Quiz

Question 1

As a pharmacy student, your knowledge in physical pharmacy may be useful in:
-an understanding of the basic physical chemistry necessary for the drug design.
-increasing the absorption of drug upon administration.
-the manufacture of medicines on both a small and a large scale.
-the cultivation, avoidance, and elimination of microorganisms in medicines.
-structural modification of drugs

Answer 1

Pharmaceutics

This requires the following:
* Understanding of the basic physical chemistry necessary for the
efficient design of dosage forms.
* The design and formulation of medicines.
* The manufacture of these medicines on both small (compound) and
large (pharmaceutical technology) scale.
* The cultivation, avoidance, and elimination of microorganisms in
medicines.

Question 2

Which of the following is/are true about the scope of definition of a drug?
Vitamins may be classified as drugs.
Drug is a chemical that affects the structure or function of animal body.
Drug may be used to poison an individual.
Drug is intended for use as a device or a component, part, or accessory of a
device.
Drug is substance recognized in the Philippine pharmacopoeia.

Answer 2

The option that is not an answer to the question would be:

Drug is intended for use as a device or a component, part, or accessory of a device.
This statement does not correctly define a drug. A drug is typically a substance used to diagnose, treat, prevent disease, or affect the body’s structure or function, but it is not classified as a device or its component. Devices are separately regulated in healthcare.

Question 3

You are a manufacturing pharmacist working in research and development department of a company. You were tasked to design a formulation with the objective of creating high-quality and efficacious dosage forms. Which of the following information is NOT a consideration formulation development?

Disease of the patients to be treated
Manufacturing capability of the company
Character and competence of the pharmacist
Physical and chemical properties of drugs and additives
None of the above

Answer 3

The correct answer is:

Character and competence of the pharmacist
While the pharmacist’s competence is important, it is not directly related to the technical aspects of formulation development. The focus in formulation development is on the disease being treated, the manufacturing capabilities of the company, and the physical and chemical properties of the drugs and additives used in the formulation.

Question 4

Which of the following is/are type/s of physical change?
Rusting of iron
Combustion of alkane gases
Vaporization of liquid water
Neutralization between acids and bases
Coloring the hair using hydrogen peroxide

Answer 4

The correct answer is:

Vaporization of liquid water
This is a physical change because it involves a change in the state of matter (from liquid to gas) without altering the chemical composition of the substance.

The other options are examples of chemical changes, where new substances with different chemical compositions are formed:

Rusting of iron (formation of iron oxide)
Combustion of alkane gases (reaction with oxygen to form CO₂ and water)
Neutralization between acids and bases (formation of water and salt)
Coloring the hair using hydrogen peroxide (oxidation process that changes hair color).

Question 5

Which of the following is/are example/s of intensive property?

Normality
solidification point
distance
particle size
color

Answer 5

The correct answers are:
Solidification point
Color
These are examples of intensive properties, which do not depend on the amount of substance present.
Normality is an extensive property (it depends on the quantity of solute in a solution).
Distance and particle size are also extensive properties (they depend on the size or extent of the material).
Intensive properties, like solidification point and color, are inherent to the substance itself, regardless of the amount.

Question 6

A student from physical pharmacy class has characterized the physical properties of water drinking fountain available in school. Upon determining the boiling point, there was an apparent difference between the actual and theoretical value. This means that:

the water is impure.
the water contains microbial contamination.
it is normal, since ideal property of substance is never achieved in the actual test.
it may be due to the temperature of the working condition.
the glassware or apparatus used was able to absorb the heat during boiling.

Question 7

From the previous question, the vapor pressure of sample was observed to decrease. This is due to:

the solubility of the solute with water.
the number of particles present in the sample.
vapor pressure being dependent on temperature.
chemical nature of solute present in the sample.
the identity of the particle present in the sample
what are the answers

Answer 7

The correct answers are:

The number of particles present in the sample.
Vapor pressure being dependent on temperature.
Chemical nature of solute present in the sample.

Question 8

Which of the following statements regarding forces of attraction is/are true?
*
Knowledge of forces of attraction is important for understanding stability of drug formulation.
Substances with stronger intermolecular attractive forces have higher boiling temperatures.
For molecules to exist as aggregates in gases, liquids, and solids, intermolecular forces must exist
The stability of the interaction will change by either a decrease in attraction or an increase in repulsion
Intermolecular forces of attraction can affect the biological activity of the drug.

Answer 8

The correct answers are:

Knowledge of forces of attraction is important for understanding stability of drug formulation.
Substances with stronger intermolecular attractive forces have higher boiling temperatures.
For molecules to exist as aggregates in gases, liquids, and solids, intermolecular forces must exist.
The stability of the interaction will change by either a decrease in attraction or an increase in repulsion.
Intermolecular forces of attraction can affect the biological activity of the drug.
All of these statements are true, as they accurately describe the role and importance of intermolecular forces in drug formulation, physical properties like boiling point, molecular aggregation, interaction stability, and the biological activity of drugs.

Question 9

What are the intermolecular forces of attraction that exist in amoxicillin

Answer 9

Amoxicillin, as a drug molecule, contains various functional groups that allow it to exhibit several intermolecular forces of attraction. The primary intermolecular forces in amoxicillin are:

Hydrogen bonding:

Amoxicillin has several polar groups, such as hydroxyl (–OH), amino (–NH₂), and carboxyl (–COOH), which can form hydrogen bonds with water molecules, other amoxicillin molecules, or biological targets like proteins.
Dipole-dipole interactions:

The polar functional groups in amoxicillin create dipoles, leading to attractions between the partially positive and negative ends of molecules. This is common in molecules with polar covalent bonds, such as the carbonyl and amine groups in amoxicillin.
London dispersion forces (van der Waals forces):

Even though amoxicillin has polar functional groups, it also exhibits weak London dispersion forces due to the temporary dipoles that can form in its nonpolar regions, such as parts of the aromatic ring and hydrocarbon chains.
These intermolecular forces contribute to amoxicillin’s solubility, stability, and interaction with biological targets like bacterial enzymes, impacting its overall pharmacological activity.

Question 10

What are the intermolecular forces of attraction that exist in cetirizine

Answer 10

Cetirizine, an antihistamine drug, exhibits several types of intermolecular forces due to its functional groups. The primary intermolecular forces in cetirizine are:

Hydrogen bonding:

Cetirizine contains hydroxyl (–OH) and amine (–NH) groups that can form hydrogen bonds with water molecules or other cetirizine molecules. This interaction is significant for its solubility and biological activity.
Dipole-dipole interactions:

The presence of polar functional groups, such as the amine group and the carboxyl group, creates dipole moments within the cetirizine molecule, leading to dipole-dipole attractions between cetirizine molecules or with other polar substances.
London dispersion forces (van der Waals forces):

Cetirizine also exhibits London dispersion forces due to temporary dipoles that occur in its nonpolar regions, such as the hydrophobic parts of the molecule, including the aromatic ring.
These intermolecular forces influence cetirizine’s solubility, stability, and interactions with biological targets, which are crucial for its efficacy as an antihistamine.

Question 11

What are the intermolecular forces of attraction that exist in dopamine

Answer 11

Dopamine, a neurotransmitter, exhibits several types of intermolecular forces due to its functional groups. The primary intermolecular forces in dopamine are:

Hydrogen bonding:

Dopamine has hydroxyl groups (–OH) and an amine group (–NH₂) that can form hydrogen bonds. The hydroxyl groups on the benzene ring and the amine group can interact with water molecules or other dopamine molecules through hydrogen bonding.
Dipole-dipole interactions:

Dopamine contains polar functional groups, such as the hydroxyl and amine groups, which create dipoles. These dipoles lead to dipole-dipole attractions between dopamine molecules or with other polar substances.
London dispersion forces (van der Waals forces):

The aromatic benzene ring in dopamine contributes to London dispersion forces. These are weak interactions that arise from temporary dipoles induced in nonpolar regions of the molecule.
Dopamine does not typically engage in ion-dipole interactions, as it is a neutral molecule and does not have ionic groups.

Question 12

What are the intermolecular forces of attraction that exist in MBS

Answer 12

Medroxyprogesterone acetate (MPA) does not typically refer to “MBS,” so I’ll assume you meant Methylbenzene sulfonate (MBS), a common chemical compound. For Methylbenzene sulfonate, the primary intermolecular forces are:

Dipole-dipole interactions:

Methylbenzene sulfonate contains a sulfonate group (–SO₃⁻) which is polar. This creates dipole-dipole interactions between molecules.
Hydrogen bonding:

The sulfonate group has the ability to act as a hydrogen bond acceptor. However, it does not have hydrogen bond donors, so hydrogen bonding will be less prominent compared to compounds with hydroxyl or amino groups.
London dispersion forces (van der Waals forces):

The benzene ring in Methylbenzene sulfonate contributes to London dispersion forces. These are weak interactions due to temporary dipoles in the nonpolar regions of the molecule.

Question 13

What are the intermolecular forces of attraction that exist in simvastatin

Answer 13

Simvastatin, a statin drug used to lower cholesterol levels, exhibits several types of intermolecular forces due to its functional groups. The primary intermolecular forces in simvastatin are:

Hydrogen bonding:

Simvastatin contains hydroxyl groups (–OH) and a lactone ring, which can participate in hydrogen bonding. The hydroxyl groups can act as hydrogen bond donors and acceptors, forming hydrogen bonds with water molecules or other simvastatin molecules.
Dipole-dipole interactions:

The presence of polar functional groups, such as the carboxyl group in the lactone ring, creates dipoles within the molecule. These dipoles lead to dipole-dipole interactions between simvastatin molecules or with other polar substances.
London dispersion forces (van der Waals forces):

The hydrocarbon portions of simvastatin, including the alkyl chains and aromatic ring, contribute to London dispersion forces. These forces arise from temporary dipoles in nonpolar regions of the molecule.
Simvastatin does not typically engage in ion-dipole interactions, as it is a neutral molecule without ionic groups. The predominant intermolecular forces in simvastatin are hydrogen bonding, dipole-dipole interactions, and London dispersion forces.

Question 14

Which of the following statements is/are TRUE regarding polar and non-polar covalent bonds?
The physical properties of substances are also affected by their polarities.
A covalent bond that has an equal sharing of electrons is called a nonpolar covalent bond.
If there is no difference in the electronegativities of atoms, this means that a molecule is nonpolar.
The polarity of a covalent bond can be judged by determining the difference in the electronegativities of the two atoms making the bond.
A nonpolar molecule is easily absorbed in the body than a polar molecule

Answer 14

The true statements regarding polar and non-polar covalent bonds are:

The physical properties of substances are also affected by their polarities.

True. The physical properties such as boiling and melting points, solubility, and interaction with other substances are influenced by the polarity of molecules.
A covalent bond that has an equal sharing of electrons is called a nonpolar covalent bond.

True. In a nonpolar covalent bond, electrons are shared equally between the two atoms, typically seen in diatomic molecules of the same element (e.g., H₂, O₂).
The polarity of a covalent bond can be judged by determining the difference in the electronegativities of the two atoms making the bond.

True. The difference in electronegativity between the two atoms helps determine the degree of polarity in the bond. A larger difference indicates a more polar bond.
The statements:

If there is no difference in the electronegativities of atoms, this means that a molecule is nonpolar.

This is generally true for diatomic molecules of the same element, but not necessarily for all molecules. For example, a molecule with multiple polar bonds could still be nonpolar overall if its geometry is symmetrical.
A nonpolar molecule is easily absorbed in the body than a polar molecule.

This statement is not universally true. The absorption of a molecule in the body depends on various factors, including its size, solubility, and how it interacts with biological membranes. Generally, nonpolar molecules may more easily diffuse through lipid membranes, while polar molecules might require specific transport mechanisms. However, the ease of absorption is context-dependent and cannot be generalized for all cases.
so are all the statements true?
ChatGPT said:
ChatGPT
No, not all the statements are true. Here’s a clarification:

The physical properties of substances are also affected by their polarities.

True.
A covalent bond that has an equal sharing of electrons is called a nonpolar covalent bond.

True.
If there is no difference in the electronegativities of atoms, this means that a molecule is nonpolar.

This statement is generally true for diatomic molecules of the same element but not necessarily for all molecules. A molecule with multiple bonds might be nonpolar overall despite having polar bonds if its geometry is symmetrical.
The polarity of a covalent bond can be judged by determining the difference in the electronegativities of the two atoms making the bond.

True.
A nonpolar molecule is easily absorbed in the body than a polar molecule.

This statement is not universally true. Absorption depends on various factors, including the nature of the molecule and the environment. Nonpolar molecules might diffuse through lipid membranes more easily, but this is context-dependent.
So, statements 1, 2, 4, and 3 (with the given context) are true, while statement 5 is not universally true.

Question 15

For a given substance, which of the following phase transitions should RELEASE the most energy.

Melting
Sublimation
Vaporization
Condensation
Deposition

Answer 15

The phase transition that releases the most energy is:

Deposition
Deposition is the process where a gas transitions directly into a solid without passing through the liquid phase. It releases a significant amount of energy because it involves both the release of latent heat from condensation and freezing.

The other processes are either energy-absorbing (endothermic) or release less energy compared to deposition:

Melting (absorbs energy)
Sublimation (absorbs energy)
Vaporization (absorbs energy)
Condensation (releases energy but less than deposition)

Question 16

in the two component system ven the diagram below, which of the statements is/are correct?
*
Captionless Image
The figure is an example representation for a ternary mixture.
The eutectic point on this phase diagram is the point at which a minimum melting point occurs.
From Gibb’s Phase Rule, the formula to determine the number of independent variables that define the system in this example is F = 1 – P + 1.
Starting from the extremes of either pure component A or pure component B, as the second component is added, the melting point of the pure component increases.
Melting temperature of the combined mixtures is always lower than their respective melting points

Answer 16

Without seeing the diagram, I’ll address each statement generally based on common phase diagram principles for two-component systems:

The figure is an example representation for a ternary mixture.

False. A two-component system typically has a binary phase diagram, not a ternary one. A ternary mixture involves three components and would be represented in a ternary phase diagram.
The eutectic point on this phase diagram is the point at which a minimum melting point occurs.

True. The eutectic point is where the mixture of components has the lowest possible melting point compared to other compositions of the two components.
From Gibb’s Phase Rule, the formula to determine the number of independent variables that define the system in this example is F = 1 – P + 1.

True. For a binary system, Gibbs’ Phase Rule is
𝐹
=
𝐶
−
𝑃
+
2
F=C−P+2. With
𝐶
=
2
C=2 components and assuming a single phase (P = 1), the formula simplifies to
𝐹
=
2
−
1
+
1
=
2
F=2−1+1=2. This indicates two degrees of freedom, but the given formula seems to suggest a simpler system with fewer phases, likely focusing on a specific example.
Starting from the extremes of either pure component A or pure component B, as the second component is added, the melting point of the pure component increases.

False. In general, the addition of a second component to a pure substance often lowers the melting point due to the formation of a solid solution or eutectic mixture.
Melting temperature of the combined mixtures is always lower than their respective melting points.

True. Typically, the melting temperature of a mixture or alloy is lower than the melting points of the pure components, particularly at the eutectic composition.
So, the correct statements are:

The eutectic point on this phase diagram is the point at which a minimum melting point occurs.
From Gibb’s Phase Rule, the formula to determine the number of independent variables that define the system in this example is F = 1 – P + 1.
Melting temperature of the combined mixtures is always lower than their respective melting points.

Question 17

Given the following phase diagram of water, which of the following choices is/are correct if the pressure of the vapor over the solid is above that of the triple point?
*
Captionless Image
Only the vapor phase will exist.
Sublimation of solid may occur.
Solid, liquid, and gas phase will co-exist.
The solid phase will be produced before transformation to vapor can proceed.
The liquid phase will be produced before transformation to vapor can proceed.

Answer 17

Based on the information about phase diagrams and the given condition (pressure of the vapor over the solid is above that of the triple point), here’s an analysis of each statement:

Only the vapor phase will exist.

False. If the pressure is above the triple point pressure, this generally means that the system is in a region where solid can coexist with vapor, but the exact phase that dominates depends on the specific pressure and temperature conditions.
Sublimation of solid may occur.

True. Sublimation (solid to vapor) can occur if the pressure over the solid is not sufficiently high to maintain the liquid phase. However, if the pressure is above the triple point pressure, the system may still contain multiple phases.
Solid, liquid, and gas phase will co-exist.

False. This is only true at the triple point, where all three phases coexist. If the pressure is above the triple point pressure, it is unlikely that all three phases will coexist simultaneously.
The solid phase will be produced before transformation to vapor can proceed.

False. This statement is not generally true. The phase diagram indicates that if pressure is above the triple point, it is more likely that the system will be in a state where the liquid phase may exist, depending on the exact temperature and pressure.
The liquid phase will be produced before transformation to vapor can proceed.

True. At pressures above the triple point, it is likely that the system will transition through a liquid phase before the vapor phase is achieved. This is because higher pressure usually stabilizes the liquid phase, and transformation from solid to vapor would typically pass through the liquid state.
So, the correct statements are:

Sublimation of solid may occur.
The liquid phase will be produced before transformation to vapor can proceed.

Question 18

Which of the following boils at the highest temperature? Justify your answer.

C2H6
C3H8
C4H10
C5H12
C6H14

Answer 18

All of these are alkanes, which are nonpolar molecules. Therefore, the primary intermolecular forces are London dispersion forces (van der Waals forces), which increase with the size of the molecule.

Justification:

London dispersion forces increase with the number of electrons and the size of the molecule. Larger molecules with more electrons have stronger dispersion forces and thus generally have higher boiling points.

Boiling Points:

C₂H₆ (Ethane): Boiling point = −88.6°C
C₃H₈ (Propane): Boiling point = −42.1°C
C₄H₁₀ (Butane): Boiling point = −0.5°C
C₅H₁₂ (Pentane): Boiling point = 36.1°C
C₆H₁₄ (Hexane): Boiling point = 68.7°C
Conclusion:

C₆H₁₄ (Hexane) boils at the highest temperature among the given compounds. This is because it is the largest molecule with the most significant London dispersion forces due to its larger number of electrons and greater surface area compared to the other alkanes.

Question 19

Which of the following are important postulates of the kinetic-molecular theory of gases?

Gases consist of large numbers of particles in rapid random motion.
There are no attractive or repulsive forces between the individual molecules.
The average kinetic energy of the molecules is inversely proportional to the absolute temperature.
The volume of the molecules of a gas is very small compared to the total volume in which the gas is contained.
The time during which a collision between two molecules occurs is negligibly short compared to the time between collisions.

Answer 19

The kinetic-molecular theory of gases is based on several key postulates that describe the behavior of gas molecules. The important postulates are:

Gases consist of large numbers of particles in rapid random motion.

True. Gas particles are in constant, random motion, which is a fundamental aspect of the kinetic-molecular theory.
There are no attractive or repulsive forces between the individual molecules.

True. According to the kinetic-molecular theory, gas molecules are considered to be non-interacting, meaning there are no forces between them except during collisions.
The average kinetic energy of the molecules is inversely proportional to the absolute temperature.

False. The correct relationship is that the average kinetic energy of gas molecules is directly proportional to the absolute temperature (Kelvin). This is expressed as
KE
avg
=
3
2
𝑘
𝐵
𝑇
KE
avg
​
=
2
3
​
k
B
​
T, where
𝑘
𝐵
k
B
​
is the Boltzmann constant and
𝑇
T is the absolute temperature.
The volume of the molecules of a gas is very small compared to the total volume in which the gas is contained.

True. In the ideal gas model, the volume of the individual gas molecules is considered negligible compared to the volume of the container.
The time during which a collision between two molecules occurs is negligibly short compared to the time between collisions.

True. Collisions between gas molecules are assumed to be instantaneous, and the time between collisions is much longer than the duration of the collisions themselves.
So, the correct postulates of the kinetic-molecular theory are:

Gases consist of large numbers of particles in rapid random motion.
There are no attractive or repulsive forces between the individual molecules.
The volume of the molecules of a gas is very small compared to the total volume in which the gas is contained.
The time during which a collision between two molecules occurs is negligibly short compared to the time between collisions.

Question 20

The SI unit for pressure is:

atm
psi
torr
mmHg
Pascal

Answer 20

The SI unit for pressure is:

Pascal (Pa)
Here’s a brief explanation of each unit listed:

atm (atmosphere): A unit of pressure commonly used in chemistry and meteorology, but not an SI unit.
psi (pounds per square inch): A unit of pressure used mainly in the United States, not an SI unit.
torr: A unit of pressure that is often used in vacuum measurements, but not an SI unit.
mmHg (millimeters of mercury): A unit of pressure used in barometry and blood pressure measurements, but not an SI unit.
Pascal (Pa): The SI unit for pressure, defined as one newton per square meter (N/m²).
So, Pascal is the correct SI unit for pressure.

Question 21

The Air Protection Laboratory in a large city isolated a new gaseous pollutant that was found to exert a pressure of 1.17 atm when 6.07 gram of the substance was confined in a 2.0-liter vessel at 28C. (a) What is the molecular weight of the pollutant? (Write only the final answer, expressed in two decimals. (b) If the pollutant is known, what do you suppose the compound might be?

Answer 21

𝑛

o find the molecular weight of the pollutant, we can use the Ideal Gas Law equation:

𝑃
𝑉
=
𝑛
𝑅
𝑇
PV=nRT

where:

𝑃
P = pressure (1.17 atm)
𝑉
V = volume (2.0 L)
𝑛
n = number of moles
𝑅
R = ideal gas constant (0.0821 L·atm/mol·K)
𝑇
T = temperature in Kelvin
First, convert the temperature from Celsius to Kelvin:
𝑇
=
28
+
273.15
=
301.15
K
T=28+273.15=301.15K

Rearrange the Ideal Gas Law to solve for
𝑛
n (number of moles):
𝑛
=
𝑃
𝑉
𝑅
𝑇
n=
RT
PV
​

Substitute the known values:
𝑛
=
(
1.17
atm
)
×
(
2.0
L
)
(
0.0821
L
\cdotp
atm/mol
\cdotp
K
)
×
(
301.15
K
)
n=
(0.0821L\cdotpatm/mol\cdotpK)×(301.15K)
(1.17atm)×(2.0L)
​

2.34
24.73
n=
24.73
2.34
​

𝑛
≈
0.0947
moles
n≈0.0947moles

Now, use the number of moles to find the molecular weight (M) using the formula:
𝑀
=
mass
𝑛
M=
n
mass
​

Substitute the known values:
𝑀
=
6.07
g
0.0947
moles
M=
0.0947moles
6.07g
​

𝑀
≈
64.0
g/mol
M≈64.0g/mol

Final answer:

(a) The molecular weight of the pollutant is 64.00 g/mol.

(b) Potential Compound: A molecular weight of 64.00 g/mol is close to that of carbon dioxide (CO₂), which has a molecular weight of approximately 44.01 g/mol. However, given the relatively higher molecular weight calculated, another compound to consider could be ethylene glycol (C₂H₄(OH)₂), which has a molecular weight of approximately 62.07 g/mol. Further analysis and context would be needed to confirm the exact identity of the pollutant.

Question 22

Nitrous oxide, USP, is used for the rapid induction of anesthesia (80% nitrous oxide, 20% oxygen gas), and at lower percentages for maintenance anesthesia. (a) Using the ideal gas equation, compute the molecular weight of this gas, given that 1 liter at 0C and 760 mmHg pressure weighs 1.97 g. (b) What is the density of this gas at the given pressure and temperature? (Type your answers in a and b, expressed in two decimals)

Answer 22

𝑃

o solve the problem, follow these steps:

(a) Compute the Molecular Weight
Convert Pressure to atm:

760
mmHg
=
1
atm
760mmHg=1atm

Convert Temperature to Kelvin:

0
∘
𝐶
=
273.15
K
0
∘
C=273.15K

Use the Ideal Gas Law to Find Molecular Weight:

The Ideal Gas Law is:
𝑃
𝑉
=
𝑛
𝑅
𝑇
PV=nRT

Where:

1
atm
P=1atm
𝑉
=
1
L
V=1L
𝑇
=
273.15
K
T=273.15K
𝑅
=
0.0821
L
\cdotp
atm/mol
\cdotp
K
R=0.0821L\cdotpatm/mol\cdotpK
Rearranging to solve for
𝑛
n:
𝑛
=
𝑃
𝑉
𝑅
𝑇
n=
RT
PV
​

(
1
atm
)
×
(
1
L
)
(
0.0821
L
\cdotp
atm/mol
\cdotp
K
)
×
(
273.15
K
)
n=
(0.0821L\cdotpatm/mol\cdotpK)×(273.15K)
(1atm)×(1L)
​

1
22.414
n=
22.414
1
​

𝑛
≈
0.0446
moles
n≈0.0446moles

The weight of 1 liter of the gas is 1.97 g. Therefore, the molecular weight
𝑀
M is:
𝑀
=
mass
𝑛
M=
n
mass
​

1.97
g
0.0446
moles
M=
0.0446moles
1.97g
​

𝑀
≈
44.2
g/mol
M≈44.2g/mol

Final answer for molecular weight:
Molecularweight
=
44.20
g/mol
Molecularweight=44.20g/mol

(b) Compute the Density
The density
𝜌
ρ of the gas can be calculated using the formula:
𝜌
=
mass
volume
ρ=
volume
mass
​

Given:

Mass of 1 liter of gas = 1.97 g
Volume = 1 L
Thus:
𝜌
=
1.97
g
1
L
ρ=
1L
1.97g
​

1.97
g/L
ρ=1.97g/L

Final answer for density:
Density
=
1.97
g/L
Density=1.97g/L