Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “Nomenclature and Structure of Carbonyl Groups – 1”.

1. Which of the following compounds do not contain a carbonyl group?
a) Alcohol
b) Aldehyde
c) Ketone
d) Carboxylic acid
Answer: a
Clarification: Carbonyl groups are those which contain a carbon-oxygen double bond. In alcohols, the bond between C and O is a single bond and hence does not contain a carbonyl group.

2. How many carbon atoms does formaldehyde have?
a) 0
b) 1
c) 2
d) 3
Answer: b
Clarification: Formaldehyde is the common name of the simplest aldehyde, methanal. Its formula is HCHO and only one carbon which is also a part of the carbonyl group.

3. What is the common name of butanal?
a) n-Butanaldehyde
b) α-Butanaldehyde
c) n-Butyraldehyde
d) α-Butyraldehyde
Answer: c
Clarification: Butanal is a straight chain aldehyde with 4 carbon atoms including the carbonyl group. There is no substituent present in the carbon chain and hence the prefix n- is given in the common name.

4. What is the common name of the compound which has a CHO group attached to the sp² hybridised carbon of a benzene ring?
a) Benzanal
b) Benzaldehyde
c) Benzenecarbaldehyde
d) Phthaldehyde
Answer: b
Clarification: The IUPAC name of the simplest aromatic aldehyde carrying the CHO group on a benzene ring is benzenecarbaldehyde. Its common name is benzaldehyde, which is also accepted by IUPAC.

5. Which of the following is the incorrect name for the following compound?

a) 3-Methylbutanal
b) β-Methylbutyraldehyde
c) γ-Methylbutyraldehyde
d) Isovaleraldehyde
Answer: c
Clarification: The compound is a 4-carbon chain aldehyde with a methyl group substituted at the beta-carbon position, i.e., the carbon next to the carbon directly linked to the aldehyde group.

6. The prefix valer- is used for generally used for naming aldehydes with how many carbon atoms in the structure?
a) 3
b) 4
c) 5
d) 6
Answer: c
Clarification: When aldehydes have exactly 5 carbon atoms in their structure, including the carbonyl group and any substituted groups, the aldehyde is given the prefix valer- in the common naming system. For example, pentanal is named as n-Valeraldehyde.

7. What is the correct IUPAC naming of the compound shown?

a) Benzenecarbaldehyde
b) Cyclohexanal
c) Cyclohexyl aldehyde
d) Cyclohexanecarbaldehyde
Answer: d
Clarification: When the CHO group is attached to a ring, the suffix carbaldehyde is added after the name of the cycloalkane. The numbering of the ring carbons starts form the carbon attached to the aldehyde group.

8. Identify the correct IUPAC name of CH₃-CH=CH-CHO.
a) But-2-enal
b) 2-Butenal
c) Buten-2-al
d) Butenal
Answer: a
Clarification: The carbon atom of the CHO group is assigned the first number. So, in this case, the double bond occurs at second carbon and hence, the prefix 2- is used for the naming of double bond -en.

9. The compound 3-Phenylprop-2-enal is also known as _________
a) Crotonaldehyde
b) Cinnamaldehyde
c) Salicylaldehyde
d) Vanillin
Answer: b
Clarification: In cinnamaldehyde, a benzene ring is attached to the end carbon of the aldehyde chain and also has a carbon double bond at the alpha carbon. Its formula is C₆H₅-CH=CH-CHO.

10. Identify the correct common name for the following compound.

a) 3-Methylcyclohexanecarbaldehyde
b) 3-Methylbenzecarbaldehyde
c) 3-Methylbenzaldehyde
d) γ-Methylcyclohexanecarbaldehyde

Answer: d
Clarification: This compound is a cyclic aldehyde where numbering starts from the carbon to which the CHO group is attached. So, the carbon to which methyl group is attached is the third carbon. In the common system, this carbon is known as the gamma carbon.

11. What is the IUPAC name of the sown compound?

a) Propane-1,2,3-trial
b) 3-Formylpentan-1,5-dial
c) 2-(Formylmethyl)butan-1,4-dial
d) Propane-1,2,3-tricarbaldehyde
View Answer

Answer: d
Clarification: There are three aldehydic groups present in the given compound and none of them have preference over the other. The CHO groups are considered as substituents with the parent chain containing only three carbon atoms. This is done to give identical treatment to all the aldehydic groups.

12. How many aldehydic groups does the compound phthaldehyde have?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: Phthaldehyde is an aromatic compound containing two CHO groups, one on each adjacent carbon of the benzene ring. To give equal importance to both the aldehydic groups, it is named as benzene-1,2-dicarbaldehyde in the IUPAC system.

13. The compound 3-Bromobenzaldehyde is named as γ-Bromobenzaldehyde in the common system.
a) True
b) False
Answer: b
Clarification: The prefix gamma- is used for substituents in a straight carbon chain. For aromatic substituted compounds, the prefixes ortho-, meta- and para- are used. Hence, 3-Bromobenzaldehyde is named as m-Bromobenzaldehyde in the common system.

14. Vanillin is an aldehydic compound.
a) True
b) False
Answer: a
Clarification: Vanillin consists of a benzene ring substituted with CHO group and OH group on opposite side. Additionally, there is a methoxy group at meta position with respect to the aldehydic group. Its IUPAC name is 4-Hydroxy-3-methoxybenzaldehyde.

Chemistry Multiple Choice Questions on “Amines Nomenclature – 1”.

1. Which of the following is the IUPAC name of the compound in which one hydrogen of ammonia is replaced by an ethyl group?
a) Ethylamine
b) Aminoethane
c) Ethanamine
d) Ethane amine
Answer: c
Clarification: Ethylamine and aminoethane are the names of CH₃CH₂NH₂ according to the common system and second system respectively. In the IUPAC system, the naming is done by replacing the ‘e’ of the alkane by amine.

2. Which of the following names of amines belong strictly to the common system?
a) Ethylmethylamine
b) Aniline
c) Benzenamine
d) Propan-1-amine
Answer: a
Clarification: In the common system, aliphatic amines are named by prefixing the alkyl group to amine, i.e., alkylamine. In case of two different alkyl groups, it is named by listing the alkyl groups in alphabetical order before the word amine, just like in ethylmethylamine. Aniline is a common name but is also accepted by IUPAC.

3. Which of the following names of aromatic amines does not belong to the common naming system?
a) Aniline
b) o-Toluidine
c) 4-Bromoaniline
d) N^–Methylaniline
Answer: c
Clarification: In the common system, the aromatic amines have special names derived from the basic compound aniline and depending upon the type of substituent group. The prefixes o-, m- and p- are used to depict the position of the substituent in common system. Whereas, the number position of carbon is used in the IUPAC system.

4. What is the correct common name of H₂N-CH₂-CH₂-NH₂?
a) Ethylenediamine
b) 1,2-Diaminoethane
c) Ethyldiamine
d) Aminoethylamine
Answer: a
Clarification: In this compound, more than one amino group is present at different positions in the parent chain. Both the amino groups are given equal importance and none of them is considered a substituent on the other. The prefix di- is used before amine, and since it is a disubstituted ethane, it is referred to as ethylene.

5. What is the correct IUPAC name of H₂N-(CH₂)₅-NH₂?
a) Pentan-1,5-diamine
b) 1,5-Diaminopentane
c) Pentamethylenediamine
d) Pentane-1,5-diamine
Answer: d
Clarification: The parent carbon chain consists of five atoms with one amino group at each end, i.e., first and fifth carbon. The prefix di is used to indicate the two amino groups and the letter ‘e’ of the suffix part of pentane is retained. Hence, the name pentane-1,5-diamine.

6. In the IUPAC system, the prefix N^– is generally used for _______ amines.
a) primary
b) secondary
c) tertiary
d) secondary and tertiary
Answer: d
Clarification: For the naming of 2° and 3° amines, the locant N is used as a prefix to identify the substituent attached to the nitrogen atom. In case of tertiary amines with same substituent groups, the prefix used is N, N^–.

7. Identify the incorrect name of the shown compound.

a) N-Propylamine
b) Isopropylamine
c) 2-Aminopropane
d) Propan-2-amine
Answer: a
Clarification: The shown compound is a primary amine and hence the locant N cannot be used as it is invalid. The amino group is attached to the second carbon of the propyl group (which is also known as isopropyl substituent).

8. Which of the following is the correct IUPAC name of (CH₃)₃N?
a) Trimethylamine
b) N-Methylethanamine
c) N,N-Dimethylmethanamine
d) N,N,N-Trimethylamine
Answer: c
Clarification: The compound is a tertiary amine with three methyl substituents. One of the methyl groups is considered as the parent chain with the N atom and the other two acts as substituents. These are represented by N, N- as they are the same groups and are directly attached to the nitrogen atom.

9. Benzylamine is the IUPAC name of aniline.
a) True
b) False
Answer: b
Clarification: Aniline is an aromatic aryl amine, whereas benzylamine is am arylalkyl amine. Aniline is both accepted as the common and IUPAC names. The IUPAC naming of aryl amines is done by replacing the suffix ‘e’ of the arene by amine. Therefore, aniline is named as benzenamine.

10. What is the correct IUPAC name of CH₃-NH-CH₂CH₃?
a) N-Methylethanamine
b) N-Ethylmethanamine
c) N-Ethyl-N-methylmethanamine
d) N-Ethyl-N-methylamine
Answer: a
Clarification: This is a secondary amine with an ethyl and a methyl group. The smaller alkyl group (CH₃) is considered as the substituent which is attached to the N atom. Hence, it is N-methyl and not N-ethyl.

11. What is the IUPAC name of the following compound?

a) N-Ethylpentan-1-amine
b) N-Ethylpentan-2-amine
c) N,N-Diethylbutan-1-amine
d) N-Ethyl-N-Butylethanamine
Answer: c
Clarification: This is a tertiary amine with two identical alkyl substituents (ethyl group) and one butyl group which acts as the major chain. The prefix N,N-Di is used for the ethyl groups and this collective group is attached at the first carbon of the butyl group.

12. The IUPAC name of allylamine is prop-2-en-1-amine.
a) True
b) False
Answer: a
Clarification: Allylamine (NH₂-CH₂-CH=CH₂) is an unsaturated amine with the amino group attached to an allyl carbon. In such compounds, the amino group is given preference and hence the double bond lies at the second carbon.

13. Identify the incorrect name of the aromatic primary amine with the formula C₇H₉N.
a) Benzenamine
b) Benzylamine
c) Phenylaminomethane
d) Phenylmethanamine
Answer: a
Clarification: Since it is an aromatic compound, the phenyl group C₆H₅ has to present with CH₄N. Since it is a primary amine, the group NH₂ can be separated leaving us with CH₂. So, the formula of the compound is C₆H₅CH₂NH₂, which is benzylamine. Benzenamine has the formula C₆H₅NH₂.

14. Hexamethylenediamine is named as __________ in the IUPAC system.
a) Hexane-1,3-diamine
b) Hexane-1,4-diamine
c) Hexane-1,5-diamine
d) Hexane-1,6-diamine
Answer: d
Clarification: The formula is NH₂-(CH₂)₆-NH₂ which is the chain of 6 carbon atoms and two amino groups at the ends of the chain, i.e., first and sixth carbon. This type of substitution is named and hexamethylene.

Chemistry Multiple Choice Questions on “Biomolecules Carbohydrates – 2”.

1. Which of the following monosaccharides is a ketohexose?
a) Glucose
b) Galactose
c) Fructose
d) Mannose
Answer: c
Clarification: Fructose is a monosaccharide that consists of six carbon atoms including a carbonyl carbon of a keto group within its chain. Glucose, galactose and mannose are examples of aldohexoses.

2. Identify the aldose form the following.
a) Arabinose
b) Xylulose
c) Ribulose
d) Sorbose
Answer: a
Clarification: Arabinose consists of an aldehydic group (CHO) at the end of its chain which has a total of five carbon atoms (including the CHO group), hence it is an aldopentose. Xylulose, ribulose and sorbose are examples of ketoses.

3. Erythrulose is a/an _______
a) aldotetrose
b) aldoheptose
c) ketotetrose
d) ketoheptose
Answer: c
Clarification: Erythrulose is a tetrose carbohydrate (because it contains 4 carbon atoms) with the formula C₄H₈O₄. It has one ketone group and hence belongs to the ketose family.

4. Identify from the following pairs, a pair of aldoses.
a) Sorbose, allose
b) Mannose, tagatose
c) Psicose, gulose
d) Talose, idose
Answer: d
Clarification: All the compounds mentioned are monosaccharides with the formula C₆H₁₂O₆. However, allose, mannose, gulose, talose and idose are aldohexoses. Sorbose, tagatose and Psicose are aldoketoses.

5. Glucose is prepared commercially from the hydrolysis of ______ by boiling it with dilute H₂SO₄ at 393K under pressure.
a) starch
b) sucrose
c) galactose
d) dextrose
Answer: a
Clarification: An aqueous solution of corn starch is acidified with dilute H₂SO₄. This is then heated under pressure (2-3 bar) for hydrolysis to take place. After this, the liquid is neutralised with sodium carbonate and the resulting solution is concentrated under reduced pressure to get glucose crystals. This process is also used for obtaining glucose from cellulose.

6. Identify ‘X’ in the following reaction.

a) sucrose
b) lactose
c) maltose
d) starch
Answer: a
Clarification: When sucrose (C₁₂H₂₂O₁₁) is boiled with dilute HCl or H₂SO₄ in an alcoholic solution, glucose and fructose are obtained in equal amounts. This is one the methods used for preparation of glucose.

7. The reaction of glucose with which of the following proves the presence of an aldehydic group?
a) Potassium iodide
b) Hydroxylamine
c) Bromine water
d) Acetic anhydride
Answer: c
Clarification: On treatment with Br₂ water, glucose gets oxidised to gluconic acid (which is a six-carbon carboxylic acid). This indicates that the carbonyl group is present as an aldehyde group.

8. Glucose on Fehling’s test gives ______
a) no reaction
b) silver mirror
c) red precipitate
d) pungent gas
Answer: c
Clarification: Since glucose is readily oxidised, it acts as a strong reducing agent and reduces Fehling’s reagent to form a carboxylate compound (in the place of CHO group which gets oxidised) along with a red precipitate of Cu₂O.

9. The reaction of glucose with hydrogen cyanide confirms the ______
a) straight chain structure of glucose
b) presence of a carbonyl group in glucose
c) presence of an aldehyde group in glucose
d) presence of a keto group in glucose
Answer: b
Clarification: Glucose reacts with hydrogen cyanide to form glucose cyanohydrin, where the carbonyl double bond gets cleaved to form one C-OH bond and one C-CN bond. This indicates the presence of a carbonyl group but does not confirm whether it is an aldehydic or a keto group.

10. The reaction of glucose with acetic anhydride conforms the presence of how many hydroxy groups in glucose?
a) 3
b) 4
c) 5
d) 6
Answer: c
Clarification: The reaction of glucose with (CH₃CO)₂O gives glucose pentaacetate, which has five acetyl groups in the structure. This reaction also proves that all the five OH groups are present at different carbon atoms, because of the stable nature of glucose.

11. Identify the product of the following reaction.

a) Gluconic acid
b) Glutaric acid
c) Glutamic acid
d) Glucaric acid
Answer: d
Clarification: Strong oxidising agents like nitric acid oxidise both the terminal groups of glucose to give a dicarboxylic acid, saccharic acid, also known as glucaric acid. This reaction also indicates the presence of a primary alcoholic group. Gluconic acid also undergoes a similar reaction.

12. Glyceraldehyde is a carbohydrate.
a) True
b) False
Answer: a
Clarification: Glyceraldehyde is the simplest of the common aldose monosaccharides. It is a triose monosaccharide with the formula C₃H₆O₃. It consists of two hydroxy groups and one aldehydic group.

13. Identify the compound from its structure as derived from that given by Fischer.

a) Glucose
b) Gluconic acid
c) Saccharic acid
d) Sorbitol
Answer: c
Clarification: Gluconic acid (as shown) is obtained when D-glucose is treated with bromine water, resulting in the oxidation of the aldehyde group to a carboxyl group. Gluconic acid on further oxidation gives saccharic acid.

14. The reaction of glucose with NH2OH gives _______
a) n-hexane
b) glucose oxime
c) glucose cyanohydrin
d) a gluconic acid
Answer: b
Clarification: Glucose reacts with hydroxylamine to form an oxime of glucose, where a carbon-nitrogen double bond is formed (C=N-OH) in place of the CHO group. This reaction establishes the presence of a carbonyl group in glucose.

15. The ‘D’ in D-(+)-glucose represents its dextrorotatory behaviour.
a) True
b) False
Answer: b
Clarification: The ‘D’ represents the correlation of the structure to that of (+)-glyceraldehyde, in which the OH attached to the carbon adjacent to the CH₂OH is on the right-hand side of the chain. Likewise, when the OH of the last asymmetric carbon in glucose lies on the left-hand side of the chain, it is known as L-glucose.

Chemistry Multiple Choice Questions on “Drug-Target Interaction”.

1. Drugs that block the binding site of an enzyme form a substrate are called ______
a) inhibitors
b) poisons
c) messengers
d) receptors
Answer: a
Clarification: Drugs that either prevent the enzyme from holding the substrate or prevent it from providing functional groups that attack the substrate, are called enzyme inhibitors. These inhibit the catalytic activity of enzymes.

2. Which of the following drugs will inhibit the activity of the shown enzyme?

a) A
b) B
c) C
d) D
Answer: a
Clarification: The shape of the active site resembles a triangle, which means that the substrate should also have a similar shape. The drug that will inhibit the activity of this enzyme should closely resemble the substrate and block the active site. These are called competitive inhibitors.

3. All drugs block enzyme activity by occupying the active site before the substrate.
a) True
b) False
Answer: b
Clarification: Drugs prevent the attachment of substrate to enzymes by two different ways. One is the competitive approach where it occupies the active site, and the other is non-competitive way in which it does not occupy the active site.

4. A certain compound X occupied a site of an enzyme exactly opposite to that of the active site. This immediately resulted in the change of shape of the active site. X is called a ______
a) competitive inhibitor
b) non-competitive inhibitor
c) competitive messenger
d) receptor
Answer: b
Clarification: Non-competitive inhibitors are drugs that affect the catalytic activity of enzymes by changing the shape of the active site by binding to the enzyme at some other site. This prevents the substrate from binding to the enzyme.

5. A certain compound occupied a site Y of an enzyme near to the active site. This immediately resulted in the change of shape of the active site. Y is called a/an ______
a) inactive site
b) binding site
c) non-competitive site
d) allosteric site
Answer: d
Clarification: When a drug binds to an enzyme from a site other than the active site, it is called an allosteric site. Binding to this site causes the change in shape of active site such that the substrate cannot recognise it.

6. If the bond between the enzyme and inhibiting drug is very strong, which of the following takes place?
a) The active site slowly regains its original shape
b) The enzyme develops a new active site
c) The enzyme is blocked temporarily
d) The body synthesizes a new enzyme
Answer: d
Clarification: When there is a strong covalent bond between the enzyme and the drug, the enzyme is blocked permanently. This results in the degradation of the entire drug-enzyme complex and the synthesis of a new enzyme.

7. Which of the following is incorrect regarding receptors?
a) They have constant shape
b) They are proteins
c) The shape of receptors binding site changes to fit the messenger
d) They are present in the cell membrane
Answer: a
Clarification: Receptors are proteins that are embedded in the cell membrane with the binding site protruding outwards. They change the shape of the binding site to accommodate the incoming messenger and facilitate communication.

8. Which of the following is not a reason for the selectivity of receptors towards messengers?
a) Shape of binding site
b) Structure
c) Amino acid composition
d) Location in the membrane
Answer: d
Clarification: There are lots of different types of receptors in the body at different locations. They interact with different kinds of chemical messengers. Their selectivity for one messenger over the other is because of the difference in their binding sites, structure and composition of amino acids.

9. The drugs that are used when there is a lack of natural chemical messenger is called ______
a) antagonists
b) agonists
c) analgesics
d) narcotics
Answer: b
Clarification: Agonists are drugs that replicate the behaviour and properties of the chemical messenger and activate the receptor. These are used when the actual messenger is not present in the body.

10. Antagonists do not allow the transfer of message through receptors.
a) True
b) False
Answer: a
Clarification: Antagonists bind to the receptor at the binding site and inhibit its natural communicative function. This prevents the message to be passed into the cell.

Chemistry Multiple Choice Questions on “Vapour Pressure of Liquid Solutions”.

1. What phenomenon occurs when a solution’s equilibrium vapor pressure equals the surrounding atmospheric pressure?
a) Boiling
b) Melting
c) Condensation
d) Sublimation
Answer: a
Clarification: Boiling, by definition, occurs when the vapor pressure of a liquid equals the surrounding atmospheric pressure. It is a very quick vaporization process which takes place at a constant temperature, referred to as the boiling point.

2. What is the boiling point of water?
a) 100 °F
b) 671.67 °R
c) 373 °C
d) 212 °r
Answer: b
Clarification: °R represents the temperature measured on Rankine scale. Typically, boiling point of water is measured as 100 °C on the Celsius scale. The conversion of Celsius to Rankine scale is R = C x 9/5 + 491.67.
Considering boiling point of water as 100 °C, R = 100 x 9/5 + 491.67 = 671.67 °R
°F, °r represent the Fahrenheit and Reaumur scale, respectively.

3. “Total pressure of gas mixture is the sum of individual pressures”. Which law is reflected in this statement?
a) Amagat’s law
b) Raoult’s law
c) Dalton’s law
d) Henry’s law
Answer: c
Clarification: Dalton’s of partial pressure states that for a non-reactive mixture of gases in a closed vessel, the total pressure of the mixture is the sum of pressures all individual gas components present. If gases A, B and C are present then, total pressure, P_total = P_A + P_B + P_C.

4. If a mixture of A and B boils at a temperature lower than the boiling point of either of the components, what kind of deviation does the mixture show?
a) No deviation
b) Maximum and minimum deviation from Raoult’s law
c) Negative deviation from Raoult’s law
d) Positive deviation from Raoult’s law
Answer: d
Clarification: Deviation is always exhibited by non-solutions. Non- ideal solutions do not obey Raoult’s behavior. This is because ΔH_mix, ΔV_mix ≠ 0. The cause for this is the intermolecular attractions. In this case, molecular attraction between A – B is weaker than that of A – A and B – B. Hence the easier it is to break the bonds and boil faster at a lower temperature than that of either of A and B.

5. What deviation is shown by a mixture of equimolar phenol and aniline?
a) Negative deviation
b) Positive deviation
c) No deviation
d) Alternating positive and negative
Answer: a
Clarification: Phenol and aniline both exhibit greater magnitude of hydrogen bonding due to the presence of hydroxyl and amine group, both possessing highly electronegative atoms of oxygen and nitrogen, respectively. The intermolecular hydrogen bonding is far greater than the degree of intramolecular hydrogen bonding which requires more thermal energy to break. Thus, there is an increase in the boiling point of the mixture.

6. If ethanol and chloroform are present in a molar ratio of 2:3 then what is the vapor pressure at 20° C if vapor pressures of pure liquids are 5.95 kPa and 21.17 kPa, respectively?
a) 16.692 kPa
b) 15.082 kPa
c) 8.731 kPa
d) 12.038 kPa
Answer: b
Clarification: Given,
P⁰_eth = 5.95 kPa
P⁰_chl = 21.17 kPa
Mole ratio of ethanol ∶ chloroform = 2 ∶ 3
Total number of parts = 2 + 3 = 5
Therefore, mole fraction of ethanol, X_eth = 2/5 = 0.4
Mole fraction of chloroform, X_chl = 3/5 = 0.6
From Raoult’s law, p_A = p⁰_A x X_A
P_eth = 5.95 x 0.4 = 2.38 kPa
P_chl = 21.17 x 0.6 = 12.702 kPa
From Dalton’s law, P_total = P_eth + P_chl
P_total = 2.38 + 12.702 = 15.082 kPa.

7. Considering a binary solution of components A and B obeys Raoult’s law, which of the following is true?
a) Total vapor pressure cannot be related to mole fraction of only one component
b) Total vapor pressure of one component varies non-linearly with another component
c) A plot of vapor pressures of both components gives a linear plot
d) Total vapor pressure of solution always decreases with increase in mole fraction of a component
Answer: c
Clarification: Raoult’s law states that the partial pressure of each component in the solution varies directly with its mole fraction in the solution. It is formulated as p_total = p_A + (p_B – p_A)x_B. From this, it is seen that when a graph is plotted, it gives a linear plot passing through origin. Total vapor pressure can be expressed as mole fraction of one component and that it varies linearly with the latter. However, total vapor solution may decrease or increase with increase in mole fraction of a component.

8. A volatile liquid with vapor pressure 85 kPa (at sea level, 25° C) is taken to the peak of Mt. Everest. Which of the following is true?
a) The vapor pressure of the solution decreases
b) The solution will condense quickly than when at sea level
c) The solution will vaporize quickly than when at sea level
d) The vapor pressure of the solution increases
Answer: d
Clarification: The vapor pressure of a liquid is independent of the pressure of surroundings. Any solution boils when its vapor pressure equals the surrounding atmospheric pressure. The atmospheric pressure decreases with increase in altitude. Due to this, the vapor pressure equals the atmospheric pressure in a shorter period of time when compared to at the solution being present at sea level. Therefore, the solution will vaporize quickly as the bowling point will reached faster.

9. Which of the following is caused by the addition of a non-volatile solute to a solvent?
a) Reduction in equilibrium of vapor pressure of solution
b) Increase in melting point of the solution
c) Decrease in the boiling point of the solution
d) Osmosis of solute in the solution
Answer: a
Clarification: This phenomenon is known as ‘relative lowering of vapor pressure’ which is a very common colligative property. When a non-volatile solute is added to a solvent, the upper surface of the solvent is covered partially. Hence, the solvent molecules do not bear maximum freedom to escape into the space as vapors. Consequently, there is a decrease in vapor pressure when compared towhat it would have been with pure solvent and no solute at all.

10. What does the vapor pressure of solvent containing a non-volatile solute, in a closed system directly vary with?
a) Mole fraction of solute
b) Mole fraction of solvent
c) Molarity of solute
d) Molarity of solvent
Answer: b
Clarification: More the solute, lower is the vapor pressure of the solvent when compared to that when it is in its purest form. In case of greater amount of non-volatile solute, more solvent can be added to raise the vapor pressure. Therefore, it is always directly proportional to the mole fraction of the solvent.

Chemistry Multiple Choice Questions on “Chemical Kinetics – Pseudo First Order Reaction”.

1. For a pseudo first-order reaction, what is the unit of the rate of the reaction?
a) s^-1
b) mol L^-1s^-1
c) mol^-1 L s^-1
d) mol^-2 L² s^-1
Answer: a
Clarification: The unit of the rate of the reaction (k) is (mol L^-1)^1-n s^-1, where n is the order of the reaction.
For a pseudo first-order reaction n=1.
So, (mol L^-1)^1-n s^-1 = (mol L^-1)^1-1 s^-1 = s^-1.

2. Reactions whose molecularity is one but order is more than one are known as pseudo first-order reactions.
a) True
b) False
Answer: b
Clarification: Pseudo first-order reactions are the reactions whose molecularity is more than one but follows first order kinetics. Acid hydrolysis of esters such as CH₃COOC₂H₅ in aqueous solution is an example for pseudo first order reaction.

3. Which of the following reactions is an example of a pseudo first-order reaction?
a) H₂ + Br₂ → 2HBr
b) CH₃CHO → CH₄ + CO
c) C₁₂H₂₂O₁₁ + H₂O → glucose + fructose
d) PCl₅ → PCl₃ + Cl₂
Answer: c
Clarification: The order of the reaction may be sometimes altered sometimes by changing the conditions, for example, taking one or more reactants in excess compared to the other. The molecularity of inversion of cane sugar is 2 but follows first order kinetics because water is present in excess.

4. Acid hydrolysis of esters such as CH₃COOC₂H₅ in aqueous solution is an example of a pseudo first-order reaction.
a) True
b) False
Answer: a
Clarification: The reaction involved in acid hydrolysis of CH₃COOC₂H₅ in aqueous solution = CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH. The rate law for the reaction is R=k[CH₃COOC₂H₅][H₂O], however, water is present in large excess and is assumed to be constant during the course of the reaction. So the rate law becomes R=k [CH₃COOC₂H₅] and follows first order kinetics.

5. The rate law for inversion of cane sugar is R=k [C₁₂H₂₂O₁₁][H₂O]. Find the concentration of sucrose if the rate of the reaction is 0.032 s^-1 and rate constant k=0.005.
a) 5.8 M
b) 6 M
c) 6.2 M
d) 6.4 M
Answer: d
Clarification: Given, rate constant k=0.005, rate of the reaction R=0.032 s^-1.
Inversion of cane sugar is a pseudo first order reaction and water is present in large excess so the rate law becomes R=k [C₁₂H₂₂O₁₁]
0.032= 0.005 [C₁₂H₂₂O₁₁]
[C₁₂H₂₂O₁₁]=6.4 M.

6. What is the rate law of hydrolysis of CH₃COOC₂H₅ in aqueous solution?
a) R=k [CH₃COOC₂H₅]²
b) R=k [CH₃COOC₂H₅]
c) R=k [CH₃COOC₂H₅] [H₂O]
d) R=kk’ [CH₃COOC₂H₅][H₂O]
Answer: b
Clarification: The reaction involved in acid hydrolysis of ester = CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH. So the rate law becomes R=k’ [CH₃COOC₂H₅][H₂O], since water is present in large excess it is assumed to be constant throughout the reaction.
R=k[CH₃COOC₂H₅], where k is k’ [H₂O].

7. Which of the following is not true for a pseudo first-order reaction?
a) The reaction follows first order kinetics
b) Molecularity of the reaction should be one
c) The reactants such as water are present in excess and are assumed to be constant throughout the reaction
d) Unit of the rate constant is s^-1
Answer: b
Clarification: The order of the reaction can be sometimes altered by taking one of the reactant in excess compared to the other. Pseudo first order reactions are those reactions in which the molecularity is more than one but follows first order kinetics.

8. Which of the following is not similar between a first order and pseudo first order reaction?
a) The molecularity is one in both the reactions
b) Both follows first order kinetics
c) The unit of rate constant is s^-1
d) The rate of the reaction depends only on one reactant

Answer: a
Clarification: Both first order and pseudo first order reactions follow first order kinetics as in the rate of the reaction depends only on one reactant and the unit for rate constant is s^-1 but the molecularity in first order reaction should be one where as in pseudo first order reaction the molecularity can be more than one. For example: Inversion of cane sugar.

9. Which of the following is not an example of a pseudo first-order reaction?
a) CH₃COOC₂H₅ + NaOH → CH₃COOH + H₂O
b) CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH
c) C₂H₅COOC₂H₅ + H₂O → C₂H₅COOH + C₂H₅OH
d) C₁₂H₂₂O₁₁ + H₂O → glucose + fructose
Answer: a
Clarification: The reactions CH₃COOC₂H₅ + H₂O → CH₃COOH + C₂H₅OH, C₂H₅COOC₂H₅ + H₂O → C₂H₅COOH + C₂H₅OH and C₁₂H₂₂O₁₁ + H₂O → glucose + fructose are examples of pseudo first order reaction because water is present in excess and is assumed to remain constant is all the reactions but the reaction CH₃COOC₂H₅ + NaOH → CH₃COOH + H₂O is an example of second order reaction and follows second order kinetics with rate law R=k [CH₃COOC₂H₅] [NaOH].

10. For the reaction C₂H₅COOC₂H₅ + H₂O → C₂H₅COOH + C₂H₅OH, find the concentration of [C₂H₅COOC₂H₅] if the rate constant k=0.2 and the rate of the reaction is 0.39 s^-1.
a) 2.87 M
b) 2.01 M
c) 1.99 M
d) 1.95 M
Answer: d
Clarification: The reaction C₂H₅COOC₂H₅ + H₂O → C₂H₅COOH + C₂H₅OH is a pseudo first order reaction and follows first order kinetics with rate law R=k[C₂H₅COOC₂H₅].
Given, rate of the reaction R=0.39 s^-1 and rate constant k=0.2
So, 0.39 = 0.2 × [C₂H₅COOC₂H₅]
[C₂H₅COOC₂H₅]=1.95 M.