CHEM 301 - ORGANIC CHEMISTRY I Solubility and polarity calculations Question Bank Set 2

CHEM 301 - ORGANIC CHEMISTRY

I - Solubility and polarity calculations

Question Bank - Set 2

Liberty University

Question 1

Question

An organic compound with the molecular formula C10H12O2is known to have a

solubility in water of 0.25 g/L at 25

C. Determine the solubility of the compound

in hexane (density = 0.654 g/mL) at the same temperature.

Solution

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

number of moles of the compound that can dissolve in water. Step 3: Determine

the molar solubility of the compound in water. Step 4: Calculate the molar

solubility of the compound in hexane.

Question 2

Question

A student wants to dissolve a compound in water in order to isolate two distinct

layers. The compound has a molecular weight of 180 g/mol and a solubility of 25

g/L in water at 25

C. The student only has 100 mL of water available. Will the

compound fully dissolve in the water? If not, how many grams of the compound

can be dissolved in the 100 mL of water at 25

Solution

Step 1: Calculate the maximum amount of compound that can be dissolved in

100 mL of water. The student has 100 mL of water, which is equivalent to 0.1 L.

Using the solubility information given, we can calculate the maximum amount

that can be dissolved:

Maximum amount of compound = solubility ×volume of solvent

Maximum amount of compound = 25 g/L ×0.1 L = 2.5 g

Step 2: Determine if the compound will fully dissolve in the water. The

compound has a molecular weight of 180 g/mol. Let’s calculate how many

moles of compound are present in 2.5 g:

moles of compound = mass of compound

molecular weight =2.5 g

180 g/mol ≈0.0139 mol

Step 3: Determine if the compound will fully dissolve in the water. To

determine if the compound will fully dissolve in water, we need to calculate the

molarity of the compound:

Molarity = moles of compound

volume of solvent (L) =0.0139 mol

0.1 L = 0.139 M

Since the molarity of the compound is less than its solubility in water (0.139

M ¡ 25 g/L), the compound will fully dissolve in the water.

Question 3

Question

An organic compound is dissolved in water to form a solution. The compound

has a solubility of 0.15 g per 100 mL of water at 25

C. Its calculated molecular

weight is 120 g/mol. Determine the polarity index of the compound.

Solution

Step 1: Calculate the molarity of the compound in the solution. Given that the

solubility of the compound is 0.15 g per 100 mL of water, we first convert this

to g/L:

0.15 g ×1 L

1000 mL = 0.0015 g/L

Now, calculate the molarity using the formula:

Molarity = moles of solute

liters of solution

Since we know the molecular weight of the compound is 120 g/mol, we can

find the moles of solute:

moles = mass

molecular weight =0.0015 g

120 g/mol = 1.25 ×10−5mol

The volume of the solution is 100 mL = 0.1 L, so:

Molarity = 1.25 ×10−5mol

0.1 L = 1.25 ×10−4M

Step 2: Calculate the polarity index. The polarity index is a measure of how

polar or nonpolar a compound is. It is defined as the Molarity divided by the

product of the molecular weight and solubility in water:

Polarity Index = Molarity

(Molecular weight ×Solubility)

Plugging in the values:

Polarity Index = 1.25 ×10−4M

(120 g/mol ×0.0015 g/L)

Polarity Index = 1.25 ×10−4

0.18

Polarity Index = 6.94 ×10−4

Therefore, the calculated polarity index of the compound is 6.94 ×10−4.

Question 4

Question

A student is given two compounds, Compound A and Compound B, and asked

to determine which compound is more soluble in water. Compound A has

a molecular weight of 180 g/mol and a logarithm of the partition coefficient

(log P) of 0.5, while Compound B has a molecular weight of 250 g/mol and a

logarithm of the partition coefficient (log P) of 1.2. Based on this information,

which compound is more soluble in water?

Solution

To determine which compound is more soluble in water, we will use the partition

coefficient (P) formula:

P=Concentration in organic phase

Concentration in aqueous phase

The partition coefficient is related to the logarithm of the partition coefficient

by the equation:

logP= log Concentration in organic phase

Concentration in aqueous phase

Taking the antilog of both sides of the equation gives:

P= 10logP

Now, we can substitute the given log P values to find the partition coefficients

for Compound A and Compound B.

Step 1: Calculate Pfor Compound A

For Compound A with log P = 0.5:

PA= 100.5= 3.162

Step 2: Calculate Pfor Compound B

For Compound B with log P = 1.2:

PB= 101.2= 15.849

Step 3: Determine which compound is more soluble in water

Since a higher partition coefficient indicates higher solubility in the organic

phase relative to the aqueous phase, Compound B, with a higher partition

coefficient (PB= 15.849), is more soluble in water compared to Compound

A (PA= 3.162).

Question 5

Question

Calculate the solubility of compound X in water at 25

C given that the solubility

product constant (Ksp) for X is 5.0×10−5mol2/L2.

Solution

Step 1: Write the solubility equilibrium for compound X: Let the solubility of

compound X be represented by the variable x. Then the equilibrium equation

for the dissolution of compound X in water can be written as:

X⇌X2+ + 2X−

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

solubility product constant (Ksp) expression for compound X can be written as:

Ksp = [X2+][X−]2=x(2x)2= 4x3

Step 3: Substitute the given Ksp and solve for x: Given that Ksp = 5.0×10−5

mol2/L2, we can set up the equation as:

5.0×10−5= 4x3

Step 4: Solve for x:

x=3

r5.0×10−5

x=3

p1.25 ×10−5

x≈0.0221 mol/L

Therefore, the solubility of compound X in water at 25

C is approximately

0.0221 mol/L.

Question 6

Question

A student performed a solubility experiment and found that 2.0 grams of a

compound dissolved in 10 mL of water at 25

C. The student then measured the

solubility of the same compound in ethanol and found that 4.0 grams dissolved

in 10 mL of ethanol at 25

C. Based on these observations, calculate the partition

coefficient (Kow) of the compound between water and ethanol at 25

Solution

Step 1: Calculate the solubility of the compound in water and ethanol. In water:

Solubility in water = 2.0 g

10 mL = 0.2 g/mL

In ethanol:

Solubility in ethanol = 4.0 g

10 mL = 0.4 g/mL

Step 2: Calculate the partition coefficient (Kow) using the formula:

Kow =Solubility in water

Solubility in ethanol

Substitute the values to find Kow:

Kow =0.2 g/mL

0.4 g/mL = 0.5

Therefore, the partition coefficient (Kow) of the compound between water

and ethanol at 25

C is 0.5.

Question 7

Question

An organic compound has a solubility of 0.35 g/100 mL in water at 25

C. Cal-

culate the solubility of the compound in grams per liter.

Solution

Step 1: Convert the given solubility from grams per 100 mL to grams per liter.

0.35 g/100 mL = 0.35 ×10

1g/L

= 3.5 g/L

Thus, the solubility of the compound in water at 25

C is 3.5 g/L.

Question 8

Question

A student performed a solubility experiment by mixing 10.0 g of compound X

in 100.0 mL of water at 25

C. After thorough mixing, only 0.25 g of compound

X dissolved in the water. Calculate the solubility of compound X in water at

C in g/L.

Solution

Step 1: Calculate the mass of compound X that remains undissolved. Given:

Mass of compound X added = 10.0 g Mass of compound X dissolved = 0.25 g

Mass of compound X remaining undissolved = 10.0 g - 0.25 g = 9.75 g

Step 2: Convert the volume of water to liters. 100.0 mL = 100.0 mL x (1 L

/ 1000 mL) = 0.1 L

Step 3: Calculate the solubility of compound X in water. Solubility (in g/L)

= (Mass of compound X dissolved / Volume of water) = (0.25 g / 0.1 L) = 2.5

g/L

Therefore, the solubility of compound X in water at 25

C is 2.5 g/L.

Question 9

Question

A student conducted an experiment to determine the solubility of two organic

compounds, Compound X and Compound Y, in three different solvents: water,

hexane, and ethanol. The results are summarized in the table below:

Compound Solvent Solubility (g/100 mL)

X Water 5.0

X Hexane 0.2

X Ethanol 10.0

Y Water 0.1

Y Hexane 15.0

Y Ethanol 2.0

Based on the solubility of Compound X and Compound Y, identify the most

polar compound and the most nonpolar compound. Justify your answer.

Solution

Step 1: Calculate the values of the solubility parameter δfor each compound in

the different solvents using the formula:

δ=qδ2

d+δ2

p+δ2

where δdis the dispersion component, δpis the polar component, and δh

is the hydrogen bonding component of the solubility parameter. The solubility

parameters for water, hexane, and ethanol are approximately 23, 7, and 12

(cal/cm3)0.5, respectively.

For Compound X in water:

δX=p(23)2+ (0)2+ (23)2= 23 (cal/cm3)0.5

For Compound X in hexane:

δX=p(7)2+ (0)2+ (0)2= 7 (cal/cm3)0.5

For Compound X in ethanol:

δX=p(12)2+ (5)2+ (15)2≈17 (cal/cm3)0.5

For Compound Y in water:

δY=p(23)2+ (0)2+ (23)2= 23 (cal/cm3)0.5

For Compound Y in hexane:

δY=p(7)2+ (0)2+ (0)2= 7 (cal/cm3)0.5

For Compound Y in ethanol:

δY=p(12)2+ (2)2+ (15)2≈20 (cal/cm3)0.5

Step 2: Analyze the calculated solubility parameters to determine the most

polar and the most nonpolar compound.

- Compound X has the highest solubility in ethanol, indicating it is the most

polar compound. - Compound Y has the highest solubility in hexane, indicating

it is the most nonpolar compound.

Question 10

Question

An organic compound has the following properties: - It is soluble in water. - It

is insoluble in hexane. - It is soluble in ethanol.

Explain the solubility behavior of this compound in terms of its polarity and

intermolecular interactions.

Solution

Step 1: Interpretation of solubility behavior - Solubility in water indicates that

the compound is capable of forming hydrogen bonds or has polar groups that

can interact with water molecules. - Insolubility in hexane suggests that the

compound lacks nonpolar groups or hydrophobic interactions necessary to dis-

solve in a nonpolar solvent like hexane. - Solubility in ethanol implies that the

compound is capable of forming hydrogen bonds or has polar groups that can

interact with ethanol molecules.

Step 2: Analysis of the compound’s polarity - The compound must contain

polar functional groups (such as hydroxyl, carbonyl, or amino groups) that

exhibit hydrogen bonding or have a high dipole moment to be soluble in water

and ethanol. - The compound must lack long nonpolar hydrocarbon chains or

rings typical of nonpolar substances to remain insoluble in hexane.

Step 3: Explanation of solubility in water - The compound is soluble in water

due to the presence of polar functional groups that can engage in hydrogen

bonding with water molecules. - These interactions disrupt the intermolecular

forces within the compound and facilitate dissolution in a polar solvent like

water.

Step 4: Explanation of insolubility in hexane - The compound is insoluble

in hexane because hexane is a nonpolar solvent that cannot effectively inter-

act with the polar functional groups present in the compound. - The lack of

compatible intermolecular forces prevents the compound from dissolving in a

nonpolar solvent like hexane.

Step 5: Explanation of solubility in ethanol - The compound is soluble in

ethanol due to the ability of polar functional groups to form hydrogen bonds

with ethanol molecules. - These interactions overcome the intermolecular forces

within the compound and allow for dissolution in a polar solvent like ethanol.

Question 11

Question

A student is given the task of determining the solubility of two compounds,

Compound A and Compound B, in different solvents. The student finds that

Compound A is soluble in both water and acetone, while Compound B is only

soluble in acetone but not in water. Based on this information, which compound

is more polar, Compound A or Compound B? Justify your answer.

Solution

To determine which compound is more polar, we need to consider the polarity

of the solvents in which they are soluble.

Step 1: Water is a polar solvent known for its ability to dissolve polar

compounds due to its high dielectric constant and polarity.

Step 2: Acetone is also a polar solvent that can dissolve polar and nonpolar

compounds due to its structure, consisting of a polar carbonyl group and a

nonpolar methyl group.

Step 3: If Compound A is soluble in both water and acetone, it indicates

that Compound A has some polar character, allowing it to interact with the

polar solvents. Since water is a more polar solvent than acetone, the fact that

Compound A is soluble in water as well indicates that it is quite polar.

Step 4: On the other hand, if Compound B is only soluble in acetone and

not in water, it suggests that Compound B has some nonpolar character that

allows it to interact favorably with the nonpolar portion of acetone but not with

water.

Step 5: Therefore, based on the solubility data provided, Compound A is

more polar than Compound B. Compound A’s ability to dissolve in water, a

highly polar solvent, indicates a higher level of polarity compared to Compound

Question 12

Question

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

of benzoic acid at this temperature is 6.54 ×10−5. Assume benzoic acid fully

dissociates in water.

Solution

Step 1: Write the dissociation equation for benzoic acid:

C7H6O2⇌C7H5O−

2+ H+

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

Ksp = [C7H5O−

2][H+]

Step 3: Since benzoic acid fully dissociates in water, the concentrations of

C7H5O−

2and H+are equal to the solubility of benzoic acid, denoted by ’S’:

Ksp = (S)(S) = S2= 6.54 ×10−5

S=p6.54 ×10−5= 0.00808 M

Therefore, the solubility of benzoic acid in water at 25◦C is 0.00808 M.

Question 13

Question

A student is given a mixture containing 5 g of biphenyl (molar mass = 154

g/mol) and 3 g of 1,4-dichlorobenzene (molar mass = 147 g/mol) in 100 mL of

water (density = 1.00 g/mL). Biphenyl is sparingly soluble in water, while 1,4-

dichlorobenzene is insoluble in water. Calculate the total mass of the mixture

that is soluble in water. Assume the densities of biphenyl and 1,4-dichlorobenzene

are both 1.00 g/mL.

Solution

Step 1: Calculate the moles of biphenyl and 1,4-dichlorobenzene.

Moles of biphenyl = Mass of biphenyl

Molar mass of biphenyl

Moles of biphenyl = 5 g

154 g/mol

Moles of biphenyl = 0.0325 mol

Moles of 1,4-dichlorobenzene = Mass of 1,4-dichlorobenzene

Molar mass of 1,4-dichlorobenzene

Moles of 1,4-dichlorobenzene = 3 g

147 g/mol

Moles of 1,4-dichlorobenzene = 0.0204 mol

Step 2: Calculate the volume of each solute in the mixture.

Volume of biphenyl = Moles of biphenyl ×Density of biphenyl

Volume of biphenyl = 0.0325 mol ×1.00 g/mL

Volume of biphenyl = 0.0325 mL

Volume of 1,4-dichlorobenzene = Moles of 1,4-dichlorobenzene ×Density of 1,4-dichlorobenzene

Volume of 1,4-dichlorobenzene = 0.0204 mol ×1.00 g/mL

Volume of 1,4-dichlorobenzene = 0.0204 mL

Step 3: Calculate the total volume of the mixture.

Total volume = Volume of biphenyl + Volume of 1,4-dichlorobenzene + Volume of water

Total volume = 0.0325 mL + 0.0204 mL + 100 mL

Total volume = 100.0529 mL

Step 4: Calculate the total mass of the mixture that is soluble in water.

Total mass of the mixture = Total volume ×Density of water

Total mass of the mixture = 100.0529 mL ×1.00 g/mL

Total mass of the mixture = 100.0529 g

Therefore, the total mass of the mixture that is soluble in water is 100.0529

Question 14

Question

A student is performing a solubility experiment in which they dissolve 5.0 g

of compound X in 100 mL of water at 25◦C. The student then measures the

solubility of compound X in water to be 1.5 g/mL at that temperature.

Calculate the molar solubility of compound X in water.

Solution

Step 1: Calculate the total volume of the solution in liters. Given that the

student dissolved 5.0 g of compound X in 100 mL of water, the total volume of

the solution is 100 mL = 0.1 L.

Step 2: Calculate the amount of compound X dissolved in the solution. The

amount of compound X dissolved in the solution is 5.0 g.

Step 3: Use the amount of compound X and the total volume of the solution

to calculate the molarity. The molarity (M) of compound X can be calculated

using the formula:

Molarity (M) = moles of solute (compound X)

liters of solution

Step 4: Calculate the moles of compound X. The moles of compound X can

be calculated using the formula:

moles = mass (g)

molar mass (g/mol)

Step 5: Find the molar mass of compound X. Without the specific molecular

formula of compound X, we cannot calculate the molar mass or the moles of

compound X. Therefore, we are unable to determine the molar solubility of

compound X in water at this time.

Question 15

Question

In a lab experiment, a student mixed 10 mL of acetone (density = 0.785 g/mL)

with 15 mL of water. The densities of acetone and water are 0.785 g/mL and

1.00 g/mL, respectively. Calculate the weight percentage of acetone in the

resulting solution.

Solution

Step 1: Calculate the mass of acetone and water used. Given that density =

mass/volume, we can rearrange the formula to find the mass: For acetone: mass

= density ×volume = 0.785 g/mL ×10 mL = 7.85 g

For water: mass = density ×volume = 1.00 g/mL ×15 mL = 15.00 g

Step 2: Calculate the total mass of the solution. Total mass = mass of

acetone + mass of water = 7.85 g + 15.00 g = 22.85 g.

Step 3: Calculate the weight percentage of acetone. Weight % of acetone =

mass of acetone

total mass ×100% Weight % of acetone = 7.85 g

22.85 g ×100% Weight % of

acetone = 7.85

22.85 ×100% Weight % of acetone ≈34.36%

Therefore, the weight percentage of acetone in the resulting solution is ap-

proximately 34.36

Question 16

Question

A student is given a sample of a white crystalline compound and asked to deter-

mine whether the compound is soluble in water or not. The student performed

the following tests and obtained the following information:

1. The compound is insoluble in water. 2. The compound is soluble in

diethyl ether and hexane. 3. The compound has a molar mass of 120 g/mol.

Based on the information provided, determine the likely functional group

present in the compound and explain the solubility results obtained.

Solution

Step 1: Calculate the degree of unsaturation (DU) of the compound based on

the molar mass. The formula for calculating the degree of unsaturation is:

DU =(2C+2+N−X−H)

Where: - C= Number of carbons - N= Number of nitrogens - X= Number

of halogens - H= Number of hydrogens

Given that the molar mass is 120 g/mol, we can deduce that the compound

likely contains 6 carbons (C), 12 hydrogens (H), and 2 oxygens (O).

By substituting these values into the formula, we can calculate the degree

of unsaturation:

DU =(2(6) + 2 −12 −24)

2=12 −12 −24

2=−24

2=−12

Step 2: Interpret the degree of unsaturation. A negative degree of unsatura-

tion indicates that the compound has more hydrogens than a saturated alkane

with the same number of carbons. This suggests the presence of functional

groups that increase the hydrogen to carbon ratio, such as halogens or oxygen-

containing functional groups.

Step 3: Based on the solubility tests results, determine the likely functional

group. - Insolubility in water suggests the absence of ionic groups like carboxylic

acids, amines, or ionic salts. It also hints at the absence of very polar groups.

- Solubility in diethyl ether and hexane indicates non-polar or slightly polar

behavior.

Step 4: Conclusion Given the insolubility in water and solubility in non-

polar solvents, the likely functional group present in the compound is an ester.

Esters are typically soluble in non-polar solvents like diethyl ether and hexane

but insoluble in water due to their limited hydrogen bonding capabilities.

Question 17

Question

A student is given a mixture of three compounds: compound A, compound B,

and compound C. They are asked to separate the compounds based on their

solubility in different solvents. The student knows that compound A is soluble in

water, compound B is soluble in ethanol, and compound C is soluble in diethyl

ether.

If the student mixes the mixture with water first, then transfers the insoluble

portion to ethanol, and finally transfers the remaining insoluble portion to di-

ethyl ether, what compounds will be separated into each solvent layer? Justify

your answer based on the solubility properties of the compounds.

Solution

Step 1: Solubility of compounds in water - Compound A is soluble in water. It

will dissolve in the water layer. - Compound B is not soluble in water. It will

remain in the insoluble portion. - Compound C is not soluble in water. It will

remain in the insoluble portion.

Step 2: Transfer insoluble portion to ethanol - The insoluble portion from the

water layer contains compounds B and C. - Compound B is soluble in ethanol.

It will dissolve in the ethanol layer. - Compound C is not soluble in ethanol. It

will remain in the insoluble portion.

Step 3: Transfer remaining insoluble portion to diethyl ether - The remaining

insoluble portion contains compound C. - Compound C is soluble in diethyl

ether. It will dissolve in the diethyl ether layer.

Therefore, compound A will be separated in the water layer, compound B

in the ethanol layer, and compound C in the diethyl ether layer based on their

solubility properties in different solvents.

Question 18

Question

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

C. The Ksp of

benzoic acid is 6.6×10−5mol2/L2and the density of water is 1.0 g/mL.

Solution

Step 1: Write the equilibrium equation for the dissolution of benzoic acid in wa-

ter. The equilibrium equation for the dissolution of benzoic acid (C6H5COOH)

in water is:

C6H5COOH ⇌C6H5COO−+H+

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

solubility product constant (Ksp) is given by:

Ksp = [C6H5COO−][H+]

Since benzoic acid is a weak acid, we can assume that the concentration of

H+ions is equal to the concentration of C6H5COO−ions.

Step 3: Convert Ksp to concentration units. Given that Ksp = 6.6×

10−5mol2/L2, we need to take the square root to get the concentration of

C6H5COO−ions:

pKsp =p6.6×10−5mol/L = 0.0081 mol/L

Step 4: Convert the concentration to g/L. Using the molar mass of benzoic

acid, C6H5COOH, which is approximately 122.12 g/mol, we can calculate the

solubility in grams per liter:

0.0081 mol/L ×122.12 g/mol = 0.98852 g/L

Therefore, the solubility of benzoic acid in water at 25

C is approximately

0.99 g/L.

Question 19

Question

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

acid has a solubility product of 6.68 ×10−3mol/L.

Solution

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

equation is:

C7H6O2(s)⇌C7H6O2(aq)

Step 2: Set up the solubility product expression.

Ksp = [C7H6O2]=6.68 ×10−3mol/L

Step 3: Determine the molar mass of benzoic acid.

Molar mass of benzoic acid (C7H6O2)=7×molar mass of carbon+6×molar mass of hydrogen+2×molar mass of oxygen

Step 4: Calculate the molar solubility of benzoic acid. Since 1 mol of benzoic

acid dissociates into 1 mol of C7H6O2ions:

6.68 ×10−3mol/L = x2

1−x

Step 5: Solve for x to find the molar solubility, which gives:

x= 0.082 mol/L

Therefore, the solubility of benzoic acid in water at 25◦C is 0.082 mol/L.

Question 20

Question

Determine the solubility of compound X in water at 25

C based on the following

data: - Solute: Compound X, molecular weight = 180 g/mol - Solvent: Water,

density = 1.00 g/mL, molar mass = 18.02 g/mol - The observed solubility of

compound X in water is 5.0 g/100 mL.

Solution

Step 1: Calculate the molar solubility of compound X in water. Step 2: Cal-

culate the molar solubility in mol/L. Step 3: Determine the solubility product

constant (Ksp) of compound X in water at 25

Step 1: To find the molar solubility of compound X, we need to convert the

observed solubility from grams per 100 mL to grams per liter:

5.0 g/100 mL ×1 mL

1.00 g = 0.050 g/L

Step 2: Next, we convert the molar solubility from grams per liter to mol/L.

Molar solubility of compound X:

0.050 g/L

180 g/mol = 2.78 ×10−4mol/L

Step 3: The solubility product constant (Ksp) is calculated as the product

of the molar concentrations of the ions in the compound at saturation. For the

compound X:

Ksp = [X]2= (2.78 ×10−4)2= 7.73 ×10−8mol2/L2

Therefore, the solubility product constant of compound X in water at 25

is 7.73 ×10−8mol2/L2.

Question 21

Question

An organic compound has the molecular formula C10H16O. When 2.5 g of the

compound is dissolved in 100 mL of water at 25◦C, a clear solution is obtained.

However, when 2.5 g of the same compound is dissolved in 10 mL of chloroform

at 25◦C, a clear solution is also obtained. Based on these observations, answer

the following questions:

a) Calculate the molar mass of the compound. b) Using the given solubility

data, discuss the compound’s polarity. c) Predict whether this compound is

more likely to be an alcohol or a ketone.

Solution

a) To calculate the molar mass of the compound, we need to first find the

number of moles of the compound dissolved in each solvent.

Step 1: Calculate the number of moles of the compound in water.

moles in water = mass

molar mass =2.5 g

molar mass

Step 2: Calculate the number of moles of the compound in chloroform.

moles in chloroform = mass

molar mass =2.5 g

molar mass

Since the number of moles of the compound is the same in both solvents,

the molar mass of the compound is the same for both cases.

b) The compound is soluble in both water and chloroform. Water is a polar

solvent, while chloroform is a nonpolar solvent. Since the compound is soluble

in both polar and nonpolar solvents, it likely has both polar and nonpolar

characteristics.

c) Based on the information given, the compound is more likely to be an

alcohol rather than a ketone. Alcohols tend to have both polar and nonpolar

characteristics, allowing them to dissolve in both polar and nonpolar solvents.

On the other hand, ketones are less likely to dissolve in water due to their

nonpolar nature.

Question 22

Question

Calculate the solubility of 2,4-dinitrophenol (C6H4N2O5) in water at 25

The partition coefficient of 2,4-dinitrophenol between water and octanol is 160.

Assume that the volume of the solution is 1 L.

Solution

Step 1: Write out the partition coefficient equation: The partition coefficient,

K, is defined as the ratio of the concentrations of a solute in the two phases.

Mathematically, this can be expressed as:

K=[Solute in octanol]

[Solute in water]

Step 2: Write the expression for the solubility of 2,4-dinitrophenol in water:

Let the solubility of 2,4-dinitrophenol in water be represented by xmol/L.

Therefore, the concentration of the solute in each phase can be represented as

follows:

[2,4-Dinitrophenol in water] = xmol/L

[2,4-Dinitrophenol in octanol] = 160xmol/L

Step 3: Write the equilibrium expression for the partitioning of 2,4-dinitrophenol:

At equilibrium, the amount of solute in water and octanol does not change. This

can be represented as:

K=160x

Step 4: Solve for xto find the solubility of 2,4-dinitrophenol in water:

160 = 160x

160 = 160

Since this is a physical impossibility, there seems to be an issue in the given

data or question formulation.

Question 23

Question

A student is given a compound with the molecular formula C6H12O6. The

compound is soluble in water and has a molar mass of 180 g/mol. The student

suspects that the compound is an aldohexose. 1. Based on the information

given, what are the potential isomers of the compound as aldohexoses? 2.

Determine the degree of unsaturation for each potential aldohexose isomer.

Solution

1. Based on the molecular formula C6H12O6, the potential aldohexose isomers

are glucose, galactose, and mannose. These isomers are all constitutional iso-

mers of each other.

2. To determine the degree of unsaturation for each potential aldohexose

isomer, we first calculate the number of hydrogen atoms in a saturated hy-

drocarbon with the same formula. The formula for counting the number of

hydrogen atoms in a saturated hydrocarbon is 2n+ 2, where nis the number

of carbon atoms. For an aldohexose with 6 carbon atoms: Number of hydrogen

atoms in a saturated hydrocarbon = 2 ×6 + 2 = 14

Next, we calculate the actual number of hydrogen atoms in each potential

aldohexose isomer using the given molecular formula. a. Glucose: C6H12O6

Number of hydrogen atoms in glucose = 12 −6 + 1

2×6 + 1 = 6 Degree of un-

saturation = Number of hydrogen atoms in a saturated hydrocarbon - Number

of hydrogen atoms in the molecule Degree of unsaturation for glucose = 14 - 6

= 8

b. Galactose: C6H12O6Number of hydrogen atoms in galactose = 12 −6 +

2×6 + 1 = 6 Degree of unsaturation for galactose = 14 - 6 = 8

c. Mannose: C6H12O6Number of hydrogen atoms in mannose = 12 −6 +

2×6 + 1 = 6 Degree of unsaturation for mannose = 14 - 6 = 8

Therefore, the degree of unsaturation for each aldohexose isomer (glucose,

galactose, and mannose) is 8.

Question 24

Question

Calculate the solubility of benzoic acid (C6H5COOH) in grams per liter at 25

in water. Given that the solubility of benzoic acid in water is 3.60 g/L at 100

and the solubility obeys Raoult’s law.

Solution

Step 1: Calculate the solubility of benzoic acid in water at 25

C using Raoult’s

law.

Raoult’s law states that the vapor pressure of a solvent above the solution

is proportional to the mole fraction of the solvent in the solution. The solution

can be represented as:

Psolution =Xsolvent ·P◦

solvent

where: Psolution = vapor pressure of the solution Xsolvent = mole fraction of

the solvent P◦

solvent = vapor pressure of the pure solvent

Since we are looking at solubility, we can relate this to the solubility of

benzoic acid.

Step 2: Calculate the mole fraction of benzoic acid (C6H5COOH) in the

solution at 25

Given that the solubility of benzoic acid in water at 100

C is 3.60 g/L and

obeys Raoult’s law, we can assume that the solubility at 25

C is lower since the

solubility generally decreases with decreasing temperature.

Step 3: Calculate the solubility of benzoic acid in water at 25

C in grams

per liter.

The formula mass of benzoic acid is calculated as follows:

C6H5COOH : 6(12.01) + 5(1.008) + 12.01 + 16.00 + 16.00 = 122.12 g/mol

Since we’ve determined the mole fraction of benzoic acid, we can now calcu-

late the solubility in grams per liter at 25

Question 25

Question

Calculate the solubility of benzoic acid (C6H5COOH) in water at 25

C. Given

that the solubility of benzoic acid in water is 3.4 g/L.

Solution

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

acid (C6H5COOH) is calculated as follows: Molar mass = (6 ×12.01) + (5 ×

1.01) + 12.01 + 16.00 + 16.00 = 122.12 g/mol

Step 2: Calculate the solubility of benzoic acid in mol/L Given that the

solubility of benzoic acid in water is 3.4 g/L, we can convert this to mol/L using

the calculated molar mass: Solubility =3.4g/L

122.12 g/mol = 0.0278 mol/L

Step 3: Calculate the solubility product constant (Ksp) of benzoic acid in

water at 25

C. The solubility product constant (Ksp) can be calculated as:

Ksp = [C6H5COO-][H+] = x2Since benzoic acid is a weak acid, it will partially

dissociate in water: C6H5COOH ⇌C6H5COO- + H+ Therefore, the expression

for the equilibrium partial pressures can be written as: Ksp =xmol/L

1L2Given

that it is a weak acid, we can make the assumption that x is small and can be

neglected compared to the initial concentration: Ksp = (0.0278)2= 7.73 ×10−4

Therefore, the solubility product constant (Ksp) of benzoic acid in water at

C is 7.73 ×10−4.

Question 26

Question

A student is trying to determine the solubility of a compound in different sol-

vents. In the first experiment, the student finds that 5 grams of the compound

dissolve in 100 mL of water at room temperature. In the second experiment, 10

grams of the compound dissolve in 100 mL of diethyl ether at room temperature.

Determine the solubility of the compound in each solvent based on the given

data. Which solvent is more polar based on the solubility of the compound?

Solution

To determine the solubility of the compound in each solvent, we need to calculate

the solubility in grams per 100 mL of solvent.

Step 1: Calculate the solubility in water Given: - Mass of compound

dissolved in water = 5 grams - Volume of water = 100 mL

The solubility of the compound in water is:

Solubility in water = 5 g

100 mL = 5 g/100 mL

Step 2: Calculate the solubility in diethyl ether Given: - Mass of

compound dissolved in diethyl ether = 10 grams - Volume of diethyl ether =

100 mL

The solubility of the compound in diethyl ether is:

Solubility in diethyl ether = 10 g

100 mL = 10 g/100 mL

Step 3: Determine the polarity of each solvent based on solubility

- Water: Solubility = 5 g/100 mL - Diethyl ether: Solubility = 10 g/100 mL

Since the compound is more soluble in diethyl ether (10 g/100 mL) compared

to water (5 g/100 mL), we can conclude that diethyl ether is more polar than

water.

Question 27

Question

A chemist is experimenting with a new compound and observes that it is soluble

in water but insoluble in hexane. The compound consists of a benzene ring

attached to a carboxylic acid functional group. Explain the solubility behavior

of this compound in water and hexane.

Solution

To understand the solubility behavior of the compound in question, we need to

consider the polarity of the compound and the solvents.

Step 1: Determine the Polarity of the Compound The compound has

a benzene ring attached to a carboxylic acid functional group. The benzene ring

is nonpolar due to the delocalized electrons, while the carboxylic acid group is

polar due to the electronegative oxygen atom. Overall, the compound is polar.

Step 2: Consider the Polarity of the Solvents Water is a highly polar

solvent due to its ability to form hydrogen bonds. Hexane, on the other hand,

is a nonpolar solvent as it only exhibits London dispersion forces.

Step 3: Explanation of Solubility in Water Since the compound is polar

and water is a polar solvent, the compound can dissolve in water through dipole-

dipole interactions or hydrogen bonding with the carboxylic acid functional

group. This explains why the compound is soluble in water.

Step 4: Explanation of Insolubility in Hexane Since hexane is a nonpo-

lar solvent and the compound is polar, there are no strong intermolecular forces

of attraction between the compound and hexane. As a result, the compound is

insoluble in hexane.

Therefore, the compound is soluble in water due to its polar nature and the

ability to form interactions with water molecules, while it is insoluble in hexane

due to the lack of strong intermolecular forces between the compound and the

nonpolar hexane molecules.

Question 28

Question

A student is experimenting with different organic compounds in the lab. They

are asked to determine the solubility of compound X in various solvents. After

testing, the student observes that compound X is soluble in hexane but insoluble

in water. Based on this information, determine the most likely functional group

present in compound X and explain the solubility behavior observed.

Solution

Step 1: Solubility in hexane - Hexane is a nonpolar solvent. - The fact that

compound X is soluble in hexane indicates that it is likely nonpolar or has a

nonpolar functional group. - Examples of nonpolar functional groups include

alkyl groups, halogens, and aromatic rings.

Step 2: Solubility in water - Water is a polar solvent. - The fact that

compound X is insoluble in water suggests that it is likely nonpolar or lacks

polar functional groups. - Polar functional groups such as hydroxyl (-OH),

carbonyl (C=O), and amino (-NH2) groups are typically required for solubility

in water due to hydrogen bonding with water molecules.

Step 3: Conclusion - Based on the solubility behavior observed, compound

X likely contains nonpolar functional groups or is overall nonpolar in nature. -

Possible functional groups in compound X could include alkyl groups, aromatic

rings, or halogens. - It is less likely to contain polar functional groups like

hydroxyl, carbonyl, or amino groups, which are required for solubility in water.

Therefore, the most likely functional group present in compound X is an

alkyl group, an aromatic ring, or a halogen.

Question 29

Question

A student is conducting an experiment to determine the solubility of caffeine in

different solvents. The student finds that caffeine is soluble in water, dichloromethane,

and ethanol, but insoluble in hexane. Based on this information, rank the sol-

vents in order of increasing polarity.

Solution

Step 1: Understand the concept of solubility and polarity - Solubility refers to

the ability of a solute to dissolve in a solvent to form a homogenous mixture.

- Polarity is a measure of the separation of charge within a molecule. Polar

molecules have an uneven distribution of electron density, leading to partially

positive and partially negative ends.

Step 2: Determine the polarity of each solvent - Water is a highly polar

solvent due to its bent molecular structure and polarity of the O-H bonds. -

Dichloromethane is a polar solvent due to its dipole moment caused by the

difference in electronegativity between C and Cl atoms. - Ethanol is polar

solvent due to the presence of the hydroxyl group, making it capable of forming

hydrogen bonds. - Hexane is a nonpolar solvent with a symmetrical molecular

structure and no polar functional groups.

Step 3: Rank the solvents in order of increasing polarity Therefore, the

solvents can be ranked in order of increasing polarity as follows: Hexane ¡

Dichloromethane ¡ Ethanol ¡ Water.

Question 30

Question

Calculate the partition coefficient (P) for the following solute between water

and diethyl ether at 25

Solute: 2,4-dimethylpentane Water solubility: 0.015 g/L Diethyl ether solu-

bility: 3.8 g/L

Solution

Step 1: Calculate the concentration of the solute in each solvent. The con-

centration of the solute in water can be calculated as follows: Concentration

in water = 0.015 g/L The concentration of the solute in diethyl ether can be

calculated as follows: Concentration in diethyl ether = 3.8 g/L

Step 2: Calculate the partition coefficient (P). The partition coefficient (P)

is defined as the ratio of the concentration of the solute in diethyl ether to the

concentration of the solute in water.

P=Concentration in diethyl ether

Concentration in water

P=3.8

0.015

P= 253.33

Therefore, the partition coefficient for 2,4-dimethylpentane between water

and diethyl ether at 25

C is 253.33.

Question 31

Question

A student is given a mixture of two compounds in a test tube. Compound A is

known to be very soluble in water, while compound B is known to be insoluble

in water. The student needs to separate the two compounds using a solvent

extraction.

If the student adds an equal volume of water and dichloromethane to the

test tube and shakes it vigorously, which compound will predominantly dissolve

in which layer? Justify your answer using principles of solubility and polarity.

Solution

To determine which compound will predominantly dissolve in which layer, we

need to consider the solubility and polarity of the compounds along with the

solvents being used.

Step 1: Compound A is very soluble in water, while compound B is insoluble

in water. Water is a polar solvent, which means it can dissolve polar compounds

well. Dichloromethane (CH2Cl2) is a nonpolar solvent.

Step 2: Based on the principles of ”like dissolves like,” we know that polar

compounds are more likely to dissolve in polar solvents, and nonpolar com-

pounds are more likely to dissolve in nonpolar solvents. Since compound A is

soluble in water (a polar solvent), it will predominantly dissolve in the water

layer.

Step 3: On the other hand, compound B, which is insoluble in water, will

not dissolve in the water layer. Instead, it will predominantly dissolve in the

nonpolar dichloromethane layer.

Therefore, after shaking the test tube vigorously and allowing it to settle, we

can expect compound A to be predominantly in the water layer and compound B

to be predominantly in the dichloromethane layer during the solvent extraction

process.

Question 32

Question

A student wants to dissolve a nonpolar compound in water. Explain whether

this process will be spontaneous or nonspontaneous based on the principles of

solubility and polarity. Defend your answer with a detailed explanation.

Solution

Step 1: First, we need to consider the polarity of the nonpolar compound and

the polarity of water. Water is a polar molecule due to its electronegativity

difference between hydrogen and oxygen atoms. Nonpolar compounds, on the

other hand, have no significant difference in electronegativity within their struc-

ture.

Step 2: The process of dissolving a nonpolar compound in water would be

nonspontaneous. This is because ”like dissolves like” – polar substances tend to

dissolve in polar solvents, while nonpolar substances tend to dissolve in nonpolar

solvents.

Step 3: In the case of dissolving a nonpolar compound in water, the interac-

tions between the nonpolar compound and water molecules would be weak due

to the difference in polarity. Therefore, the energy required to break the exist-

ing intermolecular forces in water and the nonpolar compound and the energy

released by forming new solute-solvent interactions would not be favorable.

Step 4: As a result, the process of dissolving a nonpolar compound in water

would not occur spontaneously. Additional energy input, such as mixing or

stirring, would be required to overcome the unfavorable interactions between

the nonpolar compound and water molecules.

Step 5: In conclusion, based on the principles of solubility and polarity, the

process of dissolving a nonpolar compound in water would be nonspontaneous

due to the mismatch in polarity between the solute and solvent molecules.

Question 33

Question

A student is given a mixture of three organic compounds and asked to separate

them based on their solubility in different solvents. The three compounds are

Compound A, Compound B, and Compound C. Compound A is soluble in

water but insoluble in hexane. Compound B is insoluble in water but soluble

in hexane. Compound C is soluble in both water and hexane.

If the student adds the mixture to water, hexane, and a mixture of water

and hexane, predict in which solvent each compound will dissolve. Justify your

predictions using principles of solubility and polarity calculations.

Solution

To predict the solubility of each compound in the given solvents (water, hexane,

and a mixture of water and hexane), we need to consider the polarity of the

compounds and the polarity of the solvents.

Step 1: Determine the Polarity of Each Compound: - Compound A

is soluble in water but insoluble in hexane. This suggests that Compound A

is polar since water is a polar solvent. - Compound B is soluble in hexane but

insoluble in water. This suggests that Compound B is nonpolar since hexane

is a nonpolar solvent. - Compound C is soluble in both water and hexane,

indicating that it contains both polar and nonpolar functional groups.

Step 2: Consider the Polarity of the Solvents: - Water is a polar

solvent. - Hexane is a nonpolar solvent. - A mixture of water and hexane would

be considered polar since water is polar.

Step 3: Predict the Solubility of Each Compound in the Solvents: -

Compound A: Soluble in water (polar solvent) but insoluble in hexane (nonpolar

solvent). Compound A will dissolve in the mixture of water and hexane but

preferentially dissolve in water. - Compound B: Soluble in hexane (nonpolar

solvent) but insoluble in water (polar solvent). Compound B will dissolve in the

mixture of water and hexane but preferentially dissolve in hexane. - Compound

C: Soluble in both water and hexane. Compound C will dissolve in the mixture

of water and hexane but may distribute between the two layers based on the

relative proportions of water and hexane in the mixture.

Therefore, based on the principles of solubility and polarity calculations,

the predictions for the solubility of each compound in the given solvents are:

- Compound A: Water ¿ Water/hexane ¿ Hexane - Compound B: Hexane ¿

Water/hexane ¿ Water - Compound C: Water/hexane ¿ Water, Hexane

Question 34

Question

Calculate the solubility (in g/L) of benzoic acid (C7H6O2, molar mass 122.12

g/mol) in water at 25◦C. The solubility product of benzoic acid in water is

3.3×10−3M. Assume that the density of water is 1.00 g/mL.

Solution

Step 1: Calculate the molar solubility of benzoic acid by using the solubility

product constant:

Ksp = [C7H6O2][H2O] = 3.3×10−3M

Since the moles of benzoic acid that dissolve in water are equal to the moles of

H2O formed, we can assume that [H2O] = [C7H6O2]. Therefore,

[C7H6O2]2= 3.3×10−3

[C7H6O2] = p3.3×10−3M

[C7H6O2]=5.7×10−2M

Step 2: Convert molar solubility to g/L:

[C7H6O2] = g

L×1

molar mass

5.7×10−2M = g

L×1

122.12 g/mol

g/L = 5.7×10−2×122.12

g/L = 6.95 g/L

Therefore, the solubility of benzoic acid in water at 25◦C is 6.95 g/L.

Question 35

Question

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

C, given that

its solubility in pure water is 3.5 g/L. The partition coefficient of benzoic acid

between diethyl ether and water is 10. Determine the solubility of benzoic acid

in diethyl ether at 25

Solution

Step 1: Calculate the molar mass of benzoic acid (C7H6O2). The molar mass

of benzoic acid is:

Molar Mass = 7×Atomic M ass(C)+6×Atomic Mass(H)+2×Atomic Mass(O)

Step 2: Calculate the number of moles of benzoic acid that can dissolve in

water at 25

C. Given: Solubility of benzoic acid in water = 3.5 g/L

Moles =Mass

Molar Mass

Step 3: Use the partition coefficient to determine the solubility of benzoic

acid in diethyl ether at 25

C. The partition coefficient (K) is defined as the ratio

of the solubility of the solute in two immiscible solvents, in this case diethyl ether

and water.

K=Concentration of solute in diethyl ether

Concentration of solute in water

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

Using the formula for partition coefficient:

K=[Solute]diethyl ether

[Solute]water

[Solute]diethyl ether =K×[Solute]water

Now, substitute the given values of solubility in water and partition coeffi-

cient to find the solubility in diethyl ether.

Using the solubility information given, we can calculate the maximum amount

that can be dissolved:

Maximum amount of compound = solubility ×volume of solvent

Maximum amount of compound = 25 g/L ×0.1 L = 2.5 g

Step 2: Determine if the compound will fully dissolve in the water. The

compound has a molecular weight of 180 g/mol. Let’s calculate how many

moles of compound are present in 2.5 g:

moles of compound = mass of compound

molecular weight =2.5 g

180 g/mol ≈0.0139 mol

Step 3: Determine if the compound will fully dissolve in the water. To

determine if the compound will fully dissolve in water, we need to calculate the

molarity of the compound:

Molarity = moles of compound

volume of solvent (L) =0.0139 mol

0.1 L = 0.139 M

Since the molarity of the compound is less than its solubility in water (0.139

M ¡ 25 g/L), the compound will fully dissolve in the water.

Question 3

Question

An organic compound is dissolved in water to form a solution. The compound

has a solubility of 0.15 g per 100 mL of water at 25

C. Its calculated molecular

weight is 120 g/mol. Determine the polarity index of the compound.

Solution

Step 1: Calculate the molarity of the compound in the solution. Given that the

solubility of the compound is 0.15 g per 100 mL of water, we first convert this

to g/L:

0.15 g ×1 L

1000 mL = 0.0015 g/L

Now, calculate the molarity using the formula:

Molarity = moles of solute

liters of solution

Since we know the molecular weight of the compound is 120 g/mol, we can

find the moles of solute:

moles = mass

molecular weight =0.0015 g

120 g/mol = 1.25 ×10−5mol

The volume of the solution is 100 mL = 0.1 L, so:

Molarity = 1.25 ×10−5mol

0.1 L = 1.25 ×10−4M

Step 2: Calculate the polarity index. The polarity index is a measure of how

polar or nonpolar a compound is. It is defined as the Molarity divided by the

product of the molecular weight and solubility in water:

Polarity Index = Molarity

(Molecular weight ×Solubility)

Plugging in the values:

Polarity Index = 1.25 ×10−4M

(120 g/mol ×0.0015 g/L)

Polarity Index = 1.25 ×10−4

0.18

Polarity Index = 6.94 ×10−4

Therefore, the calculated polarity index of the compound is 6.94 ×10−4.

Question 4

Question

A student is given two compounds, Compound A and Compound B, and asked

to determine which compound is more soluble in water. Compound A has

a molecular weight of 180 g/mol and a logarithm of the partition coefficient

(log P) of 0.5, while Compound B has a molecular weight of 250 g/mol and a

logarithm of the partition coefficient (log P) of 1.2. Based on this information,

which compound is more soluble in water?

Solution

To determine which compound is more soluble in water, we will use the partition

coefficient (P) formula:

P=Concentration in organic phase

Concentration in aqueous phase

The partition coefficient is related to the logarithm of the partition coefficient

by the equation:

logP= log Concentration in organic phase

Concentration in aqueous phase

Taking the antilog of both sides of the equation gives:

P= 10logP

Now, we can substitute the given log P values to find the partition coefficients

for Compound A and Compound B.

Step 1: Calculate Pfor Compound A

For Compound A with log P = 0.5:

PA= 100.5= 3.162

Step 2: Calculate Pfor Compound B

For Compound B with log P = 1.2:

PB= 101.2= 15.849

Step 3: Determine which compound is more soluble in water

Since a higher partition coefficient indicates higher solubility in the organic

phase relative to the aqueous phase, Compound B, with a higher partition

coefficient (PB= 15.849), is more soluble in water compared to Compound

A (PA= 3.162).

Question 5

Question

Calculate the solubility of compound X in water at 25

C given that the solubility

product constant (Ksp) for X is 5.0×10−5mol2/L2.

Solution

Step 1: Write the solubility equilibrium for compound X: Let the solubility of

compound X be represented by the variable x. Then the equilibrium equation

for the dissolution of compound X in water can be written as:

X⇌X2+ + 2X−

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

solubility product constant (Ksp) expression for compound X can be written as:

Ksp = [X2+][X−]2=x(2x)2= 4x3

Step 3: Substitute the given Ksp and solve for x: Given that Ksp = 5.0×10−5

mol2/L2, we can set up the equation as:

5.0×10−5= 4x3

Step 4: Solve for x:

x=3

r5.0×10−5

x=3

p1.25 ×10−5

x≈0.0221 mol/L

Therefore, the solubility of compound X in water at 25

C is approximately

0.0221 mol/L.

Question 6

Question

A student performed a solubility experiment and found that 2.0 grams of a

compound dissolved in 10 mL of water at 25

C. The student then measured the

solubility of the same compound in ethanol and found that 4.0 grams dissolved

in 10 mL of ethanol at 25

C. Based on these observations, calculate the partition

coefficient (Kow) of the compound between water and ethanol at 25

Solution

Step 1: Calculate the solubility of the compound in water and ethanol. In water:

Solubility in water = 2.0 g

10 mL = 0.2 g/mL

In ethanol:

Solubility in ethanol = 4.0 g

10 mL = 0.4 g/mL

Step 2: Calculate the partition coefficient (Kow) using the formula:

Kow =Solubility in water

Solubility in ethanol

Substitute the values to find Kow:

Kow =0.2 g/mL

0.4 g/mL = 0.5

Therefore, the partition coefficient (Kow) of the compound between water

and ethanol at 25

C is 0.5.

Question 7

Question

An organic compound has a solubility of 0.35 g/100 mL in water at 25

C. Cal-

culate the solubility of the compound in grams per liter.

Solution

Step 1: Convert the given solubility from grams per 100 mL to grams per liter.

0.35 g/100 mL = 0.35 ×10

1g/L

= 3.5 g/L

Thus, the solubility of the compound in water at 25

C is 3.5 g/L.

Question 8

Question

A student performed a solubility experiment by mixing 10.0 g of compound X

in 100.0 mL of water at 25

C. After thorough mixing, only 0.25 g of compound

X dissolved in the water. Calculate the solubility of compound X in water at

C in g/L.

Solution

Step 1: Calculate the mass of compound X that remains undissolved. Given:

Mass of compound X added = 10.0 g Mass of compound X dissolved = 0.25 g

Mass of compound X remaining undissolved = 10.0 g - 0.25 g = 9.75 g

Step 2: Convert the volume of water to liters. 100.0 mL = 100.0 mL x (1 L

/ 1000 mL) = 0.1 L

Step 3: Calculate the solubility of compound X in water. Solubility (in g/L)

= (Mass of compound X dissolved / Volume of water) = (0.25 g / 0.1 L) = 2.5

g/L

Therefore, the solubility of compound X in water at 25

C is 2.5 g/L.

Question 9

Question

A student conducted an experiment to determine the solubility of two organic

compounds, Compound X and Compound Y, in three different solvents: water,

hexane, and ethanol. The results are summarized in the table below:

Compound Solvent Solubility (g/100 mL)

X Water 5.0

X Hexane 0.2

X Ethanol 10.0

Y Water 0.1

Y Hexane 15.0

Y Ethanol 2.0

Based on the solubility of Compound X and Compound Y, identify the most

polar compound and the most nonpolar compound. Justify your answer.

Solution

Step 1: Calculate the values of the solubility parameter δfor each compound in

the different solvents using the formula:

δ=qδ2

d+δ2

p+δ2

where δdis the dispersion component, δpis the polar component, and δh

is the hydrogen bonding component of the solubility parameter. The solubility

parameters for water, hexane, and ethanol are approximately 23, 7, and 12

(cal/cm3)0.5, respectively.

For Compound X in water:

δX=p(23)2+ (0)2+ (23)2= 23 (cal/cm3)0.5

For Compound X in hexane:

δX=p(7)2+ (0)2+ (0)2= 7 (cal/cm3)0.5

For Compound X in ethanol:

δX=p(12)2+ (5)2+ (15)2≈17 (cal/cm3)0.5

For Compound Y in water:

δY=p(23)2+ (0)2+ (23)2= 23 (cal/cm3)0.5

For Compound Y in hexane:

δY=p(7)2+ (0)2+ (0)2= 7 (cal/cm3)0.5

For Compound Y in ethanol:

δY=p(12)2+ (2)2+ (15)2≈20 (cal/cm3)0.5

Step 2: Analyze the calculated solubility parameters to determine the most

polar and the most nonpolar compound.

- Compound X has the highest solubility in ethanol, indicating it is the most

polar compound. - Compound Y has the highest solubility in hexane, indicating

it is the most nonpolar compound.

Question 10

Question

An organic compound has the following properties: - It is soluble in water. - It

is insoluble in hexane. - It is soluble in ethanol.

Explain the solubility behavior of this compound in terms of its polarity and

intermolecular interactions.