CHEM 105 - ELEMENTS OF
GENERAL CHEMISTRY - Molarity
and Concentrations
Question Bank - Set 3
Liberty University
Question 1
Question
A solution is prepared by dissolving 5.00 g of glucose (C6H12O6) in enough water
to make 250.0 mL of solution. Calculate the molarity of the glucose solution.
Solution
Step 1: Calculate the molar mass of glucose. The molar mass of glucose can be
calculated by adding up the atomic masses of the elements in its formula.
Molar mass of glucose = 6×atomic mass of Carbon+12×atomic mass of Hydrogen+6×atomic mass of Oxygen
= 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol)
= 180.18 g/mol
Step 2: Calculate the number of moles of glucose in the solution. The number
of moles of glucose can be calculated using the formula:
moles of solute =mass of solute
molar mass of solute
moles of glucose =5.00 g
180.18 g/mol = 0.0278 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as the
number of moles of solute divided by the volume of solution in liters. Since the
volume of the solution is given in mL, it needs to be converted to liters.
Molarity (M) = moles of solute
volume of solution in liters
=0.0278 mol
250.0mL ×1L
1000 mL
=0.0278 mol
0.2500 L= 0.1112M
Therefore, the molarity of the glucose solution is 0.1112 M.
Question 2
Question
A solution is prepared by dissolving 8.21 g of sodium chloride (NaCl) in enough
water to make 450.0 mL of solution. Calculate the molarity of the 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).
NaCl molar mass = Na molar mass + Cl molar mass
Step 2: Calculate the number of moles of NaCl. The number of moles can
be calculated using the given mass and molar mass of NaCl.
Number of moles = Given mass
Molar mass
Step 3: Calculate the molarity of the solution. Molarity is defined as the
number of moles of solute dissolved in one liter of solution. Since we have the
volume in milliliters, we need to convert it to liters.
Molarity (M) = Number of moles of solute
Volume of solution in liters
Now, let’s solve the problem step by step.
Step 1: The molar mass of NaCl is:
Na molar mass = 22.99 g/mol
Cl molar mass = 35.45 g/mol
NaCl molar mass = 22.99 + 35.45 = 58.44 g/mol
Step 2: The number of moles of NaCl is:
Number of moles NaCl = 8.21 g
58.44 g/mol = 0.1405 mol
Step 3: The volume of the solution in liters is:
Volume of solution = 450.0 mL = 450.0×10−3L=0.4500 L
2
The molarity of the solution is:
Molarity (M) = 0.1405 mol
0.4500 L = 0.3122 M
Therefore, the molarity of the solution is 0.3122 M.
Question 3
Question
A chemist needs to prepare 500 mL of a 0.2 M sodium chloride (NaCl) solution.
The chemist has a stock solution of NaCl that is 1 M. How can the chemist
prepare the desired solution?
Solution
Step 1: Calculate the moles of NaCl needed to prepare 500 mL of a 0.2 M
solution. Given: Volume of solution needed = 500 mL = 0.5 L Molarity of NaCl
solution needed = 0.2 M
Moles of NaCl = Molarity ×Volume (in L)
Moles of NaCl = 0.2 mol/L ×0.5 L = 0.1 mol
Step 2: Determine the volume of the 1 M NaCl stock solution needed. Since
the stock solution is 1 M, which means 1 mole of NaCl is present in 1 liter of
the solution, the volume of the stock solution needed to provide 0.1 mol can be
calculated as follows:
Volume of stock solution (in L) = Moles of NaCl needed
Molarity of stock solution
Volume of stock solution (in L) = 0.1 mol
1 mol/L = 0.1 L = 100 mL
Step 3: Dilute the 100 mL of 1 M NaCl stock solution to prepare the desired
0.2 M solution. To prepare the 0.2 M solution, we need to dilute the 100 mL of
the 1 M NaCl stock solution by adding distilled water to a total volume of 500
mL:
Volume of water needed = Total volume −Volume of stock solution
Volume of water needed = 500 mL −100 mL = 400 mL
Therefore, the chemist will take 100 mL of the 1 M NaCl stock solution and
add 400 mL of distilled water to prepare 500 mL of a 0.2 M NaCl solution.
3
Question 4
Question
A chemist wants to prepare 500 mL of a 0.1 M solution of sodium chloride. The
chemist only has a stock solution of sodium chloride that is 2 M. How much of
the stock solution should the chemist use to prepare the desired solution?
Solution
Step 1: Calculate the moles of sodium chloride needed for the desired solution.
Step 2: Determine the volume of the stock solution needed.
Step 1: The formula for calculating the moles of solute is:
moles = molarity ×volume (L)
Given that the molarity of the desired solution is 0.1 M and the volume is
500 mL (which is 0.5 L), we can calculate the moles of sodium chloride needed:
moles = 0.1 mol/L ×0.5 L = 0.05 mol
Step 2: The formula for calculating the volume of stock solution needed is:
volume of stock solution (L) = moles
molarity of stock solution
Given that the molarity of the stock solution is 2 M, we can calculate the
volume needed:
volume of stock solution = 0.05 mol
2 mol/L = 0.025 L
Therefore, the chemist should use 25 mL of the stock solution to prepare
500 mL of the desired 0.1 M sodium chloride solution.
Question 5
Question
A solution is prepared by dissolving 10.0 g of glucose (C6H12O6) in enough water
to make 250.0 mL of solution. Calculate the molarity of the glucose solution.
Solution
Step 1: Calculate the molar mass of glucose (C6H12O6). The molar mass of
glucose can be calculated by adding up the atomic masses of each element
present in the compound.
Molar mass of glucose = 6×atomic mass of C+12×atomic mass of H+6×atomic mass of O
4
Molar mass of glucose = 6 ×12.01 g/mol + 12 ×1.008 g/mol + 6 ×16.00 g/mol
Molar mass of glucose = 180.18 g/mol
Step 2: Calculate the number of moles of glucose in 10.0 g. Using the
formula:
moles = mass
molar mass
moles of glucose = 10.0 g
180.18 g/mol
moles of glucose ≈0.05548 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as the
number of moles of solute per liter of solution. Since the volume of the solution
is 250.0 mL (or 0.250 L), the molarity can be calculated as:
Molarity = moles of solute
volume of solution in liters
Molarity of glucose solution = 0.05548 mol
0.250 L
Molarity of glucose solution ≈0.222 M
Therefore, the molarity of the glucose solution is approximately 0.222 M.
Question 6
Question
A solution is prepared by mixing 50 mL of a 0.4 M sulfuric acid (H2SO4) solution
with 100 mL of a 0.2 M sodium hydroxide (NaOH) solution. What is the
molarity of the resulting solution after the reaction between sulfuric acid and
sodium hydroxide is complete?
Solution
Step 1: Write the balanced chemical equation for the reaction between sulfuric
acid and sodium hydroxide. The reaction is as follows:
H2SO4+ 2NaOH →Na2SO4+ 2H2O
Step 2: Calculate the number of moles of sulfuric acid and sodium hydroxide
used in the reaction. For sulfuric acid (H2SO4):
Moles of H2SO4= Molarity ×Volume (L) = 0.4 mol/L ×0.05 L = 0.02 mol
For sodium hydroxide (NaOH):
Moles of NaOH = Molarity ×Volume (L) = 0.2 mol/L ×0.1 L = 0.02 mol
5
Step 3: Identify the limiting reactant. Since both sulfuric acid and sodium
hydroxide produce equal moles (0.02 mol), sodium hydroxide is the limiting
reactant in this case.
Step 4: Calculate the remaining excess of sulfuric acid after the reaction.
Initial moles of H2SO4= 0.02 mol
Final moles of H2SO4= Initial moles−(Moles used in reaction) = 0.02 mol−0.02 mol = 0 mol
Step 5: Calculate the total volume of the resulting solution.
Total volume = 50 mL + 100 mL = 150 mL = 0.15 L
Step 6: Calculate the molarity of the resulting solution.
Molarity of resulting solution = Total moles of solution
Total volume of solution (L) =0.02 mol
0.15 L = 0.133 M
Therefore, the molarity of the resulting solution after the reaction is complete
is 0.133 M.
Question 7
Question
A solution is prepared by dissolving 15.0 g of glucose (C6H12O6) in enough water
to make 500.0 mL of solution. Calculate the molarity of the glucose solution.
Solution
Step 1: Calculate the molar mass of glucose. Since the molecular formula of
glucose is C6H12O6, we need to find the sum of the atomic masses of carbon,
hydrogen, and oxygen in one mole of glucose.
Molar mass of glucose = 6(atomic mass of C)+12(atomic mass of H)+6(atomic mass of O)
Molar mass of glucose = 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol)
Molar mass of glucose = 180.18 g/mol
Step 2: Calculate the number of moles of glucose in the solution. Given that
the mass of glucose is 15.0 g, we can use the formula n=m
M, where nis the
number of moles, mis the mass, and Mis the molar mass.
n=15.0 g
180.18 g/mol = 0.083mol
Step 3: Calculate the molarity of the solution. Molarity (M) is defined as
the number of moles of solute per liter of solution. Since we have 0.083 moles
of glucose in 500.0 mL of solution, we first convert the volume to liters.
Volume of solution = 500.0 mL = 0.500 L
6
Now, we can calculate the molarity using the formula M=n
V.
M=0.083 mol
0.500 L = 0.166 M
Therefore, the molarity of the glucose solution is 0.166 M.
Question 8
Question
A chemist needs to prepare 500 mL of a 0.25 M solution of sodium chloride.
If the chemist has a stock solution of sodium chloride with a concentration of
2 M, what volume of the stock solution should be added to water to make the
desired solution?
Solution
Step 1: Calculate the moles of sodium chloride needed for a 0.25 M solution in
500 mL.
Given: Desired molarity, M1= 0.25 M Volume of solution, V1= 500 mL =
0.5 L
Moles of solute = M1×V1= 0.25 mol/L ×0.5 L = 0.125 mol
Step 2: Calculate the volume of the stock solution needed.
Given: Concentration of stock solution, M2= 2 M
Use the formula:
M2×V2= moles of solute
Substitute the known values:
2 mol/L ×V2= 0.125 mol
Solve for V2:
V2=0.125 mol
2 mol/L = 0.0625 L = 62.5 mL
Therefore, the chemist should add 62.5 mL of the stock solution to water to
make the desired 0.25 M sodium chloride solution.
Question 9
Question
A solution is prepared by dissolving 5.00 grams of potassium dichromate (K2Cr2O7)
in enough water to make 250.0 mL of solution. Calculate the molarity of the
potassium dichromate solution.
7
Solution
Step 1: Calculate the molar mass of potassium dichromate. The molar mass
of K2Cr2O7can be calculated by adding the molar masses of its constituents
(potassium, chromium, and oxygen).
Molar mass = 2(molar mass of K) + 2(molar mass of Cr) + 7(molar mass of O)
= 2(39.10 g/mol) + 2(52.00 g/mol) + 7(16.00 g/mol)
= 78.20 g/mol + 104.00 g/mol + 112.00 g/mol
= 294.20 g/mol
Step 2: Calculate the number of moles of potassium dichromate in the solu-
tion. Using the formula n=m
M, where nis the number of moles, mis the mass
of the substance, and Mis the molar mass:
n=5.00 g
294.20 g/mol = 0.0170 mol
Step 3: Calculate the molarity of the solution. Molarity (M) is defined as
the number of moles of solute per liter of solution. We have found the number
of moles of potassium dichromate in 250.0 mL of solution:
M=n
V=0.0170 mol
0.2500 L = 0.0680 M
Therefore, the molarity of the potassium dichromate solution is 0.0680 M.
Question 10
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of potassium iodide
(KI). The chemist has a stock solution of KI with a concentration of 0.5 M.
How many milliliters of the stock solution should the chemist use to make the
desired solution?
Solution
Step 1: Calculate the moles of potassium iodide required to make the desired
solution. Given: Volume of desired solution, Vfinal = 500 mL = 0.5 L Molarity
of the desired solution, Mfinal = 0.2 M
We can use the formula for molarity:
M=moles of solute
volume of solution in liters
Rearranging the formula to solve for moles:
moles of solute = M×Vfinal
8
Substitute the given values:
moles of solute = 0.2 mol/L ×0.5 L = 0.1 mol
Step 2: Calculate the volume of the stock solution needed. Given: Molarity
of the stock solution, Minitial = 0.5 M
We can use the same formula for molarity to find the volume of stock solution
needed:
Minitial ×Vinitial = moles of solute
Rearranging the formula to solve for Vinitial:
Vinitial =moles of solute
Minitial
Substitute the values calculated and given:
Vinitial =0.1 mol
0.5 mol/L = 0.2 L = 200 mL
Therefore, the chemist should use 200 mL of the stock solution to make 500
mL of a 0.2 M solution of potassium iodide.
Question 11
Question
A chemist needs to prepare 500 mL of a 0.2 M solution of sulfuric acid, H2SO4.
However, the only stock solution available is a 1 M solution of sulfuric acid.
How would the chemist prepare the desired solution using the stock solution?
Solution
Step 1: Calculate the amount of sulfuric acid needed from the 1 M stock solution
to prepare the desired 0.2 M solution. Let V1be the volume of the 1 M stock
solution needed. Using the formula for molarity:
M1×V1=M2×V2
where M1= 1 M (molarity of stock solution), V2= 500 mL (final volume of 0.2
M solution), and M2= 0.2 M (molarity of desired solution).
Substitute the given values into the formula:
1×V1= 0.2×500
V1=0.2×500
1
V1= 100 mL = 0.1 L
9
Step 2: Calculate the volume of solvent needed to dilute the 1 M stock
solution. The total volume of the final solution is 500 mL. Since we need 100
mL of the 1 M stock solution, the remaining volume should be the solvent
(water). Therefore, the volume of solvent required is:
500 −100 = 400 mL = 0.4 L
Step 3: Mix the calculated volumes of the 1 M stock solution and the solvent
to prepare the 0.2 M solution of sulfuric acid. The chemist should carefully
measure and mix 100 mL of the 1 M stock solution with 400 mL of solvent to
obtain the desired 0.2 M solution of sulfuric acid.
Question 12
Question
A chemist needs to prepare 1.5 L of a 0.25 M solution of potassium perman-
ganate (KMnO4). The chemist has a stock solution of potassium permanganate
with a concentration of 0.5 M. How many milliliters of the stock solution should
the chemist use to make the desired solution?
Solution
Step 1: Calculate the number of moles of potassium permanganate needed for
the desired solution. Step 2: Use the concentration of the stock solution to
determine the volume of stock solution needed.
Step 1: Given: Volume needed (Vneeded) = 1.5 L Molarity of desired solution
(Mdesired) = 0.25 M
We can use the formula for molarity to find the number of moles (n) needed:
n=Mdesired ×Vneeded
Substitute the given values:
n= 0.25 mol/L ×1.5 L
n= 0.375 mol
So, 0.375 moles of potassium permanganate are needed.
Step 2: Given: Molarity of stock solution (Mstock) = 0.5 M
We can rearrange the formula for molarity to find the volume of stock solu-
tion needed:
Vstock =n
Mstock
Substitute the calculated number of moles and the molarity of the stock
solution:
Vstock =0.375 mol
0.5 mol/L
10
Vstock = 0.75 L
Since we want the answer in milliliters, we convert 0.75 L to milliliters:
Vstock = 0.75 L ×1000 mL/L
Vstock = 750 mL
The chemist should use 750 mL of the 0.5 M stock solution to prepare a 1.5
L solution with a concentration of 0.25 M.
Question 13
Question
A chemist has a 2 L solution containing 0.5 moles of solute. She wants to dilute
this solution to make a new solution that is 0.1 M. How much solvent should
she add to achieve this new concentration?
Solution
Step 1: Calculate the initial molarity of the solution. Step 2: Use the formula
for dilution to find the volume of solvent to add.
Step 1: To find the initial molarity of the solution, we first calculate the
molarity using the formula:
M1=moles of solute
volume of solution (L)
Given that the moles of solute is 0.5 moles and the initial volume of the
solution is 2 L, we have
M1=0.5
2= 0.25 M
Step 2: To dilute the solution to 0.1 M, we use the formula for dilution:
M1V1=M2V2
where M1and V1are the initial molarity and volume of the solution, and
M2and V2are the final molarity and volume of the solution.
Substitute the known values into the formula:
(0.25)(2) = (0.1)(2 + V)
Solve for V:
0.5=0.2+0.1V
0.1V= 0.3
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V=0.3
0.1= 3 L
Therefore, the chemist should add 3 L of solvent to the 2 L solution to
achieve a final concentration of 0.1 M.
Question 14
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of potassium perman-
ganate. The chemist has a stock solution of potassium permanganate with a
concentration of 0.5 M. How many milliliters of the stock solution should the
chemist use to prepare the desired solution?
Solution
Step 1: Calculate the number of moles of potassium permanganate needed for
the desired solution. Given that the desired molarity is 0.2 M and the volume
is 500 mL, we can use the formula:
moles = molarity ×volume
Substitute the values:
moles = 0.2 mol/L ×0.5 L = 0.1 mol
Step 2: Calculate the volume of stock solution needed. We know the con-
centration of the stock solution is 0.5 M. Let the volume of the stock solution
needed be xmL. We can use the formula:
moles in stock solution = molarity of stock solution×(volume of stock solution in L)
We found that 0.1 moles of potassium permanganate are needed. Substitute
the values:
0.1 mol = 0.5 mol/L ×(x×10−3L)
Step 3: Solve for xto find the volume of stock solution needed.
xmL = 0.1 mol
0.5 mol/L ·1000 = 200 mL
Therefore, the chemist should use 200 mL of the stock solution to prepare
the desired 0.2 M potassium permanganate solution.
Question 15
Question
A solution is prepared by dissolving 10.0 g of glucose (C6H12O6) in enough water
to make 500.0 mL of solution. What is the molarity of the glucose solution?
12
Solution
Step 1: Calculate the molar mass of glucose. The molar mass of glucose is
calculated by adding up the atomic masses of its constituent elements: carbon
(C), hydrogen (H), and oxygen (O).
Molar mass C6H12O6= (6×atomic mass C)+(12×atomic mass H)+(6×atomic mass O)
= (6 ×12.01 g/mol) + (12 ×1.01 g/mol) + (6 ×16.00 g/mol)
= 72.06 g/mol + 12.12 g/mol + 96.00 g/mol
= 180.18 g/mol
Step 2: Calculate the number of moles of glucose. Given that the mass of
glucose is 10.0 g, we can calculate the number of moles using the formula:
Moles =mass
molar mass
Moles =10.0g
180.18 g/mol
Moles = 0.0555 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as the
number of moles of solute per liter of solution. In this case, the volume of the
solution is 500.0 mL or 0.5000 L.
Molarity =moles of solute
volume of solution (L)
Molarity =0.0555 mol
0.5000 L
Molarity = 0.111 mol/L
Therefore, the molarity of the glucose solution is 0.111 mol/L.
Question 16
Question
A chemist needs to prepare 500 mL of a solution with a molarity of 0.25 M.
However, the chemist only has a stock solution with a concentration of 2.0 M.
How many milliliters of the stock solution should the chemist use to make the
desired solution?
13
Solution
Step 1: Write down the given information. Let V1be the volume of the 2.0 M
stock solution to be used, M1be the molarity of the stock solution (2.0 M), V2
be the final volume of the solution (500 mL), and M2be the desired molarity
(0.25 M).
Step 2: Use the formula for dilution. The formula for dilution is given by:
M1V1=M2V2
Step 3: Fill in the values. Substitute the known values into the formula:
2.0 M ·V1= 0.25 M ·500 mL
Step 4: Solve for V1.
2.0·V1= 0.25 ·500
2.0·V1= 125
V1=125
2.0
V1= 62.5 mL
The chemist should use 62.5 mL of the 2.0 M stock solution to make 500 mL
of a 0.25 M solution.
Question 17
Question
A chemist prepared a solution by mixing 250 mL of a 0.6 M sulfuric acid (H2SO4)
solution with 500 mL of a 1.2 M sodium hydroxide (NaOH) solution. What is
the molarity of the resulting solution after the reaction H2SO4reacts with NaOH
to form water and sodium sulfate?
Solution
Step 1: Write the balanced chemical equation for the reaction between sulfuric
acid and sodium hydroxide:
H2SO4(aq) + 2NaOH(aq)→2H2O(l) + Na2SO4(aq)
Step 2: Determine the number of moles of sulfuric acid and sodium hydroxide
used in the reaction.
Moles of H2SO4= Molarity ×Volume (L)
= 0.6 mol/L ×0.25 L
= 0.15 mol
14
Moles of NaOH = Molarity ×Volume (L)
= 1.2 mol/L ×0.5 L
= 0.6 mol
Step 3: Identify the limiting reactant in the reaction. Since 1 mole of H2SO4
reacts with 2 moles of NaOH, we need to compare the moles of sulfuric acid and
sodium hydroxide to determine the limiting reactant.
The molar ratio of H2SO4to NaOH is 1 : 2, which means we would need
0.3 mol of H2SO4to completely react with 0.6 mol of NaOH. However, we only
have 0.15 mol of H2SO4, making it the limiting reactant.
Step 4: Determine the amount of sodium sulfate and water formed based on
the limiting reactant. Since 0.15 mol of H2SO4reacts, 0.15 mol of Na2SO4and
0.3 mol of water are produced.
Step 5: Calculate the total volume of the resulting solution. The total
volume of the resulting solution is 250 mL + 500 mL = 750 mL = 0.75 L.
Step 6: Calculate the molarity of the resulting solution.
Molarity of resulting solution = Total moles of solute
Total volume of solution
=0.15 mol + 0.6 mol
0.75 L
=0.75 mol
0.75 L
= 1.0 M
Therefore, the molarity of the resulting solution after the reaction is 1.0 M.
Question 18
Question
A student dilutes 50.0 mL of a 6.00 M hydrochloric acid (HCl) solution with
water to make 250.0 mL of a new solution. What is the final molarity of the
HCl solution after dilution?
Solution
Step 1: Calculate the number of moles of HCl in the initial solution. Given:
Initial volume of HCl solution (Vi) = 50.0 mL = 0.050 L Initial molarity of HCl
solution (Mi) = 6.00 M
Use the formula:
n=M×V
where nis the number of moles, Mis the molarity, and Vis the volume.
Substitute the values to find the number of moles in the initial solution:
ni= 6.00 mol/L ×0.050 L = 0.30 mol
15
Step 2: Calculate the final molarity of the HCl solution after dilution. Given:
Final volume of the new solution (Vf) = 250.0 mL = 0.250 L
Use the formula for dilution:
Mf=ni
Vf
where Mfis the final molarity, niis the number of moles in the initial solution,
and Vfis the final volume.
Substitute the values to find the final molarity:
Mf=0.30 mol
0.250 L = 1.20 M
Therefore, the final molarity of the hydrochloric acid solution after dilution
is 1.20 M.
Question 19
Question
A chemist has a solution containing 50.0 g of sodium chloride (NaCl) dissolved
in 500.0 mL of water. What is the molarity of the solution?
Solution
Step 1: Calculate the molar mass of sodium chloride (NaCl). The molar mass
of NaCl is the sum of the atomic masses of sodium (Na) and chlorine (Cl).
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. Given that the mass
of NaCl is 50.0 g, we can calculate the number of moles by dividing the mass
by the molar mass.
Number of moles of NaCl = 50.0 g
58.44 g/mol ≈0.856 mol
Step 3: Calculate the volume of the solution in liters. The volume of the
solution is 500.0 mL, which needs to be converted to liters by dividing by 1000.
Volume of the solution = 500.0 mL
1000 mL/L = 0.5000 L
Step 4: Calculate the molarity of the solution. Molarity (M) is defined as
moles of solute per liter of solution.
Molarity = Number of moles of solute
Volume of solution in liters =0.856 mol
0.5000 L = 1.71 M
Therefore, the molarity of the solution containing 50.0 g of sodium chloride
dissolved in 500.0 mL of water is 1.71 M.
16
Question 20
Question
A chemist wants to prepare 500 mL of a 0.4 M aqueous solution of potassium
permanganate (KMnO4). The chemist has a stock solution of KMnO4with a
concentration of 2.5 M. How many milliliters of the stock solution should be
used to make the desired solution?
Solution
Step 1: Let Vbe the volume of the stock solution in milliliters needed to make
the desired solution. Step 2: Calculate the amount of KMnO4in moles that are
required to make the desired solution:
Moles of KMnO4= Molarity ×Volume (L)
Moles of KMnO4= 0.4 M ×0.5 L
Moles of KMnO4= 0.2 mol
Step 3: Calculate the volume of the stock solution needed to provide 0.2 mol
of KMnO4:
Molarity = Moles of solute
Volume of solution
2.5 M = 0.2 mol
VL
V=0.2
2.5= 0.08 L = 80 mL
Therefore, the chemist should use 80 mL of the 2.5 M stock solution to
prepare the 500 mL of 0.4 M KMnO4solution.
Question 21
Question
A student wants to prepare 500 mL of a 0.2 M solution of a compound with a
molar mass of 150.3 g/mol. However, the student only has a stock solution with
a concentration of 1.5 M. How much of the stock solution should the student
use and how much water should be added to make the desired 0.2 M solution?
Solution
Step 1: Calculate the number of moles of the compound needed for the desired
solution. Let Vbe the volume of the 0.2 M solution to be prepared and M1be
the desired molarity. Given: V= 500 mL = 0.5 L M1= 0.2 M
17
Using the formula M1V1=M2V2where V1is the volume of the stock solution
to be used and M2= 1.5 M is the concentration of the stock solution:
M1V=M2V1
0.2×0.5=1.5V1
V1=0.2×0.5
1.5=0.1
1.5=1
15 = 0.06667 L = 66.67 mL
The student should use 66.67 mL of the stock solution.
Step 2: Calculate the volume of water to be added to the stock solution.
The total volume of the final solution is 500 mL. Since 66.67 mL of the stock
solution will be used, the volume of water needed is:
500 −66.67 = 433.33 mL
Therefore, the student should use 66.67 mL of the stock solution and add
433.33 mL of water to prepare the desired 0.2 M solution.
Question 22
Question
A chemist needs to prepare 500 mL of a 0.2 M solution of sodium hydroxide
(NaOH). The chemist has a stock solution of NaOH with a concentration of 1.0
M. How many milliliters of the stock solution should be diluted with water to
make the desired solution?
Solution
Step 1: Let’s calculate the number of moles of NaOH needed for the desired so-
lution. Given: Volume of solution needed (Vsolution) = 500 mL = 0.5 L Molarity
of the solution needed (Mdesired) = 0.2 M
Using the formula for molarity:
M=moles
volume (in liters)
we can rearrange the formula to find the moles:
moles = M×volume (in liters)
Therefore, the number of moles of NaOH needed is:
moles = 0.2 mol/L ×0.5 L = 0.1 mol
Step 2: Next, let’s determine the volume of the stock solution needed to
obtain 0.1 moles of NaOH. Given: Molarity of the stock solution (Mstock) = 1.0
M
Using the formula for molarity:
Mstock =moles
volume (in liters)
18
we can rearrange the formula to find the volume:
volume (in liters) = moles
Mstock
Substitute the values to find the volume of the stock solution needed:
volume (in liters) = 0.1 mol
1.0 mol/L = 0.1 L = 100 mL
Therefore, the chemist should dilute 100 mL of the stock solution with water
to make the desired 0.2 M solution of sodium hydroxide.
Question 23
Question
A chemist prepares a solution by dissolving 5.00 grams of NaCl in enough wa-
ter to make 500.0 mL of solution. The density of the solution is 1.08 g/mL.
Calculate the molarity of NaCl in the solution.
Solution
Step 1: Calculate the moles of NaCl. Given mass of NaCl = 5.00 g
Molar mass of NaCl = 58.44 g/mol
moles of NaCl = mass of NaCl
molar mass of NaCl =5.00 g
58.44 g/mol
Step 2: Calculate the volume of the solution. Given density of the solution
= 1.08 g/mL
Volume of solution = 500.0 mL
volume of solution = mass of solution
density of solution =500.0 mL
1.08 g/mL
Step 3: Calculate the molarity of NaCl in the solution. Molarity is defined
as moles of solute divided by liters of solution.
Molarity = moles of NaCl
volume of solution in liters
Make sure that the volume of solution is converted to liters before calculating
the molarity.
Question 24
Question
A chemist prepares a solution by mixing 250 mL of a 2.0 M sodium chloride
solution with 500 mL of a 1.5 M calcium chloride solution. What is the molarity
of chloride ions in the resulting solution?
19
Solution
Step 1: Calculate the moles of sodium chloride in the 250 mL solution.
Given: Volume of sodium chloride solution = 250 mL = 0.250 L
Molarity of sodium chloride solution = 2.0 M
The moles of sodium chloride can be calculated using the formula:
moles = Molarity ×Volume
moles of NaCl = 2.0 mol/L ×0.250 L
moles of NaCl = 0.500 mol
Step 2: Calculate the moles of calcium chloride in the 500 mL solution.
Given: Volume of calcium chloride solution = 500 mL = 0.500 L
Molarity of calcium chloride solution = 1.5 M
The moles of calcium chloride can be calculated using the formula:
moles = Molarity ×Volume
moles of CaCl2= 1.5 mol/L ×0.500 L
moles of CaCl2= 0.750 mol
Step 3: Calculate the total moles of chloride ions.
Since each formula unit of sodium chloride and calcium chloride has 1 chloride
ion, the total moles of chloride ions in the resulting solution is the sum of the
moles of chloride ions from each component.
Total moles of Cl−= moles of NaCl + 2 ×moles of CaCl2
Total moles of Cl−= 0.500 + 2 ×0.750
Total moles of Cl−= 2.000 mol
Step 4: Calculate the molarity of chloride ions in the resulting solution.
The volume of the resulting solution is the sum of the volumes of the two
solutions used.
Total volume = 0.250 L + 0.500 L
Total volume = 0.750 L
Finally, we can calculate the molarity of chloride ions using the formula:
Molarity of Cl−=Total moles of Cl−
Total volume
Molarity of Cl−=2.000 mol
0.750 L
Molarity of Cl−= 2.67 M
Therefore, the molarity of chloride ions in the resulting solution is 2.67 M.
20
Question 25
Question
A chemist needs to prepare 500 mL of a 0.2 M solution of potassium dichromate
and has a stock solution of 1 M potassium dichromate. How many milliliters of
the stock solution should the chemist use and how much solvent (in mL) should
be added to prepare the desired solution?
Solution
Step 1: Let Vstock be the volume of stock solution needed and Vsolvent be the
volume of solvent added to prepare the final solution.
Step 2: Use the formula for molarity to calculate the amount of solute present
in the stock solution:
M1V1=M2V2
1 M ×Vstock = 0.2 M ×500 mL
Vstock =0.2×500
1= 100 mL
Step 3: The total volume of the final solution is the sum of the volumes of
the stock solution and the solvent:
Vtotal =Vstock +Vsolvent
Vtotal = 100 mL + Vsolvent
Step 4: We know that the total volume of the final solution is 500 mL, so
we can write:
500 mL = 100 mL + Vsolvent
Step 5: Solve for Vsolvent:
Vsolvent = 500 mL −100 mL = 400 mL
Therefore, the chemist should use 100 mL of the stock solution and add 400
mL of solvent to prepare the desired 0.2 M solution of potassium dichromate.
Question 26
Question
A student wants to prepare 500 mL of a 0.2 M solution of potassium perman-
ganate. However, they only have a 750 mL stock solution with a concentration
of 0.3 M. How should the student dilute the stock solution to obtain the desired
concentration?
21
Solution
Step 1: Calculate the moles of potassium permanganate needed for the desired
solution. Let Vf inal be the final volume of the solution (500 mL) and Mfinal be
the final concentration (0.2 M). Using the formula M1V1=M2V2, we have:
moles of KMnO4=Mfinal ×Vf inal
moles of KMnO4= 0.2 mol/L ×0.5 L
moles of KMnO4= 0.1 mol
Step 2: Calculate the volume of the stock solution needed to obtain the
desired moles of potassium permanganate. Let Vstock be the volume of the
stock solution needed. Using the formula M1V1=M2V2, we have:
0.3 mol/L ×Vstock = 0.1 mol
Vstock =0.1 mol
0.3 mol/L
Vstock =1
3L
Step 3: Calculate the volume of solvent needed to dilute the stock solution.
The volume of solvent needed is:
Vsolvent = 750 mL −Vstock
Vsolvent = 750 mL −1
3L = 2250 −1
3mL
Vsolvent =2249
3mL ≈749.67 mL
Therefore, the student should measure 333.33 mL of the stock solution and
dilute it with 749.67 mL of solvent (water) to obtain 500 mL of a 0.2 M solution
of potassium permanganate.
Question 27
Question
A chemist needs to prepare 500 mL of a 0.15 M solution of potassium perman-
ganate, KMnO4. The chemist has a stock solution of KMnO4with a concen-
tration of 0.5 M. How many milliliters of the stock solution should be used and
how much water should be added to prepare the desired solution?
22
Solution
Step 1: Calculate the moles of KMnO4needed for the desired solution. Given:
Volume of desired solution = 500 mL = 0.5 L Molarity of desired solution =
0.15 M
From the formula
Molarity =moles of solute
volume of solution in liters
we can rearrange to find the moles of solute: moles of solute = Molarity ×
volume of solution moles of KMnO4= 0.15 M ×0.5 L = 0.075 moles
Step 2: Calculate the volume of stock solution needed. Given: Molarity of
stock solution = 0.5 M
Using the same formula, we can find the volume of stock solution needed:
volume of stock solution = moles of solute
Molarity of stock solution volume of stock solution =
0.075 moles
0.5 M = 0.15 L = 150 mL
Step 3: Calculate the volume of water needed to prepare the solution. The
total volume of the desired solution is 500 mL, and we are adding 150 mL of
the stock solution. Therefore, the volume of water needed is: volume of water
= total volume - volume of stock solution volume of water = 500 mL - 150 mL
= 350 mL
So, 150 mL of the stock solution should be used, and 350 mL of water should
be added to prepare the desired 0.15 M KMnO4solution.
Question 28
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?
Solution
Step 1: Calculate the molar mass of glucose. The molar mass of glucose
(C6H12O6) can be calculated by adding the atomic masses of carbon, hydro-
gen, and oxygen. - Atomic mass of carbon (C) = 12.01 g/mol - Atomic mass of
hydrogen (H) = 1.008 g/mol - Atomic mass of oxygen (O) = 16.00 g/mol
Therefore, the molar mass of glucose:
6×12.01 g/mol + 12 ×1.008 g/mol + 6 ×16.00 g/mol = 180.16 g/mol
Step 2: Calculate the number of moles of glucose. Given mass of glucose =
15.0 g
Number of moles = Mass
Molar mass =15.0 g
180.16 g/mol
Step 3: Calculate the volume of the solution in liters. Given volume of
solution = 250.0 mL = 0.250 L
23
Step 4: Calculate the molarity of the solution.
Molarity (M) = Number of moles
Volume of solution in liters =15.0 g/180.16 g/mol
0.250 L
Calculating the molarity:
Molarity = 15.0 g/180.16 g/mol
0.250 L =0.0833 mol
0.250 L = 0.333 M
Therefore, the molarity of the glucose solution is 0.333 M.
Question 29
Question
A solution is prepared by dissolving 15.0 g of glucose (C6H12O6) in enough
water to make 250.0 mL of solution.
Calculate the molarity of the solution.
Given: Molar mass of glucose = 180.0 g/mol
Solution
Step 1: Calculate the number of moles of glucose dissolved in the solution.
Number of moles of glucose = Mass of glucose
Molar mass of glucose
=15.0 g
180.0 g/mol
= 0.0833 mol
Step 2: Convert the volume of the solution from milliliters to liters.
Volume of solution = 250.0 mL = 0.250 L
Step 3: Calculate the molarity of the solution using the formula:
Molarity = Number of moles of solute
Volume of solution in liters
Substitute the values into the formula:
Therefore, the molarity of the solution is approximately 0.333 M.
24
Question 30
Question
A chemist needs to prepare 500 mL of a 0.1 M NaCl solution, but only has a
stock solution of 1 M NaCl. How many mL of the stock solution should the
chemist use and how much water should be added to make the desired solution?
Solution
Step 1: Calculate the moles of NaCl needed for the desired solution. Let’s
denote the volume of 1 M NaCl solution to be used as V1mL and the volume
of water to be added as V2mL. The moles of NaCl needed can be calculated as:
moles of NaCl = molarity ×volume (L)
moles of NaCl = 0.1 mol/L ×0.5 L
moles of NaCl = 0.05 mol
Step 2: Use the moles of NaCl to find the volume of the 1 M NaCl solution
needed. Since the stock solution is 1 M NaCl, the moles of NaCl in the stock
solution is equal to the volume of the solution in liters. Therefore, the volume
of the 1 M NaCl solution needed can be calculated as:
0.05 mol = 1 mol/L ×V1
V1= 0.05 L = 50 mL
Step 3: Calculate the volume of water needed to make the desired solution.
The total volume of the final solution is 500 mL. Therefore, the volume of water
needed is:
V2= 500 mL −50 mL = 450 mL
Thus, the chemist should use 50 mL of the 1 M NaCl stock solution and add
450 mL of water to prepare 500 mL of a 0.1 M NaCl solution.
25
Question 4
Question
A chemist wants to prepare 500 mL of a 0.1 M solution of sodium chloride. The
chemist only has a stock solution of sodium chloride that is 2 M. How much of
the stock solution should the chemist use to prepare the desired solution?
Solution
Step 1: Calculate the moles of sodium chloride needed for the desired solution.
Step 2: Determine the volume of the stock solution needed.
Step 1: The formula for calculating the moles of solute is:
moles = molarity ×volume (L)
Given that the molarity of the desired solution is 0.1 M and the volume is
500 mL (which is 0.5 L), we can calculate the moles of sodium chloride needed:
moles = 0.1 mol/L ×0.5 L = 0.05 mol
Step 2: The formula for calculating the volume of stock solution needed is:
volume of stock solution (L) = moles
molarity of stock solution
Given that the molarity of the stock solution is 2 M, we can calculate the
volume needed:
volume of stock solution = 0.05 mol
2 mol/L = 0.025 L
Therefore, the chemist should use 25 mL of the stock solution to prepare
500 mL of the desired 0.1 M sodium chloride solution.
Question 5
Question
A solution is prepared by dissolving 10.0 g of glucose (C6H12O6) in enough water
to make 250.0 mL of solution. Calculate the molarity of the glucose solution.
Solution
Step 1: Calculate the molar mass of glucose (C6H12O6). The molar mass of
glucose can be calculated by adding up the atomic masses of each element
present in the compound.
Molar mass of glucose = 6×atomic mass of C+12×atomic mass of H+6×atomic mass of O
4
Molar mass of glucose = 6 ×12.01 g/mol + 12 ×1.008 g/mol + 6 ×16.00 g/mol
Molar mass of glucose = 180.18 g/mol
Step 2: Calculate the number of moles of glucose in 10.0 g. Using the
formula:
moles = mass
molar mass
moles of glucose = 10.0 g
180.18 g/mol
moles of glucose ≈0.05548 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as the
number of moles of solute per liter of solution. Since the volume of the solution
is 250.0 mL (or 0.250 L), the molarity can be calculated as:
Molarity = moles of solute
volume of solution in liters
Molarity of glucose solution = 0.05548 mol
0.250 L
Molarity of glucose solution ≈0.222 M
Therefore, the molarity of the glucose solution is approximately 0.222 M.
Question 6
Question
A solution is prepared by mixing 50 mL of a 0.4 M sulfuric acid (H2SO4) solution
with 100 mL of a 0.2 M sodium hydroxide (NaOH) solution. What is the
molarity of the resulting solution after the reaction between sulfuric acid and
sodium hydroxide is complete?
Solution
Step 1: Write the balanced chemical equation for the reaction between sulfuric
acid and sodium hydroxide. The reaction is as follows:
H2SO4+ 2NaOH →Na2SO4+ 2H2O
Step 2: Calculate the number of moles of sulfuric acid and sodium hydroxide
used in the reaction. For sulfuric acid (H2SO4):
Moles of H2SO4= Molarity ×Volume (L) = 0.4 mol/L ×0.05 L = 0.02 mol
For sodium hydroxide (NaOH):
Moles of NaOH = Molarity ×Volume (L) = 0.2 mol/L ×0.1 L = 0.02 mol
5
Step 3: Identify the limiting reactant. Since both sulfuric acid and sodium
hydroxide produce equal moles (0.02 mol), sodium hydroxide is the limiting
reactant in this case.
Step 4: Calculate the remaining excess of sulfuric acid after the reaction.
Initial moles of H2SO4= 0.02 mol
Final moles of H2SO4= Initial moles−(Moles used in reaction) = 0.02 mol−0.02 mol = 0 mol
Step 5: Calculate the total volume of the resulting solution.
Total volume = 50 mL + 100 mL = 150 mL = 0.15 L
Step 6: Calculate the molarity of the resulting solution.
Molarity of resulting solution = Total moles of solution
Total volume of solution (L) =0.02 mol
0.15 L = 0.133 M
Therefore, the molarity of the resulting solution after the reaction is complete
is 0.133 M.
Question 7
Question
A solution is prepared by dissolving 15.0 g of glucose (C6H12O6) in enough water
to make 500.0 mL of solution. Calculate the molarity of the glucose solution.
Solution
Step 1: Calculate the molar mass of glucose. Since the molecular formula of
glucose is C6H12O6, we need to find the sum of the atomic masses of carbon,
hydrogen, and oxygen in one mole of glucose.
Molar mass of glucose = 6(atomic mass of C)+12(atomic mass of H)+6(atomic mass of O)
Molar mass of glucose = 6(12.01 g/mol) + 12(1.01 g/mol) + 6(16.00 g/mol)
Molar mass of glucose = 180.18 g/mol
Step 2: Calculate the number of moles of glucose in the solution. Given that
the mass of glucose is 15.0 g, we can use the formula n=m
M, where nis the
number of moles, mis the mass, and Mis the molar mass.
n=15.0 g
180.18 g/mol = 0.083mol
Step 3: Calculate the molarity of the solution. Molarity (M) is defined as
the number of moles of solute per liter of solution. Since we have 0.083 moles
of glucose in 500.0 mL of solution, we first convert the volume to liters.
Volume of solution = 500.0 mL = 0.500 L
6
Now, we can calculate the molarity using the formula M=n
V.
M=0.083 mol
0.500 L = 0.166 M
Therefore, the molarity of the glucose solution is 0.166 M.
Question 8
Question
A chemist needs to prepare 500 mL of a 0.25 M solution of sodium chloride.
If the chemist has a stock solution of sodium chloride with a concentration of
2 M, what volume of the stock solution should be added to water to make the
desired solution?
Solution
Step 1: Calculate the moles of sodium chloride needed for a 0.25 M solution in
500 mL.
Given: Desired molarity, M1= 0.25 M Volume of solution, V1= 500 mL =
0.5 L
Moles of solute = M1×V1= 0.25 mol/L ×0.5 L = 0.125 mol
Step 2: Calculate the volume of the stock solution needed.
Given: Concentration of stock solution, M2= 2 M
Use the formula:
M2×V2= moles of solute
Substitute the known values:
2 mol/L ×V2= 0.125 mol
Solve for V2:
V2=0.125 mol
2 mol/L = 0.0625 L = 62.5 mL
Therefore, the chemist should add 62.5 mL of the stock solution to water to
make the desired 0.25 M sodium chloride solution.
Question 9
Question
A solution is prepared by dissolving 5.00 grams of potassium dichromate (K2Cr2O7)
in enough water to make 250.0 mL of solution. Calculate the molarity of the
potassium dichromate solution.
7
Solution
Step 1: Calculate the molar mass of potassium dichromate. The molar mass
of K2Cr2O7can be calculated by adding the molar masses of its constituents
(potassium, chromium, and oxygen).
Molar mass = 2(molar mass of K) + 2(molar mass of Cr) + 7(molar mass of O)
= 2(39.10 g/mol) + 2(52.00 g/mol) + 7(16.00 g/mol)
= 78.20 g/mol + 104.00 g/mol + 112.00 g/mol
= 294.20 g/mol
Step 2: Calculate the number of moles of potassium dichromate in the solu-
tion. Using the formula n=m
M, where nis the number of moles, mis the mass
of the substance, and Mis the molar mass:
n=5.00 g
294.20 g/mol = 0.0170 mol
Step 3: Calculate the molarity of the solution. Molarity (M) is defined as
the number of moles of solute per liter of solution. We have found the number
of moles of potassium dichromate in 250.0 mL of solution:
M=n
V=0.0170 mol
0.2500 L = 0.0680 M
Therefore, the molarity of the potassium dichromate solution is 0.0680 M.
Question 10
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of potassium iodide
(KI). The chemist has a stock solution of KI with a concentration of 0.5 M.
How many milliliters of the stock solution should the chemist use to make the
desired solution?
Solution
Step 1: Calculate the moles of potassium iodide required to make the desired
solution. Given: Volume of desired solution, Vfinal = 500 mL = 0.5 L Molarity
of the desired solution, Mfinal = 0.2 M
We can use the formula for molarity:
M=moles of solute
volume of solution in liters
Rearranging the formula to solve for moles:
moles of solute = M×Vfinal
8
Substitute the given values:
moles of solute = 0.2 mol/L ×0.5 L = 0.1 mol
Step 2: Calculate the volume of the stock solution needed. Given: Molarity
of the stock solution, Minitial = 0.5 M
We can use the same formula for molarity to find the volume of stock solution
needed:
Minitial ×Vinitial = moles of solute
Rearranging the formula to solve for Vinitial:
Vinitial =moles of solute
Minitial
Substitute the values calculated and given:
Vinitial =0.1 mol
0.5 mol/L = 0.2 L = 200 mL
Therefore, the chemist should use 200 mL of the stock solution to make 500
mL of a 0.2 M solution of potassium iodide.
Question 11
Question
A chemist needs to prepare 500 mL of a 0.2 M solution of sulfuric acid, H2SO4.
However, the only stock solution available is a 1 M solution of sulfuric acid.
How would the chemist prepare the desired solution using the stock solution?
Solution
Step 1: Calculate the amount of sulfuric acid needed from the 1 M stock solution
to prepare the desired 0.2 M solution. Let V1be the volume of the 1 M stock
solution needed. Using the formula for molarity:
M1×V1=M2×V2
where M1= 1 M (molarity of stock solution), V2= 500 mL (final volume of 0.2
M solution), and M2= 0.2 M (molarity of desired solution).
Substitute the given values into the formula:
1×V1= 0.2×500
V1=0.2×500
1
V1= 100 mL = 0.1 L
9
Step 2: Calculate the volume of solvent needed to dilute the 1 M stock
solution. The total volume of the final solution is 500 mL. Since we need 100
mL of the 1 M stock solution, the remaining volume should be the solvent
(water). Therefore, the volume of solvent required is:
500 −100 = 400 mL = 0.4 L
Step 3: Mix the calculated volumes of the 1 M stock solution and the solvent
to prepare the 0.2 M solution of sulfuric acid. The chemist should carefully
measure and mix 100 mL of the 1 M stock solution with 400 mL of solvent to
obtain the desired 0.2 M solution of sulfuric acid.
Question 12
Question
A chemist needs to prepare 1.5 L of a 0.25 M solution of potassium perman-
ganate (KMnO4). The chemist has a stock solution of potassium permanganate
with a concentration of 0.5 M. How many milliliters of the stock solution should
the chemist use to make the desired solution?
Solution
Step 1: Calculate the number of moles of potassium permanganate needed for
the desired solution. Step 2: Use the concentration of the stock solution to
determine the volume of stock solution needed.
Step 1: Given: Volume needed (Vneeded) = 1.5 L Molarity of desired solution
(Mdesired) = 0.25 M
We can use the formula for molarity to find the number of moles (n) needed:
n=Mdesired ×Vneeded
Substitute the given values:
n= 0.25 mol/L ×1.5 L
n= 0.375 mol
So, 0.375 moles of potassium permanganate are needed.
Step 2: Given: Molarity of stock solution (Mstock) = 0.5 M
We can rearrange the formula for molarity to find the volume of stock solu-
tion needed:
Vstock =n
Mstock
Substitute the calculated number of moles and the molarity of the stock
solution:
Vstock =0.375 mol
0.5 mol/L
10
Vstock = 0.75 L
Since we want the answer in milliliters, we convert 0.75 L to milliliters:
Vstock = 0.75 L ×1000 mL/L
Vstock = 750 mL
The chemist should use 750 mL of the 0.5 M stock solution to prepare a 1.5
L solution with a concentration of 0.25 M.
Question 13
Question
A chemist has a 2 L solution containing 0.5 moles of solute. She wants to dilute
this solution to make a new solution that is 0.1 M. How much solvent should
she add to achieve this new concentration?
Solution
Step 1: Calculate the initial molarity of the solution. Step 2: Use the formula
for dilution to find the volume of solvent to add.
Step 1: To find the initial molarity of the solution, we first calculate the
molarity using the formula:
M1=moles of solute
volume of solution (L)
Given that the moles of solute is 0.5 moles and the initial volume of the
solution is 2 L, we have
M1=0.5
2= 0.25 M
Step 2: To dilute the solution to 0.1 M, we use the formula for dilution:
M1V1=M2V2
where M1and V1are the initial molarity and volume of the solution, and
M2and V2are the final molarity and volume of the solution.
Substitute the known values into the formula:
(0.25)(2) = (0.1)(2 + V)
Solve for V:
0.5=0.2+0.1V
0.1V= 0.3
11
V=0.3
0.1= 3 L
Therefore, the chemist should add 3 L of solvent to the 2 L solution to
achieve a final concentration of 0.1 M.
Question 14
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of potassium perman-
ganate. The chemist has a stock solution of potassium permanganate with a
concentration of 0.5 M. How many milliliters of the stock solution should the
chemist use to prepare the desired solution?
Solution
Step 1: Calculate the number of moles of potassium permanganate needed for
the desired solution. Given that the desired molarity is 0.2 M and the volume
is 500 mL, we can use the formula:
moles = molarity ×volume
Substitute the values:
moles = 0.2 mol/L ×0.5 L = 0.1 mol
Step 2: Calculate the volume of stock solution needed. We know the con-
centration of the stock solution is 0.5 M. Let the volume of the stock solution
needed be xmL. We can use the formula:
moles in stock solution = molarity of stock solution×(volume of stock solution in L)
We found that 0.1 moles of potassium permanganate are needed. Substitute
the values:
0.1 mol = 0.5 mol/L ×(x×10−3L)
Step 3: Solve for xto find the volume of stock solution needed.
xmL = 0.1 mol
0.5 mol/L ·1000 = 200 mL
Therefore, the chemist should use 200 mL of the stock solution to prepare
the desired 0.2 M potassium permanganate solution.
Question 15
Question
A solution is prepared by dissolving 10.0 g of glucose (C6H12O6) in enough water
to make 500.0 mL of solution. What is the molarity of the glucose solution?
12
Solution
Step 1: Calculate the molar mass of glucose. The molar mass of glucose is
calculated by adding up the atomic masses of its constituent elements: carbon
(C), hydrogen (H), and oxygen (O).
Molar mass C6H12O6= (6×atomic mass C)+(12×atomic mass H)+(6×atomic mass O)
= (6 ×12.01 g/mol) + (12 ×1.01 g/mol) + (6 ×16.00 g/mol)
= 72.06 g/mol + 12.12 g/mol + 96.00 g/mol
= 180.18 g/mol
Step 2: Calculate the number of moles of glucose. Given that the mass of
glucose is 10.0 g, we can calculate the number of moles using the formula:
Moles =mass
molar mass
Moles =10.0g
180.18 g/mol
Moles = 0.0555 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as the
number of moles of solute per liter of solution. In this case, the volume of the
solution is 500.0 mL or 0.5000 L.
Molarity =moles of solute
volume of solution (L)
Molarity =0.0555 mol
0.5000 L
Molarity = 0.111 mol/L
Therefore, the molarity of the glucose solution is 0.111 mol/L.
Question 16
Question
A chemist needs to prepare 500 mL of a solution with a molarity of 0.25 M.
However, the chemist only has a stock solution with a concentration of 2.0 M.
How many milliliters of the stock solution should the chemist use to make the
desired solution?
13
Solution
Step 1: Write down the given information. Let V1be the volume of the 2.0 M
stock solution to be used, M1be the molarity of the stock solution (2.0 M), V2
be the final volume of the solution (500 mL), and M2be the desired molarity
(0.25 M).
Step 2: Use the formula for dilution. The formula for dilution is given by:
M1V1=M2V2
Step 3: Fill in the values. Substitute the known values into the formula:
2.0 M ·V1= 0.25 M ·500 mL
Step 4: Solve for V1.
2.0·V1= 0.25 ·500
2.0·V1= 125
V1=125
2.0
V1= 62.5 mL
The chemist should use 62.5 mL of the 2.0 M stock solution to make 500 mL
of a 0.25 M solution.
Question 17
Question
A chemist prepared a solution by mixing 250 mL of a 0.6 M sulfuric acid (H2SO4)
solution with 500 mL of a 1.2 M sodium hydroxide (NaOH) solution. What is
the molarity of the resulting solution after the reaction H2SO4reacts with NaOH
to form water and sodium sulfate?
Solution
Step 1: Write the balanced chemical equation for the reaction between sulfuric
acid and sodium hydroxide:
H2SO4(aq) + 2NaOH(aq)→2H2O(l) + Na2SO4(aq)
Step 2: Determine the number of moles of sulfuric acid and sodium hydroxide
used in the reaction.
Moles of H2SO4= Molarity ×Volume (L)
= 0.6 mol/L ×0.25 L
= 0.15 mol
14
Moles of NaOH = Molarity ×Volume (L)
= 1.2 mol/L ×0.5 L
= 0.6 mol
Step 3: Identify the limiting reactant in the reaction. Since 1 mole of H2SO4
reacts with 2 moles of NaOH, we need to compare the moles of sulfuric acid and
sodium hydroxide to determine the limiting reactant.
The molar ratio of H2SO4to NaOH is 1 : 2, which means we would need
0.3 mol of H2SO4to completely react with 0.6 mol of NaOH. However, we only
have 0.15 mol of H2SO4, making it the limiting reactant.
Step 4: Determine the amount of sodium sulfate and water formed based on
the limiting reactant. Since 0.15 mol of H2SO4reacts, 0.15 mol of Na2SO4and
0.3 mol of water are produced.
Step 5: Calculate the total volume of the resulting solution. The total
volume of the resulting solution is 250 mL + 500 mL = 750 mL = 0.75 L.
Step 6: Calculate the molarity of the resulting solution.
Molarity of resulting solution = Total moles of solute
Total volume of solution
=0.15 mol + 0.6 mol
0.75 L
=0.75 mol
0.75 L
= 1.0 M
Therefore, the molarity of the resulting solution after the reaction is 1.0 M.
Question 18
Question
A student dilutes 50.0 mL of a 6.00 M hydrochloric acid (HCl) solution with
water to make 250.0 mL of a new solution. What is the final molarity of the
HCl solution after dilution?
Solution
Step 1: Calculate the number of moles of HCl in the initial solution. Given:
Initial volume of HCl solution (Vi) = 50.0 mL = 0.050 L Initial molarity of HCl
solution (Mi) = 6.00 M
Use the formula:
n=M×V
where nis the number of moles, Mis the molarity, and Vis the volume.
Substitute the values to find the number of moles in the initial solution:
ni= 6.00 mol/L ×0.050 L = 0.30 mol
15
Step 2: Calculate the final molarity of the HCl solution after dilution. Given:
Final volume of the new solution (Vf) = 250.0 mL = 0.250 L
Use the formula for dilution:
Mf=ni
Vf
where Mfis the final molarity, niis the number of moles in the initial solution,
and Vfis the final volume.
Substitute the values to find the final molarity:
Mf=0.30 mol
0.250 L = 1.20 M
Therefore, the final molarity of the hydrochloric acid solution after dilution
is 1.20 M.
Question 19
Question
A chemist has a solution containing 50.0 g of sodium chloride (NaCl) dissolved
in 500.0 mL of water. What is the molarity of the solution?
Solution
Step 1: Calculate the molar mass of sodium chloride (NaCl). The molar mass
of NaCl is the sum of the atomic masses of sodium (Na) and chlorine (Cl).
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. Given that the mass
of NaCl is 50.0 g, we can calculate the number of moles by dividing the mass
by the molar mass.
Number of moles of NaCl = 50.0 g
58.44 g/mol ≈0.856 mol
Step 3: Calculate the volume of the solution in liters. The volume of the
solution is 500.0 mL, which needs to be converted to liters by dividing by 1000.
Volume of the solution = 500.0 mL
1000 mL/L = 0.5000 L
Step 4: Calculate the molarity of the solution. Molarity (M) is defined as
moles of solute per liter of solution.
Molarity = Number of moles of solute
Volume of solution in liters =0.856 mol
0.5000 L = 1.71 M
Therefore, the molarity of the solution containing 50.0 g of sodium chloride
dissolved in 500.0 mL of water is 1.71 M.
16
Question 20
Question
A chemist wants to prepare 500 mL of a 0.4 M aqueous solution of potassium
permanganate (KMnO4). The chemist has a stock solution of KMnO4with a
concentration of 2.5 M. How many milliliters of the stock solution should be
used to make the desired solution?
Solution
Step 1: Let Vbe the volume of the stock solution in milliliters needed to make
the desired solution. Step 2: Calculate the amount of KMnO4in moles that are
required to make the desired solution:
Moles of KMnO4= Molarity ×Volume (L)
Moles of KMnO4= 0.4 M ×0.5 L
Moles of KMnO4= 0.2 mol
Step 3: Calculate the volume of the stock solution needed to provide 0.2 mol
of KMnO4:
Molarity = Moles of solute
Volume of solution
2.5 M = 0.2 mol
VL
V=0.2
2.5= 0.08 L = 80 mL
Therefore, the chemist should use 80 mL of the 2.5 M stock solution to
prepare the 500 mL of 0.4 M KMnO4solution.
Question 21
Question
A student wants to prepare 500 mL of a 0.2 M solution of a compound with a
molar mass of 150.3 g/mol. However, the student only has a stock solution with
a concentration of 1.5 M. How much of the stock solution should the student
use and how much water should be added to make the desired 0.2 M solution?
Solution
Step 1: Calculate the number of moles of the compound needed for the desired
solution. Let Vbe the volume of the 0.2 M solution to be prepared and M1be
the desired molarity. Given: V= 500 mL = 0.5 L M1= 0.2 M
17
Using the formula M1V1=M2V2where V1is the volume of the stock solution
to be used and M2= 1.5 M is the concentration of the stock solution:
M1V=M2V1
0.2×0.5=1.5V1
V1=0.2×0.5
1.5=0.1
1.5=1
15 = 0.06667 L = 66.67 mL
The student should use 66.67 mL of the stock solution.
Step 2: Calculate the volume of water to be added to the stock solution.
The total volume of the final solution is 500 mL. Since 66.67 mL of the stock
solution will be used, the volume of water needed is:
500 −66.67 = 433.33 mL
Therefore, the student should use 66.67 mL of the stock solution and add
433.33 mL of water to prepare the desired 0.2 M solution.
Question 22
Question
A chemist needs to prepare 500 mL of a 0.2 M solution of sodium hydroxide
(NaOH). The chemist has a stock solution of NaOH with a concentration of 1.0
M. How many milliliters of the stock solution should be diluted with water to
make the desired solution?
Solution
Step 1: Let’s calculate the number of moles of NaOH needed for the desired so-
lution. Given: Volume of solution needed (Vsolution) = 500 mL = 0.5 L Molarity
of the solution needed (Mdesired) = 0.2 M
Using the formula for molarity:
M=moles
volume (in liters)
we can rearrange the formula to find the moles:
moles = M×volume (in liters)
Therefore, the number of moles of NaOH needed is:
moles = 0.2 mol/L ×0.5 L = 0.1 mol
Step 2: Next, let’s determine the volume of the stock solution needed to
obtain 0.1 moles of NaOH. Given: Molarity of the stock solution (Mstock) = 1.0
M
Using the formula for molarity:
Mstock =moles
volume (in liters)
18
we can rearrange the formula to find the volume:
volume (in liters) = moles
Mstock
Substitute the values to find the volume of the stock solution needed:
volume (in liters) = 0.1 mol
1.0 mol/L = 0.1 L = 100 mL
Therefore, the chemist should dilute 100 mL of the stock solution with water
to make the desired 0.2 M solution of sodium hydroxide.
Question 23
Question
A chemist prepares a solution by dissolving 5.00 grams of NaCl in enough wa-
ter to make 500.0 mL of solution. The density of the solution is 1.08 g/mL.
Calculate the molarity of NaCl in the solution.
Solution
Step 1: Calculate the moles of NaCl. Given mass of NaCl = 5.00 g
Molar mass of NaCl = 58.44 g/mol
moles of NaCl = mass of NaCl
molar mass of NaCl =5.00 g
58.44 g/mol
Step 2: Calculate the volume of the solution. Given density of the solution
= 1.08 g/mL
Volume of solution = 500.0 mL
volume of solution = mass of solution
density of solution =500.0 mL
1.08 g/mL
Step 3: Calculate the molarity of NaCl in the solution. Molarity is defined
as moles of solute divided by liters of solution.
Molarity = moles of NaCl
volume of solution in liters
Make sure that the volume of solution is converted to liters before calculating
the molarity.
Question 24
Question
A chemist prepares a solution by mixing 250 mL of a 2.0 M sodium chloride
solution with 500 mL of a 1.5 M calcium chloride solution. What is the molarity
of chloride ions in the resulting solution?
19
Solution
Step 1: Calculate the moles of sodium chloride in the 250 mL solution.
Given: Volume of sodium chloride solution = 250 mL = 0.250 L
Molarity of sodium chloride solution = 2.0 M
The moles of sodium chloride can be calculated using the formula:
moles = Molarity ×Volume
moles of NaCl = 2.0 mol/L ×0.250 L
moles of NaCl = 0.500 mol
Step 2: Calculate the moles of calcium chloride in the 500 mL solution.
Given: Volume of calcium chloride solution = 500 mL = 0.500 L
Molarity of calcium chloride solution = 1.5 M
The moles of calcium chloride can be calculated using the formula:
moles = Molarity ×Volume
moles of CaCl2= 1.5 mol/L ×0.500 L
moles of CaCl2= 0.750 mol
Step 3: Calculate the total moles of chloride ions.
Since each formula unit of sodium chloride and calcium chloride has 1 chloride
ion, the total moles of chloride ions in the resulting solution is the sum of the
moles of chloride ions from each component.
Total moles of Cl−= moles of NaCl + 2 ×moles of CaCl2
Total moles of Cl−= 0.500 + 2 ×0.750
Total moles of Cl−= 2.000 mol
Step 4: Calculate the molarity of chloride ions in the resulting solution.
The volume of the resulting solution is the sum of the volumes of the two
solutions used.
Total volume = 0.250 L + 0.500 L
Total volume = 0.750 L
Finally, we can calculate the molarity of chloride ions using the formula:
Molarity of Cl−=Total moles of Cl−
Total volume
Molarity of Cl−=2.000 mol
0.750 L
Molarity of Cl−= 2.67 M
Therefore, the molarity of chloride ions in the resulting solution is 2.67 M.
20
Question 25
Question
A chemist needs to prepare 500 mL of a 0.2 M solution of potassium dichromate
and has a stock solution of 1 M potassium dichromate. How many milliliters of
the stock solution should the chemist use and how much solvent (in mL) should
be added to prepare the desired solution?
Solution
Step 1: Let Vstock be the volume of stock solution needed and Vsolvent be the
volume of solvent added to prepare the final solution.
Step 2: Use the formula for molarity to calculate the amount of solute present
in the stock solution:
M1V1=M2V2
1 M ×Vstock = 0.2 M ×500 mL
Vstock =0.2×500
1= 100 mL
Step 3: The total volume of the final solution is the sum of the volumes of
the stock solution and the solvent:
Vtotal =Vstock +Vsolvent
Vtotal = 100 mL + Vsolvent
Step 4: We know that the total volume of the final solution is 500 mL, so
we can write:
500 mL = 100 mL + Vsolvent
Step 5: Solve for Vsolvent:
Vsolvent = 500 mL −100 mL = 400 mL
Therefore, the chemist should use 100 mL of the stock solution and add 400
mL of solvent to prepare the desired 0.2 M solution of potassium dichromate.
Question 26
Question
A student wants to prepare 500 mL of a 0.2 M solution of potassium perman-
ganate. However, they only have a 750 mL stock solution with a concentration
of 0.3 M. How should the student dilute the stock solution to obtain the desired
concentration?
21
Solution
Step 1: Calculate the moles of potassium permanganate needed for the desired
solution. Let Vf inal be the final volume of the solution (500 mL) and Mfinal be
the final concentration (0.2 M). Using the formula M1V1=M2V2, we have:
moles of KMnO4=Mfinal ×Vf inal
moles of KMnO4= 0.2 mol/L ×0.5 L
moles of KMnO4= 0.1 mol
Step 2: Calculate the volume of the stock solution needed to obtain the
desired moles of potassium permanganate. Let Vstock be the volume of the
stock solution needed. Using the formula M1V1=M2V2, we have:
0.3 mol/L ×Vstock = 0.1 mol
Vstock =0.1 mol
0.3 mol/L
Vstock =1
3L
Step 3: Calculate the volume of solvent needed to dilute the stock solution.
The volume of solvent needed is:
Vsolvent = 750 mL −Vstock
Vsolvent = 750 mL −1
3L = 2250 −1
3mL
Vsolvent =2249
3mL ≈749.67 mL
Therefore, the student should measure 333.33 mL of the stock solution and
dilute it with 749.67 mL of solvent (water) to obtain 500 mL of a 0.2 M solution
of potassium permanganate.
Question 27
Question
A chemist needs to prepare 500 mL of a 0.15 M solution of potassium perman-
ganate, KMnO4. The chemist has a stock solution of KMnO4with a concen-
tration of 0.5 M. How many milliliters of the stock solution should be used and
how much water should be added to prepare the desired solution?
22
Solution
Step 1: Calculate the moles of KMnO4needed for the desired solution. Given:
Volume of desired solution = 500 mL = 0.5 L Molarity of desired solution =
0.15 M
From the formula
Molarity =moles of solute
volume of solution in liters
we can rearrange to find the moles of solute: moles of solute = Molarity ×
volume of solution moles of KMnO4= 0.15 M ×0.5 L = 0.075 moles
Step 2: Calculate the volume of stock solution needed. Given: Molarity of
stock solution = 0.5 M
Using the same formula, we can find the volume of stock solution needed:
volume of stock solution = moles of solute
Molarity of stock solution volume of stock solution =
0.075 moles
0.5 M = 0.15 L = 150 mL
Step 3: Calculate the volume of water needed to prepare the solution. The
total volume of the desired solution is 500 mL, and we are adding 150 mL of
the stock solution. Therefore, the volume of water needed is: volume of water
= total volume - volume of stock solution volume of water = 500 mL - 150 mL
= 350 mL
So, 150 mL of the stock solution should be used, and 350 mL of water should
be added to prepare the desired 0.15 M KMnO4solution.
Question 28
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?
Solution
Step 1: Calculate the molar mass of glucose. The molar mass of glucose
(C6H12O6) can be calculated by adding the atomic masses of carbon, hydro-
gen, and oxygen. - Atomic mass of carbon (C) = 12.01 g/mol - Atomic mass of
hydrogen (H) = 1.008 g/mol - Atomic mass of oxygen (O) = 16.00 g/mol
Therefore, the molar mass of glucose:
6×12.01 g/mol + 12 ×1.008 g/mol + 6 ×16.00 g/mol = 180.16 g/mol
Step 2: Calculate the number of moles of glucose. Given mass of glucose =
15.0 g
Number of moles = Mass
Molar mass =15.0 g
180.16 g/mol
Step 3: Calculate the volume of the solution in liters. Given volume of
solution = 250.0 mL = 0.250 L
23
Step 4: Calculate the molarity of the solution.
Molarity (M) = Number of moles
Volume of solution in liters =15.0 g/180.16 g/mol
0.250 L
Calculating the molarity:
Molarity = 15.0 g/180.16 g/mol
0.250 L =0.0833 mol
0.250 L = 0.333 M
Therefore, the molarity of the glucose solution is 0.333 M.
Question 29
Question
A solution is prepared by dissolving 15.0 g of glucose (C6H12O6) in enough
water to make 250.0 mL of solution.
Calculate the molarity of the solution.
Given: Molar mass of glucose = 180.0 g/mol
Solution
Step 1: Calculate the number of moles of glucose dissolved in the solution.
Number of moles of glucose = Mass of glucose
Molar mass of glucose
=15.0 g
180.0 g/mol
= 0.0833 mol
Step 2: Convert the volume of the solution from milliliters to liters.
Volume of solution = 250.0 mL = 0.250 L
Step 3: Calculate the molarity of the solution using the formula:
Molarity = Number of moles of solute
Volume of solution in liters
Substitute the values into the formula:
Therefore, the molarity of the solution is approximately 0.333 M.
24
Question 30
Question
A chemist needs to prepare 500 mL of a 0.1 M NaCl solution, but only has a
stock solution of 1 M NaCl. How many mL of the stock solution should the
chemist use and how much water should be added to make the desired solution?
Solution
Step 1: Calculate the moles of NaCl needed for the desired solution. Let’s
denote the volume of 1 M NaCl solution to be used as V1mL and the volume
of water to be added as V2mL. The moles of NaCl needed can be calculated as:
moles of NaCl = molarity ×volume (L)
moles of NaCl = 0.1 mol/L ×0.5 L
moles of NaCl = 0.05 mol
Step 2: Use the moles of NaCl to find the volume of the 1 M NaCl solution
needed. Since the stock solution is 1 M NaCl, the moles of NaCl in the stock
solution is equal to the volume of the solution in liters. Therefore, the volume
of the 1 M NaCl solution needed can be calculated as:
0.05 mol = 1 mol/L ×V1
V1= 0.05 L = 50 mL
Step 3: Calculate the volume of water needed to make the desired solution.
The total volume of the final solution is 500 mL. Therefore, the volume of water
needed is:
V2= 500 mL −50 mL = 450 mL
Thus, the chemist should use 50 mL of the 1 M NaCl stock solution and add
450 mL of water to prepare 500 mL of a 0.1 M NaCl solution.
25
