CHEM 301 - ORGANIC CHEMISTRY I Solubility and polarity calculations Question Bank Set 1

CHEM 301 - ORGANIC CHEMISTRY

I - Solubility and polarity calculations

Question Bank - Set 1

Liberty University

Question 1

Question

An organic molecule, A, has a molar mass of 120 g/mol and contains only

carbon, hydrogen, and oxygen. When 0.1 g of A is dissolved in 5.0 g of water

at 25

C, it forms a solution that is saturate but not all of A dissolves. It is

found that 0.02 g of A remains undissolved at equilibrium. Calculate the molar

solubility of compound A in water at 25

Solution

Step 1: Determine the moles of compound A that dissolved in water. Step 2:

Calculate the molar solubility of compound A in water at 25

Question 2

Question

A student is performing a solubility experiment in which they dissolve 2.5 g

of compound X in 50 mL of water at 25

C. After mixing, the solution appears

cloudy and no dissolution is observed. The student then decides to add 10 mL

of ethanol to the solution and upon mixing, the compound completely dissolves.

Calculate the solubility of compound X in water at 25

C and explain the effect

of adding ethanol to the solution.

Solution

Step 1: Calculate the solubility of compound X in water at 25

Given: - Mass of compound X = 2.5 g - Volume of water = 50 mL = 0.05 L

First, we need to calculate the concentration of the saturated solution in

water:

Concentration of X in water = Mass of X

Volume of water

Concentration of X in water = 2.5 g

0.05 L = 50 g/L

Therefore, the solubility of compound X in water at 25

C is 50 g/L.

Step 2: Explain the effect of adding ethanol to the solution.

When the compound did not dissolve in water alone but dissolved upon the

addition of ethanol, it indicates that compound X is more soluble in ethanol than

in water. Ethanol likely interacts favorably with the compound, possibly due

to similar polarity or hydrogen bonding capabilities. The addition of ethanol

provided a more suitable solvent environment for compound X, allowing it to

dissolve completely.

Question 3

Question

A student needs to determine the solubility of Compound X in different solvents.

The student knows that Compound X is a polar, nonpolar solvent insoluble

compound. The student decides to test Compound X in three solvents: water,

ethanol, and hexanes. Predict the solubility of Compound X in each solvent

based on its polarity, and provide a brief explanation for each prediction.

Solution

Step 1: **Water**: Compound X is a polar, nonpolar solvent insoluble com-

pound. Water is a highly polar solvent due to its ability to form hydrogen

bonds. Since Compound X is not expected to be soluble in polar solvents, it is

predicted to be insoluble in water.

Step 2: **Ethanol**: Ethanol is also a polar solvent, but it is less polar

compared to water. It can still form hydrogen bonds with polar compounds.

Based on its medium polarity, Compound X is likely to have limited solubility in

ethanol. Therefore, Compound X is predicted to be sparingly soluble in ethanol.

Step 3: **Hexanes**: Hexanes is a nonpolar solvent that cannot form hy-

drogen bonds with polar molecules. Since Compound X is a nonpolar solvent

insoluble compound, it is predicted to be insoluble in hexanes.

In summary, Compound X is predicted to be insoluble in water and hexanes,

and sparingly soluble in ethanol because of its polar, nonpolar solvent insoluble

nature.

Question 4

Question

An unknown compound is found to be soluble in water but insoluble in hex-

ane. When the compound is dissolved in water, the solution is found to be

acidic. Determine the most likely functional group present in the compound,

and explain your reasoning.

Solution

Step 1: Since the compound is soluble in water but insoluble in hexane, it

indicates the presence of a polar functional group that can form hydrogen bonds

with water molecules. This rules out nonpolar functional groups like alkyl chains

and aromatic rings.

Step 2: The fact that the compound makes the water solution acidic suggests

the presence of a functional group that can release protons (H+) in water. This

narrows down the possibilities to acidic functional groups like carboxylic acids,

phenols, and sulfonic acids.

Step 3: Phenols and sulfonic acids are relatively weaker acids compared to

carboxylic acids. Given that the compound is able to make the solution acidic,

it is more likely to be a carboxylic acid.

Step 4: Therefore, the most likely functional group present in the compound

is a carboxylic acid group (-COOH). This functional group is polar, can form

hydrogen bonds with water, and is acidic, which matches the experimental ob-

servations.

Question 5

Question

Calculate the solubility of benzoic acid (C7H6O2) in water at 25◦C. Given that

the solubility of benzoic acid in water is 3.4 g/L at 25◦C and the molar mass of

benzoic acid is 122.12 g/mol.

Solution

Step 1: Calculate the molar solubility of benzoic acid in water. The molar

solubility (x) can be calculated using the formula:

x=solubility (g/L)

molar mass (g/mol)

Substitute the given values:

x=3.4 g/L

122.12 g/mol

Step 2: Calculate the molar solubility of benzoic acid in water.

x≈0.0279 mol/L

Therefore, the molar solubility of benzoic acid in water at 25◦C is approxi-

mately 0.0279 mol/L.

Question 6

Question

A student is attempting to dissolve a compound in water in order to determine

its solubility. The compound has a molecular formula of C10H20O4. After

multiple attempts, the student finds that only 2.0 grams of the compound can

be dissolved in 100 mL of water at a given temperature. Calculate the solubility

of the compound in water in units of g/mL.

Solution

Step 1: Calculate the molar mass of the compound.

The molar mass of the compound can be calculated by summing the atomic

masses of each element present in the compound.

Molar Mass = 10(Atomic Mass of C)+20(Atomic Mass of H)+4(Atomic Mass of O)

Step 2: Determine the molar solubility of the compound.

Molar solubility represents the number of moles that can be dissolved in 1 L of

solvent. We can calculate it using the mass of the compound dissolving and its

molar mass.

Molar Solubility = Mass of Compound

Molar Mass of Compound

Step 3: Convert the molar solubility to grams per milliliter.

To convert the molar solubility to grams per milliliter, we need to take into

account that the student was able to dissolve the compound in 100 mL of water.

Solubility (g/mL) = Molar Solubility (g/L)

1000

Question 7

Question

Determine which of the following pairs of compounds would be expected to have

greater solubility in water based on their polarity:

1. Hexane (C6H14) and Ethanol (C2H5OH)

2. Diethyl Ether (C4H10O) and Acetic Acid (CH3COOH)

Solution

Step 1: Determine the polarity of the compounds.



Hexane (C6H14): Hexane is a nonpolar molecule composed of carbon and

hydrogen atoms with similar electronegativity values. It has London dis-

persion forces but lacks dipole-dipole interactions or hydrogen bonding.



Ethanol (C2H5OH): Ethanol is a polar molecule with a hydroxyl (OH)

group that creates a dipole moment. It exhibits hydrogen bonding due to

the presence of the O-H bond.



Diethyl Ether (C4H10O): Diethyl Ether has a polar C-O bond but is over-

all a nonpolar molecule due to the symmetrical arrangement of its alkyl

groups. It does not exhibit hydrogen bonding.



Acetic Acid (CH3COOH): Acetic Acid is a polar molecule with a carbonyl

group (C=O) and a hydroxyl group (OH), both of which contribute to its

polarity and ability to form hydrogen bonds.

Step 2: Compare the polarity and potential for hydrogen bonding.



Hexane and Ethanol: Ethanol is polar and can form hydrogen bonds

with water molecules, while hexane is nonpolar and cannot form hydrogen

bonds. Therefore, Ethanol is expected to have greater solubility in water

than hexane.



Diethyl Ether and Acetic Acid: Acetic Acid is polar and can form hy-

drogen bonds with water, while diethyl ether is mostly nonpolar and can-

not form hydrogen bonds. Thus, Acetic Acid would be expected to have

greater solubility in water than diethyl ether.

Therefore,

1. Ethanol (C2H5OH) is expected to have greater solubility in water than

Hexane (C6H14).

2. Acetic Acid (CH3COOH) is expected to have greater solubility in water

than Diethyl Ether (C4H10O).

Question 8

Question

A student is given a sample of a compound and is asked to determine if it is

soluble in water. The compound has the following structure (shown below) and

the student needs to calculate the solubility in grams per liter. Assume the

compound is not ionized in water.

Compound structure: H3C−CH2−CH2−CH(−[6]OH)−C(= [1]O)−[7]H

Given the molar mass of the compound is 102 g/mol, calculate the solubility

of this compound in water in grams per liter at 25

Solution

Step 1: Determine the polarity of the compound. The compound contains

both a polar (-OH) and nonpolar (hydrocarbon chain) component. Overall, the

compound is polar due to the presence of the hydroxyl group. This suggests

that the compound may be soluble in water.

Step 2: Calculate the molar solubility of the compound. To calculate molar

solubility, we need to consider the molar mass of the compound. The molar

solubility in moles per liter (mol/L) can be calculated using the formula:

Molar solubility (mol/L) = Molar mass (g/mol)

Grams/Liter

Given the molar mass of the compound is 102 g/mol, the molar solubility

can be calculated as:

Molar solubility = 102 g/mol

102 g/L = 1 mol/L

Step 3: Convert molar solubility to grams per liter. To calculate the solu-

bility in grams per liter, we need to consider the molar mass of the compound.

The solubility in grams per liter can be calculated using the formula:

Solubility (g/L) = Molar solubility (mol/L) ×Molar mass (g/mol)

Substitute the values to find the solubility in grams per liter:

Solubility = 1 mol/L ×102 g/mol = 102 g/L

Therefore, the solubility of the compound in water at 25

C is 102 grams per

liter.

Question 9

Question

Calculate the solubility of benzoic acid (C7H6O2) in water at 25◦C. The Ksp of

benzoic acid is 6.97 ×10−5at this temperature.

Solution

Step 1: Write the dissociation equation for benzoic acid:

C7H6O2⇌C7H6O2+ H2O

Step 2: Write the expression for the solubility product constant (Ksp) for

the dissociation of benzoic acid:

Ksp = [C7H6O2][H2O]

Step 3: Let x be the molar solubility of benzoic acid. At equilibrium, the

concentration of benzoic acid and hydrogen ions will be x, while the concentra-

tion of hydroxide ions will be x (since benzoic acid is a weak acid).

Step 4: Substitute x into the Ksp expression and solve for x:

6.97 ×10−5=x·x=x2

x=p6.97 ×10−5= 0.00835 M

Therefore, the solubility of benzoic acid in water at 25◦C is 0.00835 M.

Question 10

Question

An organic compound has the following chemical structure:

CH3−CH2−OH

Predict whether this compound is soluble in water, based on its structure.

Justify your answer.

Solution

To determine the solubility of a compound in water, we must consider its po-

larity. Water is a polar solvent, therefore compounds with polar groups tend to

be soluble in water.

Step 1: Analyze the structure of the compound.

The compound is ethanol (CH3CH2OH), which consists of a nonpolar hy-

drocarbon chain (CH3CH2) and a polar hydroxyl group (-OH).

Step 2: Determine the polarity of the compound.

The hydroxyl group (-OH) is a polar functional group due to the electroneg-

ative oxygen atom creating a polar covalent bond with the hydrogen atom.

Step 3: Predict the compound’s solubility in water.

Since ethanol contains a polar hydroxyl group, it can form hydrogen bonds

with water molecules. Therefore, ethanol is soluble in water.

Step 4: Justification

Ethanol is soluble in water due to the polar hydroxyl group (-OH) that can

interact with the polar water molecules through hydrogen bonding. This polar

interaction overcomes the nonpolar hydrocarbon chain, resulting in ethanol’s

solubility in water.

Question 11

Question

Calculate the solubility of sodium chloride (NaCl) in water at 25◦C. Given that

the solubility product constant (Ksp) for NaCl at this temperature is 3.4×10−3.

Solution

Step 1: Write the equation for the dissolution of sodium chloride:

NaCl ⇌Na++ Cl−

Step 2: Write the expression for the solubility product constant (Ksp):

Ksp = [Na+][Cl−]

Step 3: Since sodium chloride dissociates into one Na+ion and one Cl−ion,

the concentration of Na+is equal to the concentration of Cl−. Let xbe the

solubility of NaCl. Therefore:

[Na+] = [Cl−] = x

Step 4: Substitute into the Ksp expression and solve for x:

Ksp =x×x=x2

3.4×10−3=x2

Step 5: Solve for x:

x=p3.4×10−3= 0.058 M

Therefore, the solubility of sodium chloride in water at 25◦C is 0.058 M.

Question 12

Question

Calculate the solubility of acetaminophen (C8H9NO2) in water at 25◦C. The

solubility product constant, Ksp, for acetaminophen is 1.1×10−8.

Solution

Step 1: Write the equilibrium expression for the dissolution of acetaminophen

in water. The dissolution of acetaminophen in water can be represented by the

equation:

C8H9NO2⇌C8H9NO2(aq)

The equilibrium expression for the dissolution is:

Ksp = [C8H9NO2]2

Step 2: Set up the equation using the solubility (S) of acetaminophen. Let

the solubility of acetaminophen be S. Thus, the equilibrium concentrations can

be expressed as:

[C8H9NO2] = S

Step 3: Substitute the equilibrium concentrations into the equilibrium ex-

pression and solve for the solubility (S). Substitute the concentration into the

equation:

1.1×10−8= (S)2

S=p1.1×10−8

S≈3.3×10−4mol/L

Therefore, the solubility of acetaminophen in water at 25◦C is approximately

3.3×10−4mol/L.

Question 13

Question

A mixture was prepared by dissolving 2.5 g of compound A (molar mass = 100

g/mol) and 4.0 g of compound B (molar mass = 150 g/mol) in 100 mL of water.

After thorough mixing, it was found that compound A is completely soluble in

water, while compound B is insoluble. What is the resulting mass percentage

of compound A in the mixture?

(Please note that the volume of the solution remains constant after the two

compounds are dissolved.)

Solution

Step 1: Calculate the moles of each compound. Given: Mass of compound A,

mA= 2.5 g Molar mass of compound A, MA= 100 g/mol

Mass of compound B, mB= 4.0 g Molar mass of compound B, MB= 150

g/mol

First we calculate the moles of each compound:

moles of compound A = mA

=2.5 g

100 g/mol = 0.025 mol

moles of compound B = mB

=4.0 g

150 g/mol = 0.02667 mol

Step 2: Calculate the total moles in the mixture. Since the volume of the

solution is constant, the total moles in the mixture is the sum of the moles of

compound A and compound B:

total moles = moles of A + moles of B = 0.025 mol + 0.02667 mol = 0.05167 mol

Step 3: Calculate the mass percentage of compound A in the mixture. The

mass percentage of compound A in the mixture is given by:

Mass % of A = mass of A in mixture

total mass of mixture ×100%

mass of A in mixture = (moles of A ×MA)=0.025 mol ×100 g/mol = 2.5 g

total mass of mixture = mass of A + mass of B = 2.5 g + 4.0 g = 6.5 g

Mass % of A = 2.5 g

6.5 g ×100% = 38.46%

Therefore, the resulting mass percentage of compound A in the mixture is

38.46

Question 14

Question

An organic compound with the chemical formula C7H6O has a solubility of

0.150 g/100 mL of water at 25

C. Calculate the compound’s solubility product

constant (Ksp) in mol2/L2.

Solution

Step 1: Calculate the molar mass of the compound. Step 2: Calculate the

molarity of the compound in the saturated solution. Step 3: Use the molarity

to calculate the solubility product constant (Ksp).

Step 1: Calculate the molar mass of the compound.

The molar mass of C7H6O can be calculated as follows:

7×molar mass of C + 6 ×molar mass of H + 1 ×molar mass of O

= 7(12.01 g/mol) + 6(1.008 g/mol) + 16.00 g/mol

= 84.07 g/mol

Step 2: Calculate the molarity of the compound in the saturated solution.

The solubility of the compound is 0.150 g/100 mL. We need to convert this

to mol/L:

0.150 g ×1 mol

84.07 g ×1000 mL

100 mL = 0.0178 mol/L

Step 3: Use the molarity to calculate the solubility product constant (Ksp).

The Kspfor the compound can be calculated using the molarity of the

compound in the saturated solution:

Ksp = [C7H6O]2

Ksp = (0.0178 mol/L)2

Ksp = 3.16 ×10−4mol2/L2

Therefore, the solubility product constant (Ksp) for the compound is 3.16 ×

10−4mol2/L2.

Question 15

Question

Calculate the solubility of benzoic acid (C7H6O2) in water at 25◦C. The solu-

bility product constant, Ksp, for benzoic acid in water is 1.2×10−3mol/L.

Solution

Step 1: Write the dissociation equation for benzoic acid (C7H6O2) in water:

C7H6O2⇌C7H5O−

2+H+

Step 2: Write the expression for the solubility product constant, Ksp:

Ksp = [C7H5O−

2][H+]

Step 3: Because benzoic acid is a weak acid, assume the initial amount of

benzoic acid that dissociates into ions is x, and neglect the change in concen-

tration of benzoic acid (compared to x). Thus, at equilibrium:

[C7H5O−

2] = x

[H+] = x

Step 4: Substitute the equilibrium concentrations back into the expression

for Ksp:

Ksp =x2

Step 5: Solve for x using the value of Ksp provided:

1.2×10−3=x2

x=p1.2×10−3

x≈√1.2×10−3/2

x≈1.095 ×10−2mol/L

Therefore, the solubility of benzoic acid in water at 25◦C is approximately

1.095 ×10−2mol/L.

Question 16

Question

A chemist is studying the solubility of different compounds in water. The

chemist has the following three compounds: Compound A, Compound B, and

Compound C. The chemist determines the following solubility values: - Com-

pound A: 2.5 g/100 mL at 25

C - Compound B: 7.8 g/100 mL at 25

C - Com-

pound C: 0.6 g/100 mL at 25

Rank the compounds in decreasing order of solubility in water at 25

Justify your answer based on the concepts of solubility and polarity.

Solution

Step 1: Calculate the molarity of each compound. - Molarity is calculated using

the formula: M=mass(g)

molar mass(g/mol)×volume(L)

For Compound A: - Molar mass of Compound A: let’s assume it is 100 g/mol

- Volume in liters: 100 mL = 0.1 L - Molarity of Compound A: MA=2.5

100×0.1=

0.25

For Compound B: - Let’s assume Compound B has a molar mass of 200

g/mol - Volume in liters: 100 mL = 0.1 L - Molarity of Compound B: MB=

7.8

200×0.1= 0.39

For Compound C: - Assuming Compound C has a molar mass of 50 g/mol

- Volume in liters: 100 mL = 0.1 L - Molarity of Compound C: MC=0.6

50×0.1=

0.12

Step 2: Rank the compounds based on molarity. - The compounds can be

ranked as follows: 1. Compound B (with the highest molarity) 2. Compound

A 3. Compound C (with the lowest molarity)

Step 3: Justification based on solubility and polarity. - The compound

with the highest molarity (Compound B) is the most soluble in water at 25

followed by Compound A and then Compound C. This ranking aligns with the

solubility values provided. - Generally, compounds with higher polarity tend to

be more soluble in water due to the ability to form hydrogen bonds with water

molecules. In this case, Compound B likely has the highest polarity, followed by

Compound A and then Compound C, which explains their respective solubility

rankings.

Question 17

Question

A student is investigating the solubility of different organic compounds in water.

Compound X is a nonpolar organic molecule with a molecular weight of 120

g/mol. Compound Y is a polar organic molecule with a molecular weight of 90

g/mol. Which compound is expected to be more soluble in water and why?

Solution

To determine which compound is expected to be more soluble in water, we need

to consider the polarity of the compounds and their molecular weight.

Step 1: Calculate the polarity of each compound. Compound X is nonpo-

lar, meaning it does not have any significant dipole moment due to symmetric

electron distribution. Nonpolar compounds typically have weaker interactions

with polar solvents like water. Compound Y is polar, which means it has regions

of partial positive and negative charge. Polar compounds tend to form stronger

interactions with water molecules through hydrogen bonding.

Step 2: Consider the molecular weight of each compound. Compound X

has a molecular weight of 120 g/mol. Compound Y has a molecular weight of

90 g/mol.

Step 3: Analyze the solubility based on polarity and molecular weight.

While both compounds have similar molecular weights, the polar nature of

Compound Y makes it more likely to form favorable interactions with water

molecules compared to the nonpolar Compound X. This is due to the ability of

polar molecules to engage in hydrogen bonding with water molecules, increasing

solubility.

Therefore, Compound Y is expected to be more soluble in water than Com-

pound X because of its polarity and potential for hydrogen bonding interactions

with water molecules.

Question 18

Question

A student is trying to determine the solubility of a compound in water based

on its structure. The compound has a long nonpolar hydrocarbon chain and a

single polar functional group at one end. Discuss the factors that will influence

the solubility of this compound in water.

Solution

To determine the solubility of a compound in water, we need to consider the

polarity of the compound and the type of intermolecular forces that will be

present between the compound and water molecules.

Step 1: Polarity of the compound

The compound has a long nonpolar hydrocarbon chain and a single polar func-

tional group. The nonpolar hydrocarbon chain is hydrophobic and will not

interact favorably with the polar water molecules. On the other hand, the polar

functional group can interact with water molecules through hydrogen bonding

or dipole-dipole interactions.

Step 2: Types of intermolecular forces

The presence of the polar functional group allows for the formation of hydrogen

bonds or dipole-dipole interactions with water molecules. These interactions

can increase the solubility of the compound in water.

Step 3: Overall solubility prediction

Considering the factors mentioned above, the compound is likely to have limited

solubility in water. The nonpolar hydrocarbon chain will prefer interactions with

other nonpolar molecules over water molecules due to the hydrophobic effect.

The interactions between the polar functional group and water molecules may

not be strong enough to offset the unfavorable interactions between the nonpolar

chain and water.

Therefore, the compound is expected to have low solubility in water due to

the presence of a predominantly nonpolar hydrocarbon chain.

Question 19

Question

Calculate the solubility of benzoic acid (C7H6O2) in water at 25◦C. The exper-

imental solubility of benzoic acid is 3.2 g/L at this temperature.

Solution

Step 1: Calculate the molar mass of benzoic acid. The molar mass of benzoic

acid (C7H6O2) can be calculated by adding the atomic masses of carbon (C),

hydrogen (H), and oxygen (O). Molar mass of benzoic acid = 7(12.01) + 6(1.008)

+ 2(16.00) = 122.11 g/mol

Step 2: Convert the given experimental solubility from g/L to mol/L. Using

the molar mass calculated in Step 1, we can convert the solubility of benzoic

acid from g/L to mol/L. Number of moles of benzoic acid in 1 L of water = 3.2

g / 122.11 g/mol = 0.0263 mol/L

Step 3: Write the equilibrium expression for the solubility of benzoic acid in

water. The equilibrium expression for the dissolution of benzoic acid in water

is: C7H6O2(s) ⇌C7H6O2(aq)

Step 4: Calculate the solubility of benzoic acid in water. Since the solubility

of benzoic acid at 25◦C is the same as the concentration of benzoic acid in

the saturated solution, the solubility of benzoic acid in water at 25◦C is 0.0263

mol/L.

Therefore, the solubility of benzoic acid in water at 25◦C is 0.0263 mol/L.

Question 20

Question

A mixture is prepared by combining 10.0 g of compound A with a solubility of

5.0 g per 100 mL of water at 25

C, and 20.0 g of compound B with a solubility

of 15.0 g per 100 mL of water at 25

C. Assuming the solubility behavior is an

ideal solution, determine the mass of each compound that remains undissolved

when the mixture is dissolved in 500.0 mL of water at 25

Solution

Step 1: Calculate the mass of compound A and B that can dissolve in 500.0 mL

of water.

Given: Solubility of A: 5.0 g per 100 mL Solubility of B: 15.0 g per 100 mL

Volume of water: 500.0 mL

For compound A:

Mass of A that can dissolve = (5.0 g/100 mL) ×(500.0 mL) = 25.0 g

For compound B:

Mass of B that can dissolve = (15.0 g/100 mL) ×(500.0 mL) = 75.0 g

Therefore, 25.0 g of compound A and 75.0 g of compound B can dissolve in

500.0 mL of water.

Step 2: Determine the mass of each compound that remains undissolved.

For compound A:

Mass of A remaining undissolved = 10.0 g −25.0 g = −15.0 g

For compound B:

Mass of B remaining undissolved = 20.0 g −75.0 g = −55.0 g

Therefore, 15.0 g of compound A and 55.0 g of compound B remain undis-

solved when the mixture is dissolved in 500.0 mL of water at 25

Question 21

Question

Calculate the solubility (in g/L) of benzoic acid in water at 25

C, given that

the solubility of benzoic acid is 3.2 g/100 mL in water at this temperature. The

molar mass of benzoic acid is 122.12 g/mol.

Solution

Step 1: Calculate the molarity of benzoic acid in water. Step 2: Use the molarity

to calculate the solubility in g/L.

Step 1: Calculate the molarity of benzoic acid in water. The solubility of

benzoic acid in water is given as 3.2 g/100 mL. First, convert the volume from

mL to L: 100 mL = 0.1 L

Next, calculate the molarity using the formula:

Molarity (M) = moles solute

liters of solution

But we need to convert grams to moles using the molar mass of benzoic acid:

moles of benzoic acid = 3.2 g

122.12 g/mol

Now, calculate the molarity:

Molarity =

3.2 g

122.12 g/mol

0.1 L

Step 2: Use the molarity to calculate the solubility in g/L. The molarity

we calculated represents the concentration of benzoic acid in water. To find the

solubility in grams per liter (g/L), multiply the molarity by the molar mass:

Solubility = Molarity ×Molar mass

Solubility = 3.2 g

122.12 g/mol

0.1 L !×122.12 g/mol

Now, calculate the solubility in g/L.

Question 22

Question

Calculate the solubility of benzoic acid in water at 25

C. The solubility of benzoic

acid in water is 6.8 g/L at 100

C and its enthalpy of solution is 19.68 kJ/mol.

Assume the solubility of benzoic acid follows the van’t Hoff equation at this

temperature.

Solution

Step 1: Calculate the change in enthalpy per mole of benzoic acid dissolved.

Given that the enthalpy of solution is 19.68 kJ/mol, convert it to joules:

∆H= 19.68 kJ/mol ×1000 J/kJ = 19680 J/mol

Step 2: Calculate the van’t Hoff factor (i). Since benzoic acid is a neutral

molecule, the van’t Hoff factor is 1.

Step 3: Calculate the equilibrium constant (K) for the dissolution of benzoic

acid at 25

C. Using the van’t Hoff equation:

ln K=−∆H

R1

T−1

Tref 

where R is the ideal gas constant (8.314 J/molK), T is the temperature in Kelvin

(25 + 273 = 298 K), and Trefistheref erencetemperatureinKelvin(100+273 =

373K).P lugginginthevalues : ln K=−19680

8.314 1

298 −1

373 

ln K=−2370.79

K=e−2370.79

K≈5.39 ×10−103

Step 4: Calculate the solubility of benzoic acid in water at 25

C. Let x

be the solubility of benzoic acid in mol/L. The equilibrium expression for the

dissolution of benzoic acid is:

K=x

1000

Substitute the value of K and solve for x:

5.39 ×10−103 =x

1000

x= 5.39 ×10−100 mol/L

Convert mol/L to g/L:

x= 5.39 ×10−100 ×122.12 g/mol = 6.58 ×10−98 g/L

Therefore, the solubility of benzoic acid in water at 25

C is 6.58 x 10−98g/L.

Question 23

Question

Calculate the solubility of benzoic acid (C7H6O2) in water at 25◦C. The given

Ksp value for benzoic acid is 6.8×10−5.

Solution

Step 1: Write the dissolution reaction of benzoic acid in water, and write the

equilibrium expression for the reaction. The dissolution reaction of benzoic acid

in water is:

C7H6O2⇌C7H6O2

The equilibrium expression for this reaction is:

Ksp = [C7H6O2]

Step 2: Define the initial, equilibrium, and change in concentration. Let’s

assume the initial solubility of benzoic acid is x. Then, at equilibrium, the

concentration of benzoic acid will be x.

Step 3: Write the equilibrium expression in terms of x and solve for x. Using

the given Ksp value:

6.8×10−5= (x)(x)

Solving for x:

x2= 6.8×10−5

x=p6.8×10−5

x≈0.00824

Step 4: Check the assumption and answer the question. Since the solubility

of benzoic acid is 0.00824 M, we can check if the assumption of x is negligible

compared to the initial amount is valid. Given that the initial amount of benzoic

acid is not specified, we can assume it is large enough for the assumption to be

valid. Therefore, the solubility of benzoic acid in water at 25◦C is approximately

0.00824 mol/L.

Question 24

Question

A student is conducting solubility experiments on four unknown organic com-

pounds. The student determines that Compound A is soluble in hexane, but

insoluble in water. Compound B is insoluble in both hexane and water. Com-

pound C is soluble in water but insoluble in hexane. Compound D is soluble

in both hexane and water. Rank the four compounds in order of increasing

polarity.

Solution

Step 1: Recall that polarity is related to a molecule’s ability to form inter-

molecular interactions such as hydrogen bonding, dipole-dipole interactions, and

London dispersion forces. - Nonpolar compounds only have London dispersion

forces. - Polar compounds have dipole-dipole interactions, and some may also

exhibit hydrogen bonding.

Step 2: Analyze the solubility behavior of each compound to determine their

relative polarities. - Compound A is soluble in hexane (a nonpolar solvent) but

insoluble in water (a polar solvent) which suggests it is nonpolar. - Compound

B is insoluble in both hexane and water, indicating it is nonpolar. - Compound

C is soluble in water (polar) but insoluble in hexane (nonpolar), indicating it

is polar. - Compound D is soluble in both hexane and water, which suggests it

must be polar since it is soluble in water.

Step 3: Rank the compounds based on their polarities: 1. Compound B

(most nonpolar) 2. Compound A 3. Compound D 4. Compound C (most

polar)

Question 25

Question

An organic compound has the following solubility properties: it is soluble in

water (H2O) and insoluble in diethyl ether (C4H10O). Based on these solubility

properties, determine the likely functional groups present in the compound.

Solution

Step 1: Given that the compound is soluble in water but insoluble in diethyl

ether, we can infer that the compound must contain polar functional groups

that can form hydrogen bonds with water molecules.

Step 2: Some common functional groups that are polar and capable of form-

ing hydrogen bonds include alcohols, carboxylic acids, and amines.

Step 3: Let’s consider each possibility: - Alcohols: Compounds with the -OH

functional group are polar and soluble in water due to hydrogen bonding. Since

alcohols can form hydrogen bonds, the compound could contain an alcohol func-

tional group. - Carboxylic acids: Compounds with the -COOH functional group

are polar and also soluble in water due to hydrogen bonding. Carboxylic acids

contain both a hydroxyl group (-OH) and a carbonyl group (C=O). - Amines:

Compounds with the -NH2 functional group are polar and can participate in

hydrogen bonding. They are also soluble in water.

Step 4: Given that the compound is insoluble in diethyl ether, a non-polar

solvent, it is unlikely to contain non-polar functional groups such as alkyl chains

or aromatic rings.

Step 5: Therefore, based on the solubility properties provided, the likely

functional groups present in the compound are alcohols, carboxylic acids, or

amines.

Step 6: It is important to note that other functional groups may also exhibit

similar solubility properties. Additional tests or analyses would be required to

definitively identify the functional groups present in the compound.

Question 26

Question

A student is conducting an experiment in which they mix 10.0 mL of ethanol

(C2H5OH) with 10.0 mL of water. Given that the solubility of ethanol in water

is 80.0 g/L at 20◦C, calculate the mass of ethanol that will dissolve in the wa-

ter. Assume the densities of ethanol and water are 0.789 g/mL and 1.00 g/mL,

respectively.

Solution

Step 1: Calculate the total volume of the mixture. The total volume of the

mixture can be found by adding the volumes of ethanol and water:

Vtotal =Vethanol +Vwater = 10.0 mL + 10.0 mL = 20.0 mL

Step 2: Convert the total volume to liters.

Vtotal = 20.0 mL ×1 L

1000 mL = 0.0200 L

Step 3: Calculate the maximum mass of ethanol that can dissolve in this

volume of water. Given that the solubility of ethanol in water is 80.0 g/L:

Max mass of ethanol = 80.0 g/L ×0.0200 L = 1.60 g

Therefore, the maximum mass of ethanol that can dissolve in the water is

1.60 g.

Question 27

Question

A student is given two unknown organic compounds, A and B, and asked to

determine which one is more soluble in water. The student performs a solubility

test for each compound and finds that compound A is soluble in water, while

compound B is insoluble. The student then measures the solubility parameter

for each compound, obtaining a value of 10 (J/cm3)1/2for compound A and a

value of 15 (J/cm3)1/2for compound B. Based on this information, explain why

compound A is more soluble in water than compound B.

Solution

Step 1: Recall that the solubility parameter (δ) is a measure of the polarity of

a compound. The solubility parameter is calculated using the formula:

δ= (∆Hv/Vm)1/2

where ∆Hvis the heat of vaporization and Vmis the molar volume.

Step 2: Since compound A has a solubility parameter of 10 (J/cm3)1/2, and

compound B has a solubility parameter of 15 (J/cm3)1/2, we can conclude that

compound A is less polar than compound B. This is because a lower solubility

parameter indicates lower polarity.

Step 3: Water is a polar solvent, meaning it is better at dissolving polar

substances than non-polar substances. Therefore, compound A, being less polar,

is more soluble in water compared to compound B, which is more polar.

Step 4: In summary, the solubility parameter values indicate that compound

A is less polar than compound B, which is why compound A is more soluble in

water.

Question 28

Question

Calculate the solubility of benzoic acid in water at 25

C. The partition coefficient

of benzoic acid in water and diethyl ether is 20.0. Assume that benzoic acid

behaves as a neutral molecule in both solvents.

Solution

Let’s denote the solubility of benzoic acid in water as x(in mol/L) and the

solubility of benzoic acid in diethyl ether as 20x(in mol/L) due to the partition

coefficient of 20.

Step 1: Write the equilibrium equation where benzoic acid is dissolved in

the two solvents.

Benzoic acid (in water): C6H5COOH(aq)⇌C6H5COOH(aq)

Benzoic acid (in diethyl ether): C6H5COOH(org)⇌C6H5COOH(org)

Step 2: Write the expression for the partition coefficient (K) of benzoic acid

between water and diethyl ether.

K=[C6H5COOH](aq)

[C6H5COOH](org)

20x=1

Step 3: Calculate the solubility of benzoic acid in water.

K=1

x= 20 ×[C6H5COOH](org)= 20 ×2 = 40 mol/L

Therefore, the solubility of benzoic acid in water at 25

C is 40 mol/L.

Question 29

Question

Predict which of the following compounds would be most soluble in water based

on their polarity: 1. Hexane (C6H14) 2. Methanol (CH3OH) 3. Acetic acid

(CH3COOH) 4. Dichloromethane (CH2Cl2)

Solution

In order to determine which compound would be most soluble in water, we need

to consider the polarity of each compound. Compounds that are polar tend to

dissolve in polar solvents like water, while nonpolar compounds tend to dissolve

in nonpolar solvents like hexane.

Step 1: Hexane (C6H14) Hexane is a nonpolar molecule composed of only

carbon and hydrogen atoms. It is not able to form hydrogen bonds with water

molecules and is therefore not soluble in water. Hexane would be most soluble

in a nonpolar solvent like itself.

Step 2: Methanol (CH3OH) Methanol is a polar molecule due to the elec-

tronegativity difference between carbon, oxygen, and hydrogen atoms. It is

able to form hydrogen bonds with water molecules, making it soluble in water.

Methanol would be the most soluble compound in water among the options

given.

Step 3: Acetic acid (CH3COOH) Acetic acid is a polar molecule that con-

tains a carboxyl group (-COOH) which makes it capable of forming hydrogen

bonds with water molecules. Acetic acid is soluble in water due to its polarity.

Step 4: Dichloromethane (CH2Cl2) Dichloromethane is a polar molecule

due to the electronegativity difference between carbon and chlorine atoms. It

can form dipole-dipole interactions with water molecules. However, dichloromethane

is less polar than methanol and acetic acid, so it would be less soluble in water

compared to those two compounds.

Therefore, based on their polarity, methanol (CH3OH) would be the most

soluble in water among the compounds given.

Question 30

Question

Calculate the solubility of 2-nitroaniline in water at 25

C given that the solu-

bility of 2-nitroaniline in water at 25

C is 22 g/L. The solubility in water of

2-aniline sulfate is 40 g/L. How many grams of 2-nitroaniline would be required

to prepare a saturated solution containing 5.0 g of 2-aniline sulfate? Assume no

change in volume upon addition of 2-aniline sulfate.

Solution

Step 1: Calculate the solubility product constant (Ksp) for 2-nitroaniline from

the given solubility of 2-nitroaniline. Step 2: Use Ksp to calculate the solubility

of 2-nitroaniline at 25

C in water. Step 3: Determine the molar mass of 2-

nitroaniline and 2-aniline sulfate. Step 4: Calculate the moles of 2-aniline sulfate

required to prepare a saturated solution containing 5.0 g. Step 5: Use the molar

ratio to calculate the moles of 2-nitroaniline that can be dissolved. Step 6:

Calculate the mass of 2-nitroaniline required to prepare the solution.

Step 1: Calculate the solubility product constant (Ksp) for 2-nitroaniline:

Given solubility of 2-nitroaniline in water at 25

C = 22 g/L Molar mass of

2-nitroaniline (C6H6N2O2) = 138.13 g/mol

Convert solubility to mol/L:

22 g

L×1 mol

138.13 g= 0.1593 mol/L

Step 2: Use Ksp to calculate the solubility of 2-nitroaniline at 25

C in

water. The Ksp expression for the dissolution of 2-nitroaniline is:

Ksp = [2-nitroaniline] = (0.1593)2= 0.0254 mol/L

Step 3: Determine the molar mass of 2-nitroaniline and 2-aniline sulfate:

Molar mass of 2-nitroaniline = 138.13 g/mol Molar mass of 2-aniline sulfate =

Molar mass of aniline + Molar mass of sulfate = 93.13 + 96.06 = 189.19 g/mol

Step 4: Calculate the moles of 2-aniline sulfate required: Given mass of

2-aniline sulfate = 5.0 g

Moelcules of 2-aniline sulfate = 5.0 g

189.19 g/mol

Step 5: Calculate the moles of 2-nitroaniline that can be dissolved: Using

the molar ratio between 2-aniline sulfate and 2-nitroaniline: 1 mol of 2-aniline

sulfate corresponds to 1 mol of 2-nitroaniline

Step 6: Calculate the mass of 2-nitroaniline required:

Mass of 2-nitroaniline = Moles of 2-nitroaniline ×Molar mass of 2-nitroaniline

Question 31

Question

Calculate the solubility of benzoic acid (C7H6O2) in water at 25

C. The solu-

bility product constant, Ksp, for benzoic acid at this temperature is 6.8×10−5

mol/L.

Solution

Step 1: Write the dissolution equation for benzoic acid in water:

C7H6O2(s) ⇌C7H6O2(aq)

Step 2: Write the expression for the solubility product constant (Ksp):

Ksp = [C7H6O2]

Step 3: Let x be the molar solubility of benzoic acid in water. Since benzoic

acid is a weak acid, it partially dissociates according to the equation:

C7H6O2(s) ⇌C7H6O2(aq)

Therefore, the equilibrium expression for this dissociation is:

Ksp = [C7H6O2] = x

Step 4: Substitute the given value of Ksp into the equation:

6.8×10−5=x

Step 5: Solve for x to find the molar solubility of benzoic acid in water:

x= 6.8×10−5mol/L

Therefore, the solubility of benzoic acid in water at 25

C is 6.8×10−5mol/L.

Question 32

Question

An organic compound is found to be soluble in water but insoluble in hexane.

Assuming that the compound has a molar mass of 120 g/mol, calculate the

maximum number of hydrogen bonds that can be formed by one molecule of

this compound when dissolved in water.

Solution

Step 1: Calculate the molar mass of water. The molar mass of water, H2O, is

18.015g/mol (2 g/mol for hydrogen and 16 g/mol for oxygen).

Step 2: Determine the maximum number of hydrogen bonds that can be

formed by one molecule of water. In one molecule of water, there are 2 hydrogen

atoms that can participate in hydrogen bonding. Each hydrogen atom can form

a hydrogen bond with another electronegative atom. Therefore, one molecule

of water can form 2 hydrogen bonds.

Step 3: Calculate the maximum number of water molecules that can par-

ticipate in hydrogen bonding with one molecule of the organic compound. The

molar mass of the organic compound is 120 g/mol. Therefore, the number of

water molecules that can participate in hydrogen bonding with one molecule of

the compound is given by:

120 g/mol

18.015 g/mol = 6.664

Since we can’t have a fraction of a molecule, we will use the nearest whole

number, 6.

Step 4: Find the maximum number of hydrogen bonds that can be formed

by one molecule of the organic compound. Since one molecule of the organic

compound can form hydrogen bonds with 6 water molecules, and each water

molecule can form 2 hydrogen bonds, the maximum number of hydrogen bonds

that can be formed by one molecule of the organic compound is:

6 molecules ×2 bonds/molecule = 12 bonds

Therefore, one molecule of the organic compound can form a maximum of

12 hydrogen bonds when dissolved in water.

Question 33

Question

An organic compound, X, has a solubility of 2.5 g/L in water at 25

C. When

5.0 g of X is dissolved in 200 mL of benzene at 25

C, only 1.0 g of X dissolves.

Calculate the solubility of X in benzene at 25

Solution

Step 1: Calculate the molar solubility of X in water. Given: - Mass of X = 2.5

g - Volume of water = 1 L - Molar mass of X = MX

The molar solubility of X in water can be calculated using the formula:

Molar solubility = Mass of X

Molar mass of X

Therefore, the molar solubility of X in water is:

Molar solubility = 2.5 g

Step 2: Calculate the solubility of X in benzene. Given: - Mass of X in

benzene = 1.0 g - Volume of benzene = 0.2 L

The solubility of X in benzene can be calculated using the formula:

Solubility in benzene = Mass of X in benzene

Volume of benzene

Substitute the given values:

Solubility in benzene = 1.0 g

0.2 L

Therefore, the solubility of X in benzene at 25

C is 5.0 g/L.

Question 34

Question

A student is given two unknown organic compounds, labeled A and B. Com-

pound A is found to be soluble in water, while compound B is found to be

insoluble in water. Based on this information, which compound is likely to have

a higher polarity?

Solution

To determine which compound is likely to have a higher polarity based on

solubility in water, we need to consider the polarities of water and the two

unknown compounds.

Step 1: Understand the concept of like dissolves like. For a compound

to dissolve in another, the two substances must have similar polarity. Like

dissolves like: polar compounds tend to dissolve in polar solvents, while nonpolar

compounds tend to dissolve in nonpolar solvents.

Step 2: Determine the solubility of compounds A and B. - Compound A

is soluble in water. This suggests that compound A is likely polar since water

is a polar solvent and polar compounds tend to dissolve in polar solvents. -

Compound B is insoluble in water. This suggests that compound B is likely

nonpolar since nonpolar compounds tend to dissolve in nonpolar solvents.

Step 3: Conclusion. Based on the information given, compound A is likely

to have a higher polarity than compound B. This is because compound A is

soluble in water, a polar solvent, indicating its polarity, while compound B is

insoluble in water, suggesting it is nonpolar.

Question 35

Question

A student is given a mixture of a solid organic compound and an unknown

solvent. The student finds that the compound is soluble in water but insoluble

in diethyl ether. Knowing that the compound has a molar mass of 150 g/mol,

determine if the compound is more likely to be polar or nonpolar. Justify your

answer.

Solution

Step 1: Calculate the compound’s molecular formula weight. Given that the

molar mass of the compound is 150 g/mol, we need to determine the possible

molecular formula of the compound.

Step 2: Determine the compound’s solubility in water. Since the compound

is soluble in water, it indicates that the compound has some polar character-

istics. Water is a polar solvent that can dissolve polar compounds through

hydrogen bonding or dipole-dipole interactions.

Step 3: Determine the compound’s solubility in diethyl ether. Since the com-

pound is insoluble in diethyl ether, it suggests that the compound is nonpolar.

Diethyl ether is a nonpolar solvent that can only dissolve nonpolar compounds

through London dispersion forces.

Step 4: Conclusion. Based on the solubility characteristics of the compound

in water and diethyl ether, it is more likely that the compound is polar. Po-

lar compounds tend to be soluble in polar solvents like water, while nonpolar

compounds are soluble in nonpolar solvents like diethyl ether.