, the molarity of the solution is 0.0948 M.
Question 24
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of a compound, but
only has a stock solution that is 1 M. What volume of the stock solution should
be used and how should the solution be prepared?
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Solution
Step 1: Use the formula for Molarity to find the volume of the stock solution
needed. The formula for Molarity (M) is:
M=moles of solute
liters of solution
Given that the chemist wants to prepare 500 mL of a 0.2 M solution, we can
rearrange the formula to solve for the moles of solute:
moles of solute = M×liters of solution
moles of solute = 0.2 M ×0.5 L = 0.1 moles
Step 2: Use the moles of solute to find the volume of the stock solution
needed. Since the stock solution is 1 M, the moles of solute present in 1 L of
the solution is equal to the Molarity of the solution. So, the volume of the stock
solution needed can be calculated using the formula:
moles of solute = M×volume of solution =⇒0.1 moles = 1 M×volume of solution
Volume of solution = 0.1 moles
1 M = 0.1 liters = 100 mL
Step 3: Prepare the solution. To prepare the 0.2 M solution, the chemist
should measure out 100 mL of the 1 M stock solution using a volumetric flask and
then dilute it with water to reach a total volume of 500 mL. Additional water
should be added up to the 500 mL mark on the flask and then the solution
should be thoroughly mixed to ensure homogeneity.
Question 25
Question
A chemist wants to prepare a solution by diluting 50.0 mL of a 3.00 M sulfuric
acid (H2SO4) solution to a final volume of 250.0 mL. What will be the molarity
of the final solution?
Solution
Step 1: Calculate the number of moles of sulfuric acid in the initial solution.
mol = M ×L = 3.00 mol/L ×0.0500 L = 0.150 mol
Step 2: Use the moles of sulfuric acid and final volume to determine the
molarity of the final solution.
Mfinal =mol
Lfinal
=0.150 mol
0.250 L = 0.60 M
Answer: The molarity of the final solution will be 0.60 M.
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Question 26
Question
A chemist has a solution containing 25.0 g of NaCl dissolved in 500.0 mL of
water. What is the molarity of the solution? (Molar mass of NaCl is 58.44
g/mol)
Solution
Step 1: Calculate the number of moles of NaCl in the solution.
Moles of NaCl = Mass of NaCl
Molar mass of NaCl =25.0 g
58.44 g/mol
Moles of NaCl = 0.428 mol
Step 2: Calculate the volume of the solution in liters.
Volume of solution = 500.0 mL = 0.500 L
Step 3: Calculate the molarity of the solution.
Molarity = Moles of solute
Volume of solution in liters =0.428 mol
0.500 L
Molarity = 0.856 M
Therefore, the molarity of the solution is 0.856 M.
Question 27
Question
A solution is prepared by dissolving 25.0 g of potassium chloride (KCl) in enough
water to make 500.0 mL of solution. What is the molarity of the solution?
Solution
Step 1: Calculate the molar mass of potassium chloride (KCl). The molar mass
of KCl is the sum of the atomic masses of potassium (K) and chlorine (Cl).
Molar mass of KCl = Atomic mass of K + Atomic mass of Cl
The atomic mass of potassium (K) is 39.10 g/mol, and the atomic mass of
chlorine (Cl) is 35.45 g/mol.
Molar mass of KCl = 39.10 g/mol + 35.45 g/mol = 74.55 g/mol
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Step 2: Calculate the number of moles of potassium chloride dissolved. Given
that 25.0 g of KCl is dissolved, we can calculate the number of moles using the
formula:
Number of moles = Mass
Molar mass
Number of moles = 25.0 g
74.55 g/mol = 0.335 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as the
number of moles of solute per liter of solution. To calculate molarity, we use
the formula:
Molarity = Number of moles of solute
Volume of solution in liters
Given that the volume of the solution is 500.0 mL (or 0.500 L), we can
calculate the molarity:
Molarity = 0.335 mol
0.500 L = 0.670 M
Therefore, the molarity of the solution prepared is 0.670 M.
Question 28
Question
A solution is prepared by dissolving 5.00 grams of sodium chloride (NaCl) in
enough water to make 250.0 mL of solution. Calculate the molarity (M) of the
resulting solution.
Solution
Step 1: Calculate the molar mass of NaCl. The molar mass of NaCl is the sum
of the atomic masses of sodium (Na) and chlorine (Cl). Molar mass of NaCl =
Molar mass of Na + Molar mass of Cl = 22.99 g/mol+35.45 g/mol = 58.44 g/mol
Step 2: Calculate the number of moles of NaCl. Given mass of NaCl =
5.00 grams and molar mass of NaCl = 58.44 g/mol Number of moles of NaCl =
5.00 g
58.44 g/mol = 0.0856 mol
Step 3: Calculate the molarity of the solution. Molarity (M) is defined as
the number of moles of solute per liter of solution. Volume of solution = 250.0
mL = 0.250 L Molarity = 0.0856 mol
0.250 L = 0.3424 M
Therefore, the molarity of the resulting solution is 0.3424 M.
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Question 29
Question
A chemist wants to prepare 2.00 L of a 0.100 M solution of sodium chloride
(NaCl). The only stock solution available is a 2.00 M solution of NaCl. How
many milliliters of the stock solution should the chemist use to make the desired
solution?
Solution
Let’s denote the volume of the 2.00 M stock solution that the chemist needs to
use as xmL.
Step 1: Calculate the amount of solute in the desired solution. The
amount of solute in the final solution can be calculated using the formula:
Amount of solute = Molarity ×Volume
For the final solution:
Amount of solute in final solution = 0.100 mol/L ×2.00 L = 0.200 mol
Step 2: Calculate the amount of solute in the stock solution. Using
the same formula as Step 1:
Amount of solute in stock solution = 2.00 mol/L ×(xmL/1000) = 2x/1000 mol
Step 3: Set up the equation and solve for x.Since the amount of solute
in the final solution should equal the amount of solute in the stock solution, we
have:
0.200 mol = 2x/1000 mol
x= (0.200 ×1000)/2 = 100 mL
Therefore, the chemist should use 100 mL of the 2.00 M stock solution to
prepare the desired 0.100 M solution of NaCl.
Question 30
Question
A solution is prepared by dissolving 15.0 g of glucose (C6H12O6) in enough water
to make 250.0 mL of solution. What is the molarity of the glucose solution?
(Molar mass of glucose = 180.16 g/mol)
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Solution
Step 1: Calculate the number of moles of glucose in the solution.
Moles of glucose = Mass of glucose
Molar mass of glucose
Moles of glucose = 15.0 g
180.16 g/mol ≈0.083moles
Step 2: Convert the volume of the solution to liters.
Volume of solution in liters = 250.0 mL
1000 mL/L = 0.250L
Step 3: Calculate the molarity of the solution.
Molarity = Moles of solute
Volume of solution in liters
Molarity = 0.083 moles
0.250L= 0.332 M
Therefore, the molarity of the glucose solution is 0.332 M.
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