CHEM 105 - ELEMENTS OF
GENERAL CHEMISTRY - Molarity
and Concentrations
Question Bank - Set 1
Liberty University
Question 1
Question
A solution is prepared by dissolving 8.50 grams of glucose (C6H12O6) in enough
water to make 250.0 mL of solution. Calculate the molarity of the glucose
solution. (Molar mass of glucose = 180.16 g/mol)
Solution
Step 1: Calculate the number of moles of glucose present in the solution. Given:
Mass of glucose, m= 8.50 g Molar mass of glucose, M= 180.16 g/mol
Using the formula for calculating moles:
moles of glucose = mass of glucose
molar mass of glucose
moles of glucose = 8.50 g
180.16 g/mol = 0.0471 mol
Step 2: Calculate the volume of the solution in liters. Volume of solution,
V= 250.0 mL = 250.0×10−3L = 0.2500 L
Step 3: Calculate the molarity of the glucose solution. The definition of
molarity is:
Molarity (M) = moles of solute
volume of solution in liters
Molarity (M) = 0.0471 mol
0.2500 L = 0.1884 M
Therefore, the molarity of the glucose solution is 0.1884 M.
Question 2
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of sodium chloride
(NaCl). If the molar mass of NaCl is 58.44 g/mol, how many grams of NaCl
should the chemist dissolve in water?
Solution
Step 1: Calculate the number of moles of NaCl needed.
Molarity = moles of solute
volume of solution in liters
⇒moles of solute = Molarity ×volume of solution in liters
⇒moles of NaCl = 0.2 mol/L ×0.5 L = 0.1 moles
Step 2: Calculate the mass of NaCl needed.
Mass = moles ×molar mass
⇒Mass of NaCl = 0.1 moles ×58.44 g/mol = 5.844 g
Therefore, the chemist should dissolve 5.844 grams of NaCl in water to
prepare 500 mL of a 0.2 M NaCl solution.
Question 3
Question
A chemist is preparing a solution by mixing 50 mL of a 2 M HCl solution with
100 mL of a 3 M HCl solution. What is the molarity of the final solution?
Solution
Step 1: Find the total moles of HCl in each solution. For the first solution:
0.050L×2mol/L = 0.100mol of HCl
For the second solution: 0.100L×3mol/L = 0.300mol of HCl
Step 2: Find the total moles of HCl in the final solution. 0.100mol +
0.300mol = 0.400mol of HCl
Step 3: Find the total volume of the final solution. 0.050L+0.100L= 0.150L
or 150 mL
Step 4: Calculate the molarity of the final solution using the total moles and
volume. Molarity =0.400mol
0.150L=40
15 M= 2.67Mor 2.67 mol/L
Therefore, the molarity of the final solution is 2.67 M.
2
Question 4
Question
A chemist wants to prepare 500 mL of a 0.2 M hydrochloric acid (HCl) solution.
The chemist has a stock solution of 6 M HCl. How many milliliters of the stock
solution should be used, and how much water should be added to make the
desired solution?
Solution
Step 1: Calculate the moles of HCl 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 (L)
Converting the volume to liters:
500 mL = 500 ×10−3L=0.5 L
Therefore, the moles of HCl needed is:
moles = 0.2 M ×0.5 L = 0.1 moles
Step 2: Calculate the volume of the stock solution needed. Using the equa-
tion:
moles = molarity ×volume (L)
We can rearrange the formula to solve for the volume of the stock solution:
volume (L) = moles
molarity =0.1 moles
6 M
volume (L) = 0.1
6= 0.01667 L = 16.67 mL
Step 3: Calculate the volume of water needed. The total volume of the final
solution is 500 mL. Therefore, the volume of water needed is:
volume of water (mL) = total volume (mL) −volume of stock solution (mL)
volume of water (mL) = 500 mL −16.67 mL = 483.33 mL
Thus, the chemist should add 16.67 mL of the 6 M HCl stock solution and
dilute it with 483.33 mL of water to prepare 500 mL of a 0.2 M HCl solution.
Question 5
Question
A chemist needs to prepare 500 mL of a 0.25 M solution of sulfuric acid, H2SO4.
The chemist has a stock solution of sulfuric acid that is labeled as 1.0 M. De-
termine how many milliliters of the 1.0 M solution should be used to prepare
the desired 0.25 M solution.
3
Solution
Step 1: Calculate the moles of sulfuric acid needed for the desired solution.
Given: Volume of final solution (Vf) = 500 mL = 0.5 L Molarity of final solution
(Mf) = 0.25 M Using the formula:
Moles = Molarity ×Volume (in liters)
we have:
Moles of H2SO4= 0.25 mol/L ×0.5 L = 0.125 mol
Step 2: Determine how many milliliters of the 1.0 M solution contain 0.125
moles of sulfuric acid. Given: Molarity of stock solution (Ms) = 1.0 M By
rearranging the formula, we find:
Volume (in liters) = Moles
Molarity
Volume (in liters) = 0.125 mol
1.0 mol/L = 0.125 L
Converting liters to milliliters:
0.125 L ×1000 mL/L = 125 mL
Answer: The chemist should use 125 mL of the 1.0 M sulfuric acid solution
to prepare 500 mL of a 0.25 M solution.
Question 6
Question
A solution is prepared by dissolving 5.00 g of glucose (C6H12O6) in enough
water to make 250.0 mL of solution.
Part (a)
Calculate the molarity of the glucose solution.
Part (b)
If 20.0 mL of this solution is diluted to 100.0 mL, what will be the molarity of
the diluted solution?
4
Solution
Part (a)
Step 1: Calculate the molar mass of glucose (C6H12O6). The molar mass of
glucose can be calculated by adding the atomic masses of carbon, hydrogen,
and oxygen.
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 ×1.008 g/mol + 6 ×16.00 g/mol
= 180.16 g/mol
Step 2: Calculate the number of moles of glucose in the solution.
Number of moles of glucose = Mass of glucose
Molar mass of glucose =5.00 g
180.16 g/mol = 0.0278 mol
Step 3: Calculate the molarity of the glucose solution.
Molarity = Number of moles of solute
Volume of solution in liters =0.0278 mol
0.2500 L = 0.111 M
Therefore, the molarity of the glucose solution is 0.111 M.
Part (b)
Step 1: Calculate the number of moles of glucose in the 20.0 mL portion of the
solution.
Number of moles in 20.0 mL = Molarity×Volume in liters = 0.111 M×0.0200 L = 0.00222 mol
Step 2: Calculate the molarity of the diluted solution. Since the 20.0 mL
portion is diluted to a total volume of 100.0 mL, we need to calculate the total
number of moles in 100.0 mL before finding the molarity.
Total moles in 100.0 mL = 0.00222 mol
0.0200 L ×0.1000 L = 0.0111 mol
Molarity = 0.0111 mol
0.1000 L = 0.111 M
Therefore, the molarity of the diluted solution is 0.111 M.
Question 7
Question
A solution is prepared by dissolving 10.0 g of glucose (C6H12O6) in enough wa-
ter to make 500.0 mL of solution. What is the molarity of the glucose solution?
(Molar mass of glucose = 180.16 g/mol)
5
Solution
Step 1: Calculate the number of moles of glucose in the solution.
To find the number of moles of glucose, we use the formula:
moles = mass
molar mass
Given: mass of glucose = 10.0 g, molar mass of glucose = 180.16 g/mol,
volume of solution = 500.0 mL.
Substitute the values into the formula:
moles = 10.0 g
180.16 g/mol
moles = 0.0555 mol
Step 2: Calculate the molarity of the solution.
Molarity (M) is defined as the number of moles of solute per liter of solution.
Molarity = moles of solute
liters of solution
Given: moles of glucose = 0.0555 mol, volume of solution = 500.0 mL = 0.5
L.
Substitute the values into the formula:
Molarity = 0.0555 mol
0.5 L
Molarity = 0.111 M
Therefore, the molarity of the glucose solution is 0.111 M.
Question 8
Question
A chemist wants to prepare 500 mL of a 0.1 M solution of hydrochloric acid
(HCl) from a stock solution of 2 M. How many milliliters of the 2 M solution
should be used?
Solution
Step 1: Let Vbe the volume of 2 M solution needed to prepare the desired 0.1
M solution.
Step 2: From the formula for molarity, we have:
M1V1=M2V2
6
Substitute M1= 2 M, V1=VmL, M2= 0.1 M, and V2= 500 mL into the
formula:
2V= 0.1×500
2V= 50
Step 3: Solve for V:
V=50
2
V= 25
Therefore, the chemist should use 25 mL of the 2 M solution to prepare 500
mL of a 0.1 M solution of hydrochloric acid.
Question 9
Question
A chemist prepares a solution by dissolving 5.00 g of sodium chloride (NaCl) in
enough water to make 250.0 mL of solution. What is the molarity of the NaCl
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 = Atomic mass of Na + Atomic mass of Cl
Step 2: Calculate the number of moles of NaCl. Since molarity is defined
as the number of moles of solute per liter of solution, we need to calculate the
number of moles of NaCl.
Number of moles = Mass of solute (g)
Molar mass of solute (g/mol)
Step 3: Calculate the volume of the solution in liters. Since the volume of
the solution is given in milliliters, we need to convert it to liters.
Volume of solution (L) = Volume of solution (mL)
1000
Step 4: Calculate the molarity of the NaCl solution. Molarity (M) is defined
as the number of moles of solute per liter of solution.
M=Number of moles of solute
Volume of solution (L)
7
Question 10
Question
A chemist has a stock solution of hydrochloric acid (HCl) with a concentration
of 10.0 M. The chemist needs to prepare 250 mL of a solution that has a con-
centration of 2.00 M. How much of the stock solution should the chemist dilute
with water to prepare the desired solution?
Solution
Step 1: Let’s calculate the number of moles of hydrochloric acid (HCl) needed to
prepare the desired solution. Given: Initial concentration of stock solution, C1=
10.0 M Volume of stock solution to be used, V1=? (in mL) Final concentration
of desired solution, C2= 2.00 M Volume of desired solution, V2= 250 mL
Step 2: Use the formula for molarity to calculate the moles of HCl in the
desired solution.
C1×V1=C2×V2
10.0 M ×V1mL = 2.00 M ×250 mL
V1=2.00 M ×250 mL
10.0 M
V1= 50.0 mL
Therefore, the chemist should dilute 50.0 mL of the stock solution with water
to prepare the desired 250 mL, 2.00 M solution of hydrochloric acid.
Question 11
Question
A solution is prepared by dissolving 5.00 g of potassium permanganate (KMnO4)
in enough water to make 250.0 mL of solution. What is the molarity of the
potassium permanganate 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.
Molar mass of KMnO4= (39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol)) =
158.04 g/mol
Step 2: Determine the number of moles of KMnO4in the solution. Given
mass of KMnO4= 5.00 g, Number of moles of KMnO4=5.00 g
158.04 g/mol =
0.0316 mol
8
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 = 0.2500 L Molarity M=Number of moles of solute
Volume of solution in liters
Substitute the values into the formula: M=0.0316 mol
0.2500 L = 0.1264 M
Therefore, the molarity of the potassium permanganate solution is 0.1264
M.
Question 12
Question
A student wants to prepare 500 mL of a 0.2 M calcium chloride solution. The
student has a stock solution of calcium chloride (CaCl2) with a concentration
of 1 M. How much of the stock solution should the student use to make the
desired solution?
Solution
Step 1: Calculate the moles of calcium chloride needed to make a 0.2 M solution.
Step 2: Use the formula Molarity =moles solute
volume solution (in liters) to find the volume
of stock solution needed. Step 3: Convert the volume of stock solution from
liters to milliliters for the final answer.
Step 1: Given: - Desired molarity (M1) = 0.2 M - Desired volume (V1) =
500 mL = 0.5 L - Stock molarity (M2) = 1 M
The formula for molarity (M) is: M=moles solute
volume solution (in liters)
Rearranging the formula, we get: moles solute = M x volume solution
Substitute the values into the formula: moles of CaCl2= 0.2 M x 0.5 L
moles of CaCl2= 0.1 moles
Step 2: Now, use the formula for molarity to find the volume of stock solution
needed: M2V2=M1V1
Substitute the known values: 1M ×V2= 0.2M ×0.5L
Solve for V2:V2=0.2mol×0.5L
1mol V2= 0.1L
Step 3: Convert the volume from liters to milliliters: 0.1L ×1000mL/L =
100mL
Therefore, the student should use 100 mL of the stock solution to prepare
500 mL of a 0.2 M calcium chloride solution.
Question 13
Question
A student wants to prepare 500.0 mL of a 0.250 M solution of NaCl. However,
they only have a stock solution of NaCl that is 6.00 M. What volume of the
stock solution should the student use to make the desired solution?
9
Solution
Step 1: Write down the given information.
Volume of desired solution, Vdesired = 500.0 mL = 0.5000 L
Molarity of desired solution, Mdesired = 0.250 M
Molarity of stock solution, Mstock = 6.00 M
Step 2: Use the formula for dilution to find the volume of stock solution
required.
The formula for dilution is:
MstockVstock =MdesiredVdesired
Step 3: Substitute the given values into the formula and solve for Vstock.
6.00 M ×Vstock = 0.250 M ×0.5000 L
Vstock =0.250 ×0.5000
6.00
Vstock = 0.0208 L = 20.8 mL
Answer: The student should use 20.8 mL of the 6.00 M NaCl solution to
prepare 500.0 mL of a 0.250 M NaCl solution.
Question 14
Question
A chemist has a solution of hydrochloric acid with a concentration of 6.0 M.
How many milliliters of this solution must be added to water to make 500 mL
of a 2.0 M solution?
Solution
Step 1: Let’s denote the volume of the 6.0 M solution to be added as VmL.
The final volume of the solution will be 500 mL, and the concentration of the
final solution will be 2.0 M.
Step 2: The amount of hydrochloric acid in the 6.0 M solution before dilution
is given by 6.0 mol/L ×VL. The amount of hydrochloric acid in the final
solution after dilution is given by 2.0 mol/L ×500 mL.
Step 3: Since the amount of hydrochloric acid remains constant before and
after dilution, we can set up the equation:
6.0 mol/L ×VL=2.0 mol/L ×500 mL
Step 4: Let’s convert 500 mL to liters: 500 mL = 0.5 L.
Step 5: Substituting the values into the equation, we get:
6.0V= 2 ×0.5
Step 6: Solve for V:
V=2×0.5
6.0
10
V=1
6L
Step 7: Convert the volume to milliliters:
V=1
6×1000 mL
V= 166.67 mL
So, approximately 166.67 mL of the 6.0 M solution must be added to water
to make 500 mL of a 2.0 M solution.
Question 15
Question
A chemist prepares a solution 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?
Solution
Step 1: Calculate the molar mass of glucose. The molar mass of glucose
(C6H12O6) can be calculated by adding the atomic masses of each element
present: 6 carbons, 12 hydrogens, and 6 oxygens.
Molar mass of C6H12O6= 6(Atomic mass of C)+12(Atomic mass of H)+6(Atomic mass of 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 in the solution. First,
convert the mass of glucose to moles using the formula:
moles = mass (g)
molar mass (g/mol)
moles of glucose = 10.0 g
180.18 g/mol ≈0.0554 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as moles
of solute per liter of solution:
Molarity = moles of solute
volume of solution (L)
Since the volume of the solution is 500.0 mL, we need to convert it to liters:
Volume of solution = 500.0 mL ×1 L
1000 mL = 0.500 L
11
Now calculate the molarity of the glucose solution:
Molarity = 0.0554 mol
0.500 L = 0.111 M
Therefore, the molarity of the glucose solution is 0.111 M.
Question 16
Question
A chemist prepares a solution by dissolving 25.0 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. The molar mass of glucose
(C6H12O6) can be calculated by summing the atomic masses of the elements
present:
Molar mass of glucose = 6×Molar mass of C+12×Molar mass of H+6×Molar mass of O
The molar masses are: - Molar mass of C: 12.01 g/mol - Molar mass of H:
1.008 g/mol - Molar mass of O: 16.00 g/mol
Plugging the values into the formula:
Molar mass of glucose = 6(12.01) + 12(1.008) + 6(16.00)
= 72.06 + 12.096 + 96.00
= 180.156 g/mol
Step 2: Calculate the number of moles of glucose in the solution. Given that
25.0 g of glucose is dissolved in the solution, we can calculate the number of
moles of glucose using the formula:
Number of moles = Mass
Molar mass
Substitute the values:
Number of moles = 25.0
180.156
Number of moles ≈0.1387 mol
Step 3: Calculate the molarity of the solution. The molarity of a solution is
defined as the number of moles of solute per liter of solution.
12
Given that the final volume of the solution is 500.0 mL (0.5000 L), we can
calculate the molarity using the formula:
Molarity = Number of moles
Volume (L)
Substitute the values:
Molarity = 0.1387
0.5000
Molarity = 0.277 M
Therefore, the molarity of the glucose solution is 0.277 M.
Question 17
Question
A chemist wants to prepare 500 mL of a 0.25 M sulfuric acid (H2SO4) solution.
However, the only sulfuric acid available is a concentrated solution that is 98
Solution
Step 1: Determine the molar mass of H2SO4.
Molar mass(H2SO4)=2×molar mass(H)+1×molar mass(S) + 4 ×molar mass(O)
= 2(1.01 g/mol) + 1(32.07 g/mol) + 4(16.00 g/mol)
= 98.09 g/mol
Step 2: Calculate the molarity of the concentrated sulfuric acid solution.
Molarity = moles of solute
volume of solution in liters
Moles of H2SO4=mass of H2SO4
molar mass(H2SO4)
=0.98 ×500 g
98.09 g/mol
= 5 mol
Volume of concentrated solution (V) = moles of solute
molarity
=5 mol
0.98 M
= 5.10 L (5100 mL)
Therefore, the chemist should use 5100 mL of the concentrated sulfuric acid
solution to make 500 mL of a 0.25 M sulfuric acid solution.
13
Question 18
Question
A student wants to prepare 500 mL of a 0.2 M solution of potassium perman-
ganate (KMnO4). The student has a stock solution of KMnO4with a concen-
tration of 1.0 M. How many milliliters of the stock solution should the student
use to prepare the desired solution?
Solution
Step 1: Determine the number of moles required to prepare the desired solution.
Given: Volume of solution (V) = 500 mL = 0.5 L Molarity of solution (M) =
0.2 M
The formula for calculating the number of moles is:
moles = Molarity ×Volume (in L)
Substitute the given values:
moles = 0.2 mol/L ×0.5 L = 0.1 mol
Step 2: Calculate the volume of the stock solution needed. Given: Molarity
of stock solution = 1.0 M
The formula for calculating the volume of stock solution needed is:
Volume (stock solution) = moles
Molarity of stock solution
Substitute the values calculated in Step 1:
Volume (stock solution) = 0.1 mol
1.0 mol/L = 0.1 L = 100 mL
Therefore, the student should use 100 mL of the 1.0 M KMnO4stock solution
to prepare 500 mL of a 0.2 M KMnO4solution.
Question 19
Question
A chemist needs to prepare 500 mL of a 0.2 M solution of sulfuric acid, H2SO4.
They have a stock solution of sulfuric acid with a concentration of 6 M. How
many milliliters of the stock solution should the chemist use to make the desired
solution?
14
Solution
Step 1: Calculate the moles of sulfuric acid needed for the desired solution.
Given that the chemist needs to prepare 500 mL of a 0.2 M solution of sulfuric
acid, the moles of sulfuric acid required can be calculated as:
Moles = Molarity ×Volume (L)
Moles = 0.2 mol/L ×0.5 L
Moles = 0.1 mol
Step 2: Calculate the volume of the stock solution required. Since the stock
solution has a concentration of 6 M, the volume of the stock solution needed
can be calculated using the formula:
Moles = Molarity ×Volume (L)
0.1 mol = 6 mol/L ×Volume (L)
Volume (L) = 0.1 mol
6 mol/L
Volume (L) = 0.0167 L
Step 3: Convert the volume to milliliters. To find the volume in milliliters,
we can multiply the volume in liters by 1000:
Volume (mL) = 0.0167 L ×1000
Volume (mL) = 16.7 mL
The chemist should use 16.7 mL of the 6 M stock solution to prepare 500
mL of a 0.2 M solution of sulfuric acid.
Question 20
Question
A chemist has a solution of sulfuric acid with a molarity of 2.5 M. She wants
to dilute this solution to make 500 mL of a new solution with a molarity of 1.5
M. What volume of the original 2.5 M sulfuric acid solution should she use to
make the new solution?
15
Solution
Step 1: Let the volume of the original 2.5 M sulfuric acid solution to be used
be VmL. We can set up the dilution equation as follows:
M1V1=M2V2
where M1= 2.5 M, V1=VmL, M2= 1.5 M, and V2= 500 mL = 0.5 L.
Step 2: Plug in the values and solve for V:
(2.5 M)(VL) = (1.5 M)(0.5 L)
2.5V= 0.75
V=0.75
2.5= 0.3 L = 300 mL
So, the chemist should use 300 mL of the original 2.5 M sulfuric acid solution
to make the new solution.
Question 21
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?
(The molar mass of glucose is 180.16 g/mol.)
Solution
Step 1: Calculate the number of moles of glucose in the solution. Given: Mass
of glucose = 10.0 g Molar mass of glucose = 180.16 g/mol
Use the formula:
moles of solute = mass of solute
molar mass of solute
Plugging in the values:
moles of solute = 10.0 g
180.16 g/mol
moles of solute = 0.0555 mol
Step 2: Calculate the molarity of the solution. Given: Volume of solution =
500.0 mL = 0.5000 L
Use the formula:
Molarity(M) = moles of solute
volume of solution in liters
16
Plugging in the values:
M=0.0555 mol
0.5000 L
M= 0.111 M
Therefore, the molarity of the glucose solution is 0.111 M.
Question 22
Question
A solution is prepared by mixing 50.0 mL of a 0.200 M potassium iodide (KI)
solution with 100.0 mL of a 0.500 M lead(II) nitrate (Pb(NO3)2) solution. Cal-
culate the molarity of potassium ions (K+) and iodide ions (I−) in the final
solution.
Solution
Step 1: Calculate the moles of potassium iodide (KI) and lead(II) nitrate
(Pb(NO3)2) used.
The moles of KI can be calculated as:
moles of KI = Molarity ×Volume (L)
Given that the volume of KI solution is 50.0 mL and the molarity is 0.200
M:
moles of KI = 0.200 mol/L ×0.0500 L = 0.0100 mol
The moles of Pb(NO3)2can be calculated in a similar manner:
moles of Pb(NO3)2)=0.500 mol/L ×0.100 L = 0.0500 mol
Step 2: Determine the limiting reactant.
From the balanced chemical equation for the reaction between KI and Pb(NO3)2,
we see that a 1:1 molar ratio exists between KI and Pb(NO3)2. Therefore, the
limiting reactant will be the one with fewer moles.
Since 0.0100 mol of KI and 0.0500 mol of Pb(NO3)2are used, KI is the
limiting reactant.
Step 3: Calculate the moles of potassium ions (K+) and iodide ions (I−) in
the final solution.
The molarity of potassium ions (K+) in the final solution can be calculated
using the total volume of the final solution (150.0 mL or 0.1500 L):
Molarity of K+in the final solution = moles of K+
Volume of solution (L)
Since KI provides 1 mol of K+ions for every 1 mol of KI:
moles of K+= 0.0100 mol
17
Thus, the molarity of K+in the final solution is:
Molarity of K+in the final solution = 0.0100 mol
0.1500 L = 0.067 M
Similarly, the molarity of iodide ions (I−) in the final solution can be calcu-
lated as:
Molarity of I−in the final solution = 0.0100 mol
0.1500 L = 0.067 M
Question 23
Question
A chemist wants to prepare 500 mL of a 0.25 M solution of potassium perman-
ganate (KMnO4). The chemist realizes that the only stock solution available is
1 M KMnO4. How can the chemist prepare the desired solution using the stock
solution?
Solution
Step 1: Calculate the number of moles of KMnO4needed to prepare the desired
solution.
Given:
Desired molarity, Mdesired = 0.25 M
Desired volume, Vdesired = 500 mL = 0.5 L
First, calculate the number of moles needed using the formula:
Moles = Mdesired ×Vdesired
Moles = 0.25 mol/L ×0.5 L
Moles = 0.125 mol
The chemist will need 0.125 moles of KMnO4to prepare the desired solution.
Step 2: Calculate the volume of the 1 M stock solution needed to obtain the
required moles of KMnO4.
Given:
Stock molarity, Mstock = 1 M
Use the formula:
Vstock =Moles needed
Mstock
Vstock =0.125 mol
1 mol/L
Vstock = 0.125 L = 125 mL
The chemist should measure out 125 mL of the 1 M KMnO4stock solution
and then dilute it with water to 500 mL to obtain the desired 0.25 M solution.
18
Question 24
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of sulfuric acid (H2SO4).
The chemist has a stock solution of sulfuric acid that is 12 M. How many
milliliters of the 12 M solution should be used to make the desired solution?
Solution
Let Vbe the volume of the 12 M solution needed to make the desired 0.2 M
solution.
Step 1: Calculate the moles of sulfuric acid needed for the desired solution.
First, determine the number of moles of H2SO4required:
moles = Molarity ×Volume (L)
moles = 0.2 mol/L ×0.5 L
moles = 0.1 mol
Step 2: Use the definition of molarity to find the volume of the 12 M
solution required. The formula for molarity is:
M1V1= M2V2
where M is the molarity and V is the volume. Substitute the given values into
the formula:
12 mol/L ×V= 0.1 mol
V=0.1 mol
12 mol/L
V≈0.008 L = 8 mL
Step 3: Convert the volume to milliliters. Since 1 L = 1000 mL, we have:
8 mL = 8 ×1000 mL = 8000 mL
The chemist should use 8000 mL of the 12 M sulfuric acid solution to prepare
500 mL of a 0.2 M solution.
Question 25
Question
A chemist has a stock solution of hydrochloric acid with a molarity of 12.5 M.
How many milliliters of this stock solution must be diluted with water to prepare
500 mL of a 3.0 M hydrochloric acid solution?
19
Solution
Step 1: Let V1be the volume of the stock solution (in mL) and V2be the volume
of water added (in mL) to prepare the final solution.
Step 2: Write the equation based on the principle of conservation of moles:
V1+V2= 500
Step 3: Calculate the number of moles of HCl before and after dilution.
Before dilution: nHCl =Mstock ×V1
After dilution: nHCl =Mfinal ×500
Step 4: Set up the equation for the conservation of moles of HCl:
Mstock ×V1=Mfinal ×500
Step 5: Substitute the given values into the equation:
12.5×V1= 3.0×500
Step 6: Solve for V1:
V1=3.0×500
12.5
Step 7: Calculate the volume of the stock solution V1:
V1=1500
12.5= 120 mL
Answer: The chemist must dilute 120 mL of the stock solution with water
to prepare 500 mL of a 3.0 M hydrochloric acid solution.
Question 26
Question
A chemist wants to prepare 500.0 mL of a solution with a molarity of 0.250 M.
The chemist has a stock solution of the solute that has a molarity of 1.000 M.
How much of the stock solution should the chemist use to prepare the desired
solution?
Solution
Step 1: Calculate the moles of solute needed to prepare the desired solution.
Given: Volume of final solution, Vfinal = 500.0 mL = 0.5000 L Molarity of final
solution, Mfinal = 0.250 M
The formula relating molarity, moles, and volume is:
M=n
V
where Mis the molarity, nis the moles, and Vis the volume.
20
Rearranging the formula to solve for moles of solute:
n=M×V
Substitute the given values:
n= (0.250 M) ×(0.5000 L)
n= 0.1250 moles
Therefore, the chemist needs 0.1250 moles of the solute for the desired solu-
tion.
Step 2: Calculate the volume of stock solution needed. Given: Molarity of
stock solution, Mstock = 1.000 M
The formula relating moles, molarity, and volume is the same:
M=n
V
Solving for volume:
V=n
M
Substitute the moles needed and the molarity of the stock solution:
V=0.1250 moles
1.000 M
V= 0.1250 L = 125.0 mL
Therefore, the chemist should use 125.0 mL of the 1.000 M stock solution to
prepare the desired solution.
Question 27
Question
A solution is prepared by dissolving 15.0 grams of glucose (C6H12O6) in enough
water to make 500.0 mL of solution. Calculate the molarity of the solution.
Solution
Step 1: Calculate the molar mass of glucose. The molar mass of glucose
(C6H12O6) is:
6×molar mass of C+12×molar mass of H+6×molar mass of O = 6(12.01 g/mol)+12(1.008 g/mol)+6(16.00 g/mol)
= 72.06 g/mol + 12.09 g/mol + 96.00 g/mol = 180.15 g/mol
Step 2: Calculate the number of moles of glucose in the solution. Given mass
of glucose = 15.0 g Molar mass of glucose = 180.15 g/mol Number of moles of
glucose = 15.0 g
180.15 g/mol ≈0.0833 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as moles
of solute per liter of solution. Volume of solution = 500.0 mL = 0.500 L Molarity
=0.0833 mol
0.500 L = 0.1666 M
Therefore, the molarity of the solution is 0.1666 M.
21
Question 28
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of potassium perman-
ganate, KMnO4. However, they only have a 0.5 M stock solution of KMnO4
available. How many milliliters of the stock solution should they use to make
the desired solution?
Solution
Step 1: Calculate the moles of KMnO4needed for the desired solution.
Given: Volume of final solution, V = 500 mL = 0.5 L
Molarity of final solution, Mf= 0.2 M
According to the formula for molarity:
Mf=n
V
Solving for moles (n):
n=Mf×V= 0.2 mol/L ×0.5 L = 0.1 mol
Step 2: Calculate the volume of the stock solution needed.
Given: Molarity of stock solution, Ms= 0.5 M
Let the volume of stock solution needed be x mL.
According to the formula for molarity:
Ms=n
V
Solving for volume (V):
V=n
Ms
=0.1 mol
0.5 mol/L = 0.2 L = 200 mL
Therefore, the chemist should use 200 mL of the 0.5 M stock solution to
prepare 500 mL of the 0.2 M solution of potassium permanganate.
Question 29
Question
A chemist prepares a solution by dissolving 10 g of glucose (molar mass = 180
g/mol) in enough water to make 500 mL of solution. Calculate the molarity of
the solution.
22
Solution
Step 1: Calculate the number of moles of glucose. Given: Mass of glucose = 10
g Molar mass of glucose = 180 g/mol
Using the formula:
Number of moles = Mass
Molar mass
we have:
Number of moles = 10 g
180 g/mol = 0.0556 mol
Step 2: Calculate the volume of the solution in liters. Given: Volume of
solution = 500 mL = 0.5 L
Step 3: Calculate the molarity of the solution. Using the formula for molar-
ity:
Molarity = Number of moles solute
Volume of solution in liters
Substitute the values:
Molarity = 0.0556 mol
0.5 L = 0.1112 M
Therefore, the molarity of the solution is 0.1112 M.
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. Calculate the molarity of the solution.
Solution
Step 1: Calculate the molar mass of glucose. The molar mass of glucose
(C6H12O6) can be calculated as follows: M olarM ass = 6 ×M olarM ass(C) +
12 ×MolarM ass(H)+6×M olarM ass(O)
Substitute the molar masses of the elements: MolarMass = 6 ×12.01 +
12 ×1.01 + 6 ×16.00
Calculate the molar mass: MolarM ass = 72.06 + 12.12 + 96.00
Add the values: MolarM ass = 180.18 g/mol
Step 2: Calculate the number of moles of glucose. The number of moles of
glucose can be calculated using the formula: moles =mass
MolarM ass
Substitute the values: moles =15.0
180.18
Calculate the number of moles: moles ≈0.083moles
Step 3: Calculate the molarity of the solution. Molarity is defined as moles
of solute per liter of solution. Molarity =moles
volume(in L)
23
Question 2
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of sodium chloride
(NaCl). If the molar mass of NaCl is 58.44 g/mol, how many grams of NaCl
should the chemist dissolve in water?
Solution
Step 1: Calculate the number of moles of NaCl needed.
Molarity = moles of solute
volume of solution in liters
⇒moles of solute = Molarity ×volume of solution in liters
⇒moles of NaCl = 0.2 mol/L ×0.5 L = 0.1 moles
Step 2: Calculate the mass of NaCl needed.
Mass = moles ×molar mass
⇒Mass of NaCl = 0.1 moles ×58.44 g/mol = 5.844 g
Therefore, the chemist should dissolve 5.844 grams of NaCl in water to
prepare 500 mL of a 0.2 M NaCl solution.
Question 3
Question
A chemist is preparing a solution by mixing 50 mL of a 2 M HCl solution with
100 mL of a 3 M HCl solution. What is the molarity of the final solution?
Solution
Step 1: Find the total moles of HCl in each solution. For the first solution:
0.050L×2mol/L = 0.100mol of HCl
For the second solution: 0.100L×3mol/L = 0.300mol of HCl
Step 2: Find the total moles of HCl in the final solution. 0.100mol +
0.300mol = 0.400mol of HCl
Step 3: Find the total volume of the final solution. 0.050L+0.100L= 0.150L
or 150 mL
Step 4: Calculate the molarity of the final solution using the total moles and
volume. Molarity =0.400mol
0.150L=40
15 M= 2.67Mor 2.67 mol/L
Therefore, the molarity of the final solution is 2.67 M.
2
Question 4
Question
A chemist wants to prepare 500 mL of a 0.2 M hydrochloric acid (HCl) solution.
The chemist has a stock solution of 6 M HCl. How many milliliters of the stock
solution should be used, and how much water should be added to make the
desired solution?
Solution
Step 1: Calculate the moles of HCl 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 (L)
Converting the volume to liters:
500 mL = 500 ×10−3L=0.5 L
Therefore, the moles of HCl needed is:
moles = 0.2 M ×0.5 L = 0.1 moles
Step 2: Calculate the volume of the stock solution needed. Using the equa-
tion:
moles = molarity ×volume (L)
We can rearrange the formula to solve for the volume of the stock solution:
volume (L) = moles
molarity =0.1 moles
6 M
volume (L) = 0.1
6= 0.01667 L = 16.67 mL
Step 3: Calculate the volume of water needed. The total volume of the final
solution is 500 mL. Therefore, the volume of water needed is:
volume of water (mL) = total volume (mL) −volume of stock solution (mL)
volume of water (mL) = 500 mL −16.67 mL = 483.33 mL
Thus, the chemist should add 16.67 mL of the 6 M HCl stock solution and
dilute it with 483.33 mL of water to prepare 500 mL of a 0.2 M HCl solution.
Question 5
Question
A chemist needs to prepare 500 mL of a 0.25 M solution of sulfuric acid, H2SO4.
The chemist has a stock solution of sulfuric acid that is labeled as 1.0 M. De-
termine how many milliliters of the 1.0 M solution should be used to prepare
the desired 0.25 M solution.
3
Solution
Step 1: Calculate the moles of sulfuric acid needed for the desired solution.
Given: Volume of final solution (Vf) = 500 mL = 0.5 L Molarity of final solution
(Mf) = 0.25 M Using the formula:
Moles = Molarity ×Volume (in liters)
we have:
Moles of H2SO4= 0.25 mol/L ×0.5 L = 0.125 mol
Step 2: Determine how many milliliters of the 1.0 M solution contain 0.125
moles of sulfuric acid. Given: Molarity of stock solution (Ms) = 1.0 M By
rearranging the formula, we find:
Volume (in liters) = Moles
Molarity
Volume (in liters) = 0.125 mol
1.0 mol/L = 0.125 L
Converting liters to milliliters:
0.125 L ×1000 mL/L = 125 mL
Answer: The chemist should use 125 mL of the 1.0 M sulfuric acid solution
to prepare 500 mL of a 0.25 M solution.
Question 6
Question
A solution is prepared by dissolving 5.00 g of glucose (C6H12O6) in enough
water to make 250.0 mL of solution.
Part (a)
Calculate the molarity of the glucose solution.
Part (b)
If 20.0 mL of this solution is diluted to 100.0 mL, what will be the molarity of
the diluted solution?
4
Solution
Part (a)
Step 1: Calculate the molar mass of glucose (C6H12O6). The molar mass of
glucose can be calculated by adding the atomic masses of carbon, hydrogen,
and oxygen.
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 ×1.008 g/mol + 6 ×16.00 g/mol
= 180.16 g/mol
Step 2: Calculate the number of moles of glucose in the solution.
Number of moles of glucose = Mass of glucose
Molar mass of glucose =5.00 g
180.16 g/mol = 0.0278 mol
Step 3: Calculate the molarity of the glucose solution.
Molarity = Number of moles of solute
Volume of solution in liters =0.0278 mol
0.2500 L = 0.111 M
Therefore, the molarity of the glucose solution is 0.111 M.
Part (b)
Step 1: Calculate the number of moles of glucose in the 20.0 mL portion of the
solution.
Number of moles in 20.0 mL = Molarity×Volume in liters = 0.111 M×0.0200 L = 0.00222 mol
Step 2: Calculate the molarity of the diluted solution. Since the 20.0 mL
portion is diluted to a total volume of 100.0 mL, we need to calculate the total
number of moles in 100.0 mL before finding the molarity.
Total moles in 100.0 mL = 0.00222 mol
0.0200 L ×0.1000 L = 0.0111 mol
Molarity = 0.0111 mol
0.1000 L = 0.111 M
Therefore, the molarity of the diluted solution is 0.111 M.
Question 7
Question
A solution is prepared by dissolving 10.0 g of glucose (C6H12O6) in enough wa-
ter to make 500.0 mL of solution. What is the molarity of the glucose solution?
(Molar mass of glucose = 180.16 g/mol)
5
Solution
Step 1: Calculate the number of moles of glucose in the solution.
To find the number of moles of glucose, we use the formula:
moles = mass
molar mass
Given: mass of glucose = 10.0 g, molar mass of glucose = 180.16 g/mol,
volume of solution = 500.0 mL.
Substitute the values into the formula:
moles = 10.0 g
180.16 g/mol
moles = 0.0555 mol
Step 2: Calculate the molarity of the solution.
Molarity (M) is defined as the number of moles of solute per liter of solution.
Molarity = moles of solute
liters of solution
Given: moles of glucose = 0.0555 mol, volume of solution = 500.0 mL = 0.5
L.
Substitute the values into the formula:
Molarity = 0.0555 mol
0.5 L
Molarity = 0.111 M
Therefore, the molarity of the glucose solution is 0.111 M.
Question 8
Question
A chemist wants to prepare 500 mL of a 0.1 M solution of hydrochloric acid
(HCl) from a stock solution of 2 M. How many milliliters of the 2 M solution
should be used?
Solution
Step 1: Let Vbe the volume of 2 M solution needed to prepare the desired 0.1
M solution.
Step 2: From the formula for molarity, we have:
M1V1=M2V2
6
Substitute M1= 2 M, V1=VmL, M2= 0.1 M, and V2= 500 mL into the
formula:
2V= 0.1×500
2V= 50
Step 3: Solve for V:
V=50
2
V= 25
Therefore, the chemist should use 25 mL of the 2 M solution to prepare 500
mL of a 0.1 M solution of hydrochloric acid.
Question 9
Question
A chemist prepares a solution by dissolving 5.00 g of sodium chloride (NaCl) in
enough water to make 250.0 mL of solution. What is the molarity of the NaCl
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 = Atomic mass of Na + Atomic mass of Cl
Step 2: Calculate the number of moles of NaCl. Since molarity is defined
as the number of moles of solute per liter of solution, we need to calculate the
number of moles of NaCl.
Number of moles = Mass of solute (g)
Molar mass of solute (g/mol)
Step 3: Calculate the volume of the solution in liters. Since the volume of
the solution is given in milliliters, we need to convert it to liters.
Volume of solution (L) = Volume of solution (mL)
1000
Step 4: Calculate the molarity of the NaCl solution. Molarity (M) is defined
as the number of moles of solute per liter of solution.
M=Number of moles of solute
Volume of solution (L)
7
Question 10
Question
A chemist has a stock solution of hydrochloric acid (HCl) with a concentration
of 10.0 M. The chemist needs to prepare 250 mL of a solution that has a con-
centration of 2.00 M. How much of the stock solution should the chemist dilute
with water to prepare the desired solution?
Solution
Step 1: Let’s calculate the number of moles of hydrochloric acid (HCl) needed to
prepare the desired solution. Given: Initial concentration of stock solution, C1=
10.0 M Volume of stock solution to be used, V1=? (in mL) Final concentration
of desired solution, C2= 2.00 M Volume of desired solution, V2= 250 mL
Step 2: Use the formula for molarity to calculate the moles of HCl in the
desired solution.
C1×V1=C2×V2
10.0 M ×V1mL = 2.00 M ×250 mL
V1=2.00 M ×250 mL
10.0 M
V1= 50.0 mL
Therefore, the chemist should dilute 50.0 mL of the stock solution with water
to prepare the desired 250 mL, 2.00 M solution of hydrochloric acid.
Question 11
Question
A solution is prepared by dissolving 5.00 g of potassium permanganate (KMnO4)
in enough water to make 250.0 mL of solution. What is the molarity of the
potassium permanganate 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.
Molar mass of KMnO4= (39.10 g/mol + 54.94 g/mol + 4(16.00 g/mol)) =
158.04 g/mol
Step 2: Determine the number of moles of KMnO4in the solution. Given
mass of KMnO4= 5.00 g, Number of moles of KMnO4=5.00 g
158.04 g/mol =
0.0316 mol
8
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 = 0.2500 L Molarity M=Number of moles of solute
Volume of solution in liters
Substitute the values into the formula: M=0.0316 mol
0.2500 L = 0.1264 M
Therefore, the molarity of the potassium permanganate solution is 0.1264
M.
Question 12
Question
A student wants to prepare 500 mL of a 0.2 M calcium chloride solution. The
student has a stock solution of calcium chloride (CaCl2) with a concentration
of 1 M. How much of the stock solution should the student use to make the
desired solution?
Solution
Step 1: Calculate the moles of calcium chloride needed to make a 0.2 M solution.
Step 2: Use the formula Molarity =moles solute
volume solution (in liters) to find the volume
of stock solution needed. Step 3: Convert the volume of stock solution from
liters to milliliters for the final answer.
Step 1: Given: - Desired molarity (M1) = 0.2 M - Desired volume (V1) =
500 mL = 0.5 L - Stock molarity (M2) = 1 M
The formula for molarity (M) is: M=moles solute
volume solution (in liters)
Rearranging the formula, we get: moles solute = M x volume solution
Substitute the values into the formula: moles of CaCl2= 0.2 M x 0.5 L
moles of CaCl2= 0.1 moles
Step 2: Now, use the formula for molarity to find the volume of stock solution
needed: M2V2=M1V1
Substitute the known values: 1M ×V2= 0.2M ×0.5L
Solve for V2:V2=0.2mol×0.5L
1mol V2= 0.1L
Step 3: Convert the volume from liters to milliliters: 0.1L ×1000mL/L =
100mL
Therefore, the student should use 100 mL of the stock solution to prepare
500 mL of a 0.2 M calcium chloride solution.
Question 13
Question
A student wants to prepare 500.0 mL of a 0.250 M solution of NaCl. However,
they only have a stock solution of NaCl that is 6.00 M. What volume of the
stock solution should the student use to make the desired solution?
9
Solution
Step 1: Write down the given information.
Volume of desired solution, Vdesired = 500.0 mL = 0.5000 L
Molarity of desired solution, Mdesired = 0.250 M
Molarity of stock solution, Mstock = 6.00 M
Step 2: Use the formula for dilution to find the volume of stock solution
required.
The formula for dilution is:
MstockVstock =MdesiredVdesired
Step 3: Substitute the given values into the formula and solve for Vstock.
6.00 M ×Vstock = 0.250 M ×0.5000 L
Vstock =0.250 ×0.5000
6.00
Vstock = 0.0208 L = 20.8 mL
Answer: The student should use 20.8 mL of the 6.00 M NaCl solution to
prepare 500.0 mL of a 0.250 M NaCl solution.
Question 14
Question
A chemist has a solution of hydrochloric acid with a concentration of 6.0 M.
How many milliliters of this solution must be added to water to make 500 mL
of a 2.0 M solution?
Solution
Step 1: Let’s denote the volume of the 6.0 M solution to be added as VmL.
The final volume of the solution will be 500 mL, and the concentration of the
final solution will be 2.0 M.
Step 2: The amount of hydrochloric acid in the 6.0 M solution before dilution
is given by 6.0 mol/L ×VL. The amount of hydrochloric acid in the final
solution after dilution is given by 2.0 mol/L ×500 mL.
Step 3: Since the amount of hydrochloric acid remains constant before and
after dilution, we can set up the equation:
6.0 mol/L ×VL=2.0 mol/L ×500 mL
Step 4: Let’s convert 500 mL to liters: 500 mL = 0.5 L.
Step 5: Substituting the values into the equation, we get:
6.0V= 2 ×0.5
Step 6: Solve for V:
V=2×0.5
6.0
10
V=1
6L
Step 7: Convert the volume to milliliters:
V=1
6×1000 mL
V= 166.67 mL
So, approximately 166.67 mL of the 6.0 M solution must be added to water
to make 500 mL of a 2.0 M solution.
Question 15
Question
A chemist prepares a solution 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?
Solution
Step 1: Calculate the molar mass of glucose. The molar mass of glucose
(C6H12O6) can be calculated by adding the atomic masses of each element
present: 6 carbons, 12 hydrogens, and 6 oxygens.
Molar mass of C6H12O6= 6(Atomic mass of C)+12(Atomic mass of H)+6(Atomic mass of 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 in the solution. First,
convert the mass of glucose to moles using the formula:
moles = mass (g)
molar mass (g/mol)
moles of glucose = 10.0 g
180.18 g/mol ≈0.0554 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as moles
of solute per liter of solution:
Molarity = moles of solute
volume of solution (L)
Since the volume of the solution is 500.0 mL, we need to convert it to liters:
Volume of solution = 500.0 mL ×1 L
1000 mL = 0.500 L
11
Now calculate the molarity of the glucose solution:
Molarity = 0.0554 mol
0.500 L = 0.111 M
Therefore, the molarity of the glucose solution is 0.111 M.
Question 16
Question
A chemist prepares a solution by dissolving 25.0 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. The molar mass of glucose
(C6H12O6) can be calculated by summing the atomic masses of the elements
present:
Molar mass of glucose = 6×Molar mass of C+12×Molar mass of H+6×Molar mass of O
The molar masses are: - Molar mass of C: 12.01 g/mol - Molar mass of H:
1.008 g/mol - Molar mass of O: 16.00 g/mol
Plugging the values into the formula:
Molar mass of glucose = 6(12.01) + 12(1.008) + 6(16.00)
= 72.06 + 12.096 + 96.00
= 180.156 g/mol
Step 2: Calculate the number of moles of glucose in the solution. Given that
25.0 g of glucose is dissolved in the solution, we can calculate the number of
moles of glucose using the formula:
Number of moles = Mass
Molar mass
Substitute the values:
Number of moles = 25.0
180.156
Number of moles ≈0.1387 mol
Step 3: Calculate the molarity of the solution. The molarity of a solution is
defined as the number of moles of solute per liter of solution.
12
Given that the final volume of the solution is 500.0 mL (0.5000 L), we can
calculate the molarity using the formula:
Molarity = Number of moles
Volume (L)
Substitute the values:
Molarity = 0.1387
0.5000
Molarity = 0.277 M
Therefore, the molarity of the glucose solution is 0.277 M.
Question 17
Question
A chemist wants to prepare 500 mL of a 0.25 M sulfuric acid (H2SO4) solution.
However, the only sulfuric acid available is a concentrated solution that is 98
Solution
Step 1: Determine the molar mass of H2SO4.
Molar mass(H2SO4)=2×molar mass(H)+1×molar mass(S) + 4 ×molar mass(O)
= 2(1.01 g/mol) + 1(32.07 g/mol) + 4(16.00 g/mol)
= 98.09 g/mol
Step 2: Calculate the molarity of the concentrated sulfuric acid solution.
Molarity = moles of solute
volume of solution in liters
Moles of H2SO4=mass of H2SO4
molar mass(H2SO4)
=0.98 ×500 g
98.09 g/mol
= 5 mol
Volume of concentrated solution (V) = moles of solute
molarity
=5 mol
0.98 M
= 5.10 L (5100 mL)
Therefore, the chemist should use 5100 mL of the concentrated sulfuric acid
solution to make 500 mL of a 0.25 M sulfuric acid solution.
13
Question 18
Question
A student wants to prepare 500 mL of a 0.2 M solution of potassium perman-
ganate (KMnO4). The student has a stock solution of KMnO4with a concen-
tration of 1.0 M. How many milliliters of the stock solution should the student
use to prepare the desired solution?
Solution
Step 1: Determine the number of moles required to prepare the desired solution.
Given: Volume of solution (V) = 500 mL = 0.5 L Molarity of solution (M) =
0.2 M
The formula for calculating the number of moles is:
moles = Molarity ×Volume (in L)
Substitute the given values:
moles = 0.2 mol/L ×0.5 L = 0.1 mol
Step 2: Calculate the volume of the stock solution needed. Given: Molarity
of stock solution = 1.0 M
The formula for calculating the volume of stock solution needed is:
Volume (stock solution) = moles
Molarity of stock solution
Substitute the values calculated in Step 1:
Volume (stock solution) = 0.1 mol
1.0 mol/L = 0.1 L = 100 mL
Therefore, the student should use 100 mL of the 1.0 M KMnO4stock solution
to prepare 500 mL of a 0.2 M KMnO4solution.
Question 19
Question
A chemist needs to prepare 500 mL of a 0.2 M solution of sulfuric acid, H2SO4.
They have a stock solution of sulfuric acid with a concentration of 6 M. How
many milliliters of the stock solution should the chemist use to make the desired
solution?
14
Solution
Step 1: Calculate the moles of sulfuric acid needed for the desired solution.
Given that the chemist needs to prepare 500 mL of a 0.2 M solution of sulfuric
acid, the moles of sulfuric acid required can be calculated as:
Moles = Molarity ×Volume (L)
Moles = 0.2 mol/L ×0.5 L
Moles = 0.1 mol
Step 2: Calculate the volume of the stock solution required. Since the stock
solution has a concentration of 6 M, the volume of the stock solution needed
can be calculated using the formula:
Moles = Molarity ×Volume (L)
0.1 mol = 6 mol/L ×Volume (L)
Volume (L) = 0.1 mol
6 mol/L
Volume (L) = 0.0167 L
Step 3: Convert the volume to milliliters. To find the volume in milliliters,
we can multiply the volume in liters by 1000:
Volume (mL) = 0.0167 L ×1000
Volume (mL) = 16.7 mL
The chemist should use 16.7 mL of the 6 M stock solution to prepare 500
mL of a 0.2 M solution of sulfuric acid.
Question 20
Question
A chemist has a solution of sulfuric acid with a molarity of 2.5 M. She wants
to dilute this solution to make 500 mL of a new solution with a molarity of 1.5
M. What volume of the original 2.5 M sulfuric acid solution should she use to
make the new solution?
15
Solution
Step 1: Let the volume of the original 2.5 M sulfuric acid solution to be used
be VmL. We can set up the dilution equation as follows:
M1V1=M2V2
where M1= 2.5 M, V1=VmL, M2= 1.5 M, and V2= 500 mL = 0.5 L.
Step 2: Plug in the values and solve for V:
(2.5 M)(VL) = (1.5 M)(0.5 L)
2.5V= 0.75
V=0.75
2.5= 0.3 L = 300 mL
So, the chemist should use 300 mL of the original 2.5 M sulfuric acid solution
to make the new solution.
Question 21
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?
(The molar mass of glucose is 180.16 g/mol.)
Solution
Step 1: Calculate the number of moles of glucose in the solution. Given: Mass
of glucose = 10.0 g Molar mass of glucose = 180.16 g/mol
Use the formula:
moles of solute = mass of solute
molar mass of solute
Plugging in the values:
moles of solute = 10.0 g
180.16 g/mol
moles of solute = 0.0555 mol
Step 2: Calculate the molarity of the solution. Given: Volume of solution =
500.0 mL = 0.5000 L
Use the formula:
Molarity(M) = moles of solute
volume of solution in liters
16
Plugging in the values:
M=0.0555 mol
0.5000 L
M= 0.111 M
Therefore, the molarity of the glucose solution is 0.111 M.
Question 22
Question
A solution is prepared by mixing 50.0 mL of a 0.200 M potassium iodide (KI)
solution with 100.0 mL of a 0.500 M lead(II) nitrate (Pb(NO3)2) solution. Cal-
culate the molarity of potassium ions (K+) and iodide ions (I−) in the final
solution.
Solution
Step 1: Calculate the moles of potassium iodide (KI) and lead(II) nitrate
(Pb(NO3)2) used.
The moles of KI can be calculated as:
moles of KI = Molarity ×Volume (L)
Given that the volume of KI solution is 50.0 mL and the molarity is 0.200
M:
moles of KI = 0.200 mol/L ×0.0500 L = 0.0100 mol
The moles of Pb(NO3)2can be calculated in a similar manner:
moles of Pb(NO3)2)=0.500 mol/L ×0.100 L = 0.0500 mol
Step 2: Determine the limiting reactant.
From the balanced chemical equation for the reaction between KI and Pb(NO3)2,
we see that a 1:1 molar ratio exists between KI and Pb(NO3)2. Therefore, the
limiting reactant will be the one with fewer moles.
Since 0.0100 mol of KI and 0.0500 mol of Pb(NO3)2are used, KI is the
limiting reactant.
Step 3: Calculate the moles of potassium ions (K+) and iodide ions (I−) in
the final solution.
The molarity of potassium ions (K+) in the final solution can be calculated
using the total volume of the final solution (150.0 mL or 0.1500 L):
Molarity of K+in the final solution = moles of K+
Volume of solution (L)
Since KI provides 1 mol of K+ions for every 1 mol of KI:
moles of K+= 0.0100 mol
17
Thus, the molarity of K+in the final solution is:
Molarity of K+in the final solution = 0.0100 mol
0.1500 L = 0.067 M
Similarly, the molarity of iodide ions (I−) in the final solution can be calcu-
lated as:
Molarity of I−in the final solution = 0.0100 mol
0.1500 L = 0.067 M
Question 23
Question
A chemist wants to prepare 500 mL of a 0.25 M solution of potassium perman-
ganate (KMnO4). The chemist realizes that the only stock solution available is
1 M KMnO4. How can the chemist prepare the desired solution using the stock
solution?
Solution
Step 1: Calculate the number of moles of KMnO4needed to prepare the desired
solution.
Given:
Desired molarity, Mdesired = 0.25 M
Desired volume, Vdesired = 500 mL = 0.5 L
First, calculate the number of moles needed using the formula:
Moles = Mdesired ×Vdesired
Moles = 0.25 mol/L ×0.5 L
Moles = 0.125 mol
The chemist will need 0.125 moles of KMnO4to prepare the desired solution.
Step 2: Calculate the volume of the 1 M stock solution needed to obtain the
required moles of KMnO4.
Given:
Stock molarity, Mstock = 1 M
Use the formula:
Vstock =Moles needed
Mstock
Vstock =0.125 mol
1 mol/L
Vstock = 0.125 L = 125 mL
The chemist should measure out 125 mL of the 1 M KMnO4stock solution
and then dilute it with water to 500 mL to obtain the desired 0.25 M solution.
18
Question 24
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of sulfuric acid (H2SO4).
The chemist has a stock solution of sulfuric acid that is 12 M. How many
milliliters of the 12 M solution should be used to make the desired solution?
Solution
Let Vbe the volume of the 12 M solution needed to make the desired 0.2 M
solution.
Step 1: Calculate the moles of sulfuric acid needed for the desired solution.
First, determine the number of moles of H2SO4required:
moles = Molarity ×Volume (L)
moles = 0.2 mol/L ×0.5 L
moles = 0.1 mol
Step 2: Use the definition of molarity to find the volume of the 12 M
solution required. The formula for molarity is:
M1V1= M2V2
where M is the molarity and V is the volume. Substitute the given values into
the formula:
12 mol/L ×V= 0.1 mol
V=0.1 mol
12 mol/L
V≈0.008 L = 8 mL
Step 3: Convert the volume to milliliters. Since 1 L = 1000 mL, we have:
8 mL = 8 ×1000 mL = 8000 mL
The chemist should use 8000 mL of the 12 M sulfuric acid solution to prepare
500 mL of a 0.2 M solution.
Question 25
Question
A chemist has a stock solution of hydrochloric acid with a molarity of 12.5 M.
How many milliliters of this stock solution must be diluted with water to prepare
500 mL of a 3.0 M hydrochloric acid solution?
19
Solution
Step 1: Let V1be the volume of the stock solution (in mL) and V2be the volume
of water added (in mL) to prepare the final solution.
Step 2: Write the equation based on the principle of conservation of moles:
V1+V2= 500
Step 3: Calculate the number of moles of HCl before and after dilution.
Before dilution: nHCl =Mstock ×V1
After dilution: nHCl =Mfinal ×500
Step 4: Set up the equation for the conservation of moles of HCl:
Mstock ×V1=Mfinal ×500
Step 5: Substitute the given values into the equation:
12.5×V1= 3.0×500
Step 6: Solve for V1:
V1=3.0×500
12.5
Step 7: Calculate the volume of the stock solution V1:
V1=1500
12.5= 120 mL
Answer: The chemist must dilute 120 mL of the stock solution with water
to prepare 500 mL of a 3.0 M hydrochloric acid solution.
Question 26
Question
A chemist wants to prepare 500.0 mL of a solution with a molarity of 0.250 M.
The chemist has a stock solution of the solute that has a molarity of 1.000 M.
How much of the stock solution should the chemist use to prepare the desired
solution?
Solution
Step 1: Calculate the moles of solute needed to prepare the desired solution.
Given: Volume of final solution, Vfinal = 500.0 mL = 0.5000 L Molarity of final
solution, Mfinal = 0.250 M
The formula relating molarity, moles, and volume is:
M=n
V
where Mis the molarity, nis the moles, and Vis the volume.
20
Rearranging the formula to solve for moles of solute:
n=M×V
Substitute the given values:
n= (0.250 M) ×(0.5000 L)
n= 0.1250 moles
Therefore, the chemist needs 0.1250 moles of the solute for the desired solu-
tion.
Step 2: Calculate the volume of stock solution needed. Given: Molarity of
stock solution, Mstock = 1.000 M
The formula relating moles, molarity, and volume is the same:
M=n
V
Solving for volume:
V=n
M
Substitute the moles needed and the molarity of the stock solution:
V=0.1250 moles
1.000 M
V= 0.1250 L = 125.0 mL
Therefore, the chemist should use 125.0 mL of the 1.000 M stock solution to
prepare the desired solution.
Question 27
Question
A solution is prepared by dissolving 15.0 grams of glucose (C6H12O6) in enough
water to make 500.0 mL of solution. Calculate the molarity of the solution.
Solution
Step 1: Calculate the molar mass of glucose. The molar mass of glucose
(C6H12O6) is:
6×molar mass of C+12×molar mass of H+6×molar mass of O = 6(12.01 g/mol)+12(1.008 g/mol)+6(16.00 g/mol)
= 72.06 g/mol + 12.09 g/mol + 96.00 g/mol = 180.15 g/mol
Step 2: Calculate the number of moles of glucose in the solution. Given mass
of glucose = 15.0 g Molar mass of glucose = 180.15 g/mol Number of moles of
glucose = 15.0 g
180.15 g/mol ≈0.0833 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as moles
of solute per liter of solution. Volume of solution = 500.0 mL = 0.500 L Molarity
=0.0833 mol
0.500 L = 0.1666 M
Therefore, the molarity of the solution is 0.1666 M.
21
Question 28
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of potassium perman-
ganate, KMnO4. However, they only have a 0.5 M stock solution of KMnO4
available. How many milliliters of the stock solution should they use to make
the desired solution?
Solution
Step 1: Calculate the moles of KMnO4needed for the desired solution.
Given: Volume of final solution, V = 500 mL = 0.5 L
Molarity of final solution, Mf= 0.2 M
According to the formula for molarity:
Mf=n
V
Solving for moles (n):
n=Mf×V= 0.2 mol/L ×0.5 L = 0.1 mol
Step 2: Calculate the volume of the stock solution needed.
Given: Molarity of stock solution, Ms= 0.5 M
Let the volume of stock solution needed be x mL.
According to the formula for molarity:
Ms=n
V
Solving for volume (V):
V=n
Ms
=0.1 mol
0.5 mol/L = 0.2 L = 200 mL
Therefore, the chemist should use 200 mL of the 0.5 M stock solution to
prepare 500 mL of the 0.2 M solution of potassium permanganate.
Question 29
Question
A chemist prepares a solution by dissolving 10 g of glucose (molar mass = 180
g/mol) in enough water to make 500 mL of solution. Calculate the molarity of
the solution.
22
Solution
Step 1: Calculate the number of moles of glucose. Given: Mass of glucose = 10
g Molar mass of glucose = 180 g/mol
Using the formula:
Number of moles = Mass
Molar mass
we have:
Number of moles = 10 g
180 g/mol = 0.0556 mol
Step 2: Calculate the volume of the solution in liters. Given: Volume of
solution = 500 mL = 0.5 L
Step 3: Calculate the molarity of the solution. Using the formula for molar-
ity:
Molarity = Number of moles solute
Volume of solution in liters
Substitute the values:
Molarity = 0.0556 mol
0.5 L = 0.1112 M
Therefore, the molarity of the solution is 0.1112 M.
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. Calculate the molarity of the solution.
Solution
Step 1: Calculate the molar mass of glucose. The molar mass of glucose
(C6H12O6) can be calculated as follows: M olarM ass = 6 ×M olarM ass(C) +
12 ×MolarM ass(H)+6×M olarM ass(O)
Substitute the molar masses of the elements: MolarMass = 6 ×12.01 +
12 ×1.01 + 6 ×16.00
Calculate the molar mass: MolarM ass = 72.06 + 12.12 + 96.00
Add the values: MolarM ass = 180.18 g/mol
Step 2: Calculate the number of moles of glucose. The number of moles of
glucose can be calculated using the formula: moles =mass
MolarM ass
Substitute the values: moles =15.0
180.18
Calculate the number of moles: moles ≈0.083moles
Step 3: Calculate the molarity of the solution. Molarity is defined as moles
of solute per liter of solution. Molarity =moles
volume(in L)
23
Question 24
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of sulfuric acid (H2SO4).
The chemist has a stock solution of sulfuric acid that is 12 M. How many
milliliters of the 12 M solution should be used to make the desired solution?
Solution
Let Vbe the volume of the 12 M solution needed to make the desired 0.2 M
solution.
Step 1: Calculate the moles of sulfuric acid needed for the desired solution.
First, determine the number of moles of H2SO4required:
moles = Molarity ×Volume (L)
moles = 0.2 mol/L ×0.5 L
moles = 0.1 mol
Step 2: Use the definition of molarity to find the volume of the 12 M
solution required. The formula for molarity is:
M1V1= M2V2
where M is the molarity and V is the volume. Substitute the given values into
the formula:
12 mol/L ×V= 0.1 mol
V=0.1 mol
12 mol/L
V≈0.008 L = 8 mL
Step 3: Convert the volume to milliliters. Since 1 L = 1000 mL, we have:
8 mL = 8 ×1000 mL = 8000 mL
The chemist should use 8000 mL of the 12 M sulfuric acid solution to prepare
500 mL of a 0.2 M solution.
Question 25
Question
A chemist has a stock solution of hydrochloric acid with a molarity of 12.5 M.
How many milliliters of this stock solution must be diluted with water to prepare
500 mL of a 3.0 M hydrochloric acid solution?
19
Solution
Step 1: Let V1be the volume of the stock solution (in mL) and V2be the volume
of water added (in mL) to prepare the final solution.
Step 2: Write the equation based on the principle of conservation of moles:
V1+V2= 500
Step 3: Calculate the number of moles of HCl before and after dilution.
Before dilution: nHCl =Mstock ×V1
After dilution: nHCl =Mfinal ×500
Step 4: Set up the equation for the conservation of moles of HCl:
Mstock ×V1=Mfinal ×500
Step 5: Substitute the given values into the equation:
12.5×V1= 3.0×500
Step 6: Solve for V1:
V1=3.0×500
12.5
Step 7: Calculate the volume of the stock solution V1:
V1=1500
12.5= 120 mL
Answer: The chemist must dilute 120 mL of the stock solution with water
to prepare 500 mL of a 3.0 M hydrochloric acid solution.
Question 26
Question
A chemist wants to prepare 500.0 mL of a solution with a molarity of 0.250 M.
The chemist has a stock solution of the solute that has a molarity of 1.000 M.
How much of the stock solution should the chemist use to prepare the desired
solution?
Solution
Step 1: Calculate the moles of solute needed to prepare the desired solution.
Given: Volume of final solution, Vfinal = 500.0 mL = 0.5000 L Molarity of final
solution, Mfinal = 0.250 M
The formula relating molarity, moles, and volume is:
M=n
V
where Mis the molarity, nis the moles, and Vis the volume.
20
Rearranging the formula to solve for moles of solute:
n=M×V
Substitute the given values:
n= (0.250 M) ×(0.5000 L)
n= 0.1250 moles
Therefore, the chemist needs 0.1250 moles of the solute for the desired solu-
tion.
Step 2: Calculate the volume of stock solution needed. Given: Molarity of
stock solution, Mstock = 1.000 M
The formula relating moles, molarity, and volume is the same:
M=n
V
Solving for volume:
V=n
M
Substitute the moles needed and the molarity of the stock solution:
V=0.1250 moles
1.000 M
V= 0.1250 L = 125.0 mL
Therefore, the chemist should use 125.0 mL of the 1.000 M stock solution to
prepare the desired solution.
Question 27
Question
A solution is prepared by dissolving 15.0 grams of glucose (C6H12O6) in enough
water to make 500.0 mL of solution. Calculate the molarity of the solution.
Solution
Step 1: Calculate the molar mass of glucose. The molar mass of glucose
(C6H12O6) is:
6×molar mass of C+12×molar mass of H+6×molar mass of O = 6(12.01 g/mol)+12(1.008 g/mol)+6(16.00 g/mol)
= 72.06 g/mol + 12.09 g/mol + 96.00 g/mol = 180.15 g/mol
Step 2: Calculate the number of moles of glucose in the solution. Given mass
of glucose = 15.0 g Molar mass of glucose = 180.15 g/mol Number of moles of
glucose = 15.0 g
180.15 g/mol ≈0.0833 mol
Step 3: Calculate the molarity of the solution. Molarity is defined as moles
of solute per liter of solution. Volume of solution = 500.0 mL = 0.500 L Molarity
=0.0833 mol
0.500 L = 0.1666 M
Therefore, the molarity of the solution is 0.1666 M.
21
Question 28
Question
A chemist wants to prepare 500 mL of a 0.2 M solution of potassium perman-
ganate, KMnO4. However, they only have a 0.5 M stock solution of KMnO4
available. How many milliliters of the stock solution should they use to make
the desired solution?
Solution
Step 1: Calculate the moles of KMnO4needed for the desired solution.
Given: Volume of final solution, V = 500 mL = 0.5 L
Molarity of final solution, Mf= 0.2 M
According to the formula for molarity:
Mf=n
V
Solving for moles (n):
n=Mf×V= 0.2 mol/L ×0.5 L = 0.1 mol
Step 2: Calculate the volume of the stock solution needed.
Given: Molarity of stock solution, Ms= 0.5 M
Let the volume of stock solution needed be x mL.
According to the formula for molarity:
Ms=n
V
Solving for volume (V):
V=n
Ms
=0.1 mol
0.5 mol/L = 0.2 L = 200 mL
Therefore, the chemist should use 200 mL of the 0.5 M stock solution to
prepare 500 mL of the 0.2 M solution of potassium permanganate.
Question 29
Question
A chemist prepares a solution by dissolving 10 g of glucose (molar mass = 180
g/mol) in enough water to make 500 mL of solution. Calculate the molarity of
the solution.
22
Solution
Step 1: Calculate the number of moles of glucose. Given: Mass of glucose = 10
g Molar mass of glucose = 180 g/mol
Using the formula:
Number of moles = Mass
Molar mass
we have:
Number of moles = 10 g
180 g/mol = 0.0556 mol
Step 2: Calculate the volume of the solution in liters. Given: Volume of
solution = 500 mL = 0.5 L
Step 3: Calculate the molarity of the solution. Using the formula for molar-
ity:
Molarity = Number of moles solute
Volume of solution in liters
Substitute the values:
Molarity = 0.0556 mol
0.5 L = 0.1112 M
Therefore, the molarity of the solution is 0.1112 M.
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. Calculate the molarity of the solution.
Solution
Step 1: Calculate the molar mass of glucose. The molar mass of glucose
(C6H12O6) can be calculated as follows: M olarM ass = 6 ×M olarM ass(C) +
12 ×MolarM ass(H)+6×M olarM ass(O)
Substitute the molar masses of the elements: MolarMass = 6 ×12.01 +
12 ×1.01 + 6 ×16.00
Calculate the molar mass: MolarM ass = 72.06 + 12.12 + 96.00
Add the values: MolarM ass = 180.18 g/mol
Step 2: Calculate the number of moles of glucose. The number of moles of
glucose can be calculated using the formula: moles =mass
MolarM ass
Substitute the values: moles =15.0
180.18
Calculate the number of moles: moles ≈0.083moles
Step 3: Calculate the molarity of the solution. Molarity is defined as moles
of solute per liter of solution. Molarity =moles
volume(in L)
23
Convert the volume from mL to L: 250.0mL = 0.250 L
Now, substitute the values: Molarity =0.083
0.250
Calculate the molarity: Molarity = 0.332 M
Therefore, the molarity of the solution is 0.332 M.
24
