250+ TOP MCQs on Surface Chemistry – Colloids and Answers

Chemistry Multiple Choice Questions on “Surface Chemistry – Colloids”.

1. Which of the following colloidal system represents a gel?
a) Solid in liquid
b) Solid in gas
c) Liquid in solid
d) Liquid in gas
Answer: c
Clarification: A gel is a colloidal system in which the dispersed phase is a liquid and the dispersion medium is a solid. Toothpaste, jam, cheese, rubber and gelatin (animal protein) are some of the examples of colloidal systems which are gels.

2. How are colloidal solutions of gold prepared by different colours?
a) Different diameters of colloidal gold particles
b) Variable valency of gold
c) Different concentration of gold particles
d) Impurities produced by different methods
Answer: a
Clarification: Colloidal solutions of gold prepared by different methods are of different colours because of different diameters of colloidal gold particles. The colour of colloidal solutions depends upon the size of the colloidal particles.

3. What are the dispersed phase and dispersion medium in alcohol respectively?
a) Alcohol, water
b) Solid, water
c) Water, alcohol
d) Solid, alcohol
Answer: d
Clarification: A colloidal solution having alcohol as the dispersion medium and a solid substance as the dispersed phase is called an alcohol. For example: A colloidal solution of cellulose nitrate in ethyl alcohol is an alcohol.

4. What is the range of the size of colloidal particles?
a) 1 to 100 nm
b) 10 to 100 pm
c) 1 to 100 µm
d) 1 to 10 mm
Answer: a
Clarification: A colloid is typically a two-phase system consisting of a continuous phase (the dispersion medium) and dispersed phase (the particles or emulsion droplets). The particle size of the dispersed phase typically ranges from 1 nanometre to 1 micrometre.

5. By using what can the colloidal particles can be separated from particles of true solution?
a) Parchment paper
b) An ultracentrifuge machine
c) An electrolyte
d) Ordinary cloth
Answer: a
Clarification: Centrifugation is a process which involves the use of the centrifugal force for the sedimentation of heterogeneous mixtures with a centrifuge used in industry and in laboratory settings. This process is used to separate two immiscible liquids. Particles of colloids are big enough to be blocked by parchment paper or animal membrane.

6. What isthe order of diameter of colloidal particles?
a) 10-3 m
b) 10-6 m
c) 10-15 m
d) 10-7 m
Answer: d
Clarification: Colloidal state of matter is, therefore, a state in which the size of the particles is such (1 to 1000 nm) that they can pass through filter paper but not through animal or vegetable membrane. Thus, every substance can be brought into the colloidal state by adopting suitable methods.

7. Dust is a colloid.
a) True
b) False
Answer: a
Clarification: Yes, the statement is true. Dust is a colloid if suspended in the air. It consists of a solid in a gas which is present in the atmosphere and it comes under the category of aerosols (contains small particles of liquid or solid dispersed in a gas).

8. When hit by light, what happens to a colloidal mixture?
a) Absorbed
b) Reflected
c) Diffracted
d) Passes through
Answer: c
Clarification: When light strikes a colloidal mixture, it is reflected off the large particles and spreads out. It is so because the colloidal particles move rapidly and randomly. This is what happens when light hits a mixture.

9. Under which category is colloidal system?
a) Homogeneous mixture
b) Heterogeneous mixture
c) Suspensions
d) True solution
View Answer

Answer: b
Clarification: Colloidal sols form heterogeneous mixtures consisting of particles of dispersed phase and the dispersion medium. The dispersed particles are spread evenly throughout the dispersion medium, which can be a solid, liquid, or gas.

10. What is the colloidal solution of a gas in liquid called?
a) Aerosol
b) Gel
c) Foam
d) Solution
Answer: c
Clarification: Depending upon whether the dispersed phase and the dispersion medium are solids, liquids or gases, eight types of colloidal systems are possible. The colloidal solution wherein gas is the dispersed phase and liquid is the dispersion medium is called foam.

250+ TOP MCQs on P-Block Elements – Nitric Acid and Answers

Chemistry Multiple Choice Questions on “P-Block Elements – Nitric Acid”.

1. Which of the following is not an oxo-acid of nitrogen?
a) Hyponitric acid
b) Hyponitrous acid
c) Nitrous acid
d) Nitric acid
Answer: a
Clarification: Hyponitric acid does not exist. The rest three mentioned are commonly occurring oxoacids of nitrogen. Hyponitrous acid, H2N2O2 is an isomer tautomer of nitramide, with the structure of the former being HON = NOH. Nitrous acid, HNO2 is usually formed in the atmosphere prior conversion to nitric acid. It is highly unstable.

2. Which of the following is true regarding nitric acid?
a) It is a strong reducing agent
b) It is a weak oxidizing agent
c) Its basicity is unity
d) It is non-planar in gaseous state

Answer: b
Clarification: Nitric acid is a very weak reducing agent since it has a polar O – H bond. This breaks to donate the H+ ion which is why it is a strong oxidizing agent and a strong acid. Since there is only one cleavable O – H bond, the basicity of nitric acid is unity (one HNO3 molecule can donate only 1 H+ ion). It exists as a planar molecule in vapor phase.

3. Which of the following reactions best represents lab scale preparation of nitric acid?
a) 3HNO2 → HNO3 + H2O + 2NO
b) NO2 + O2 → NO3
c) NaNO3 + H2SO4 → NaHSO4 + HNO3
d) 3NO2 + H2O → 2HNO3 + NO

Answer: c
Clarification: The most appropriate lab scale preparation method of nitric acid, HNO3 is using an alkali nitrate salt and react it with concentration nitric acid in a glass retort. Nitrous acid being highly unstable decomposes into nitric acid. The other two sets of reaction represent the industrial process of manufacturing nitric acid i.e. Ostwald’s process.

4. What is the name of the industrial process to manufacture nitric acid?
a) Contact process
b) Haber-Bosch process
c) Solvay process
d) Ostwald’s process

Answer: d
Clarification: Ostwald’s process is the name of the industrial process to manufacture nitric acid in bulk. It involves the oxidation of ammonia which forms nitric oxide. This is then reacted with more oxygen to produce nitrogen dioxide. Subsequently, nitrogen dioxide is dissolved in water to produce adequate concentrations of nitric acid. Contact process is used to produce sulfuric acid. Solvay is used to obtain sodium carbonate and Haber-Bosch to obtain ammonia.

5. What is the catalyst used in the industrial manufacture of nitric acid?
a) Powdered iron (III) oxide
b) Vanadium (V) oxide
c) Zinc-mercury amalgam
d) Platinum-Rhodium gauze sheet
Answer: d
Clarification: Pt-Rh gauze sheet is widely used as the catalyst in ammonic oxidation, the first step of Ostwald’s process. Fe2O3 is used in Haber’s process; V2O5 in contact process and Zn (Hg) is used in Clemmensen reduction of aldehydes.

6. What is the nitric acid – water composition by mass, respectively, for the components to form an azeotrope?
a) 70% – 30%
b) 68% – 32%
c) 30% – 70%
d) 32% – 68%
Answer: b
Clarification: Experimentally, it is determined that nitric acid and water form a constant boiling azeotrope at 68% – 32% by mass composition, respectively. Here, it becomes impossible to separate water and nitric acid by distillation methods. Thus, concentrated sulfuric acid is used for dehydration and removal of water.

7. Which of these gases is released upon treating zinc with diluted and then concentrated nitric acid?
a) Nitrogen dioxide and nitrous oxide
b) Nitric oxide and nitrous oxide
c) Nitrous oxide and nitrogen dioxide
d) Nitrous oxide and nitric oxide
Answer: c
Clarification: The products released depend on the concentration of nitric acid. In case of zinc metal, diluted nitric acid treatment release nitrous oxide and concentrated nitric acid causes the release of nitrogen dioxide.

8. What product(s) is/are formed when aluminum metal is treated with concentrated nitric acid?
a) Al (NO3) 3
b) Al (NO2) 3 + H2
c) Al2O3
d) Al4O3
Answer: c
Clarification: Aluminum does not dissolve in nitric acid. This is because treatment with nitric acid results in the formation of a tough oxide layer. This oxide layer prevents it from further reacting with the oxide. Hence, the compound formed is Al2O3 i.e. aluminum (III) oxide.

9. Which reagent is predominantly used in pickling of stainless steel?
a) Iodic acid
b) Nitric acid
c) Phosphoric acid
d) Sulfuric acid
Answer: b
Clarification: Pickling of stainless steel is the process of removal of a thin layer of the alloyed metal from the surface. The common reagent used is nitric acid along with calculated amounts of hydrofluoric acid.

10. How many moles of nitric acid is required to convert 1 mole of sulfur to sulfuric acid?
a) 10
b) 4
c) 48
d) 20
Answer: c
Clarification: 1 mole of sulfur, S8 requires 48 moles of concentrated nitric acid. The reaction is given by S8 + 48HNO3 → 8H2SO4 + 48NO2 + 16H2O. 10, 4 and 20 moles of concentrated nitric acid is required to produce iodic acid, carbon dioxide and phosphoric acid from 1 mole of iodine, carbon and phosphorus, respectively.

250+ TOP MCQs on P-Block Elements – Oxoacids of Halogens and Answers

Chemistry Problems for Schools on “P-Block Elements – Oxoacids of Halogens”.

1. Why does fluorine form only one oxoacid?
a) High electronegativity of fluorine
b) Large radius of fluorine
c) Presence of a single valence electron
d) High electropositivity of fluorine
Answer: a
Clarification: Fluorine is the most electronegative element on the periodic table. Due to its high electronegativity and small atomic radius, fluorine only forms one oxoacid, HOF known as hypofluorous acid.

2. All halogens form only one oxoacid due to their high electronegativity and small size.
a) True
b) False
Answer: b
Clarification: Only fluorine forms one oxoacid due to their high electronegativity and small size. The other halogens form several oxoacids, most of them cannot be isolated in pure state. They are stable only in aqueous solutions or in the form of their salts.

3. Which of the following is not an oxoacid of chlorine?
a) HCl
b) HOCl
c) HOClO
d) HOClO2
Answer: a
Clarification: HCl, known as hydrochloric acid and is not an oxoacid. All oxoacids have the acidic hydrogen bound to an oxygen atom, in HCl the hydrogen is bound to the chlorine atom making it a Lewis acid.

4. Which of the following is a hypohalous acid?
a) HOClO
b) HOBrO3
c) HOF
d) HOIO3
Answer: c
Clarification: Hypohalous acids are oxoacids in which the hydroxyl group is singly bonded to a halogen. In HOF the hydroxyl group is singly bonded to fluorine making it a hypohalous acid named hypofluorous acid.

5. What is the bond angle of hypochlorous acid?
a) 108°
b) 108.5°
c) 109°
d) 109.5°
Answer: d
Clarification: Hypochlorous acid is an oxoacid of chlorine in which the hydrogen atom is singly bonded to the oxygen atom and chlorine is also singly bonded to oxygen atom forming a bend structure with bond angle 109.5°.

6. How many bonds does chlorine make with oxygen in perchloric acid?
a) 3 double bonds and a single bond
b) 7 single bonds
c) 2 single bonds and 2 double bonds
d) 3 single bonds and a double bond
Answer: a
Clarification: The formula for perchloric acid is HOClO3, in which the chlorine group is attached to 3 oxygen atoms and a hydroxyl group. The 3 oxygen atoms are attached to chlorine through a double bond and the hydroxyl group is singly bonded to chlorine. Thus, there are 3 double bonds and a single bond between oxygen and chlorine.

7. Which of the following shows the correct sequence of acidic strength?
a) HClO4 > HBrO4 > HIO4
b) HClO44 4
c) HClO4 = HBrO4 > HIO4
d) HClO4 > HIO4 > HBrO4
Answer: a
Clarification: As the electronegativity of the halogen decreases, the tendency of the XO3 group to withdraw electrons of the O—H bond towards itself decreases and hence the acid strength of the perhalic acid decreases in the same order.

8. Which the following is the correct order of oxidizing power of perhalates?
a) BrO444
b) IO4 > BrO4 > ClO4
c) IO444
d) BrO4 > IO4 > ClO4
Answer: d
Clarification: Perhalates are strong oxidizing agents, their oxidizing power decreases in the order: BrO4 > IO4 > ClO4. This can be explained on the basis of their electrode potentials. Although among perhalates BrO4 is the strongest oxidizing agent, yet it is weaker oxidizing agent than F2.

9. Which of the following is the correct order of acidic strength?
a) HClO44 2
b) HClO4 = HClO3 = HClO2
c) HClO4> HClO3 > HClO2
d) HClO2 > HClO4 > HClO3
Answer: c
Clarification: Acidic strength of oxoacids of the same halogen increases with increase in the oxidation number of the halogen. The oxidation number of halogens in HClO4, HClO3 and HClO2 is 7, 5 and 3 so the Acidic strength of HClO4 is greatest followed by HClO4 and HClO2.

10. The acidic strength of HClO4 is lesser than HClO.
a) False
b) True
Answer: a
Clarification: The acidic strength of the oxoacids of the same halogen can be determined by the oxidation number of the halogen in the compound. In HClO4 the oxidation number of chlorine is +7 and in HClO the oxidation number of chlorine is +1. So, the acidic strength of HClO4 is greater than HClO.

Chemistry Problems for Schools,

250+ TOP MCQs on Isomerism in Coordination Compounds and Answers

Chemistry Multiple Choice Questions and Answers for Class 12 on “Isomerism in Coordination Compounds – 2”.

1. Which of the following compounds has enantiomers?
a) K3[Fe(CN)6]
b) K3[Al(C2O4)3]
c) K2[Zn(OH)4]
d) K2[PdCl4]

Answer: b
Clarification: K3[Fe(CN)6], K2[Zn(OH)4] and K2[PdCl4] are compounds with unidentate ligands and CN of 6 (octahedral), 4 and 4 respectively. Whereas, K3[Al(C2O4)3] is an octahedral compound with a didentate ligand oxalate and shows optical isomerism.

2. How many stereoisomers does the following compound have?

a) 0
b) 2
c) 3
d) 4

Answer: c
Clarification: Firstly, the compound can have 2 geometric isomers, cis and trans, depending on whether similar ligands are placed adjacent to or opposite each other. Furthermore, one of them, i.e. the cis form is optically active and has two non-superimposable forms. This means that the cis form is both a geometric isomers as well as an optical isomer. Hence, there are 3 stereoisomers (1 geometric, 1 optical, 1 both geometric and optical).

3. The coordination entity [CrCl2(ox)2]3- is optically active.
a) True
b) False

Answer: b
Clarification: [CrCl2(ox)2]3- has two geometric isomers, cis and trans. The trans isomer is achiral and only the cis isomer is optically active. Thus, the statement is not always true.

4. Identify the type of isomerism exhibited by the following structures.

a) Geometric isomerism
b) Optical isomerism
c) Linkage isomerism
d) Coordination isomerism

Answer: c
Clarification: Geometric and optical isomerism are types of stereoisomerism in which the chemical bonds remain same. The two structures shown have different chemical bonds between the central atom (Co) and the ligand group (NO2). Nitrate is an ambidentate ligand which can result in two forms in the entities it is present in. If it links through N atom, it forms nitrito-N form (NO2) as in figure i), and if it binds through O atom, it forms nitrito-O form (ONO) as in figure ii). This type of isomerism is called linkage isomerism.

5. Linkage isomerism is seen in compounds having ________ ligand.
a) unidentate
b) polydentate
c) chelate
d) ambidentate

Answer: d
Clarification: Ambidentate ligands can bind through either one of their two donor atoms and can result in different forms. These forms are called linkage isomers.

6. Which of the following compounds is not a linkage isomer?
a) Hg[Co(SCN)4]
b) [Cr(H2O)5(NO2)]Cl2
c) [CoCl2(en)2]Cl
d) K[Cr(NH3)2(ONO)4]
View Answer

Answer: c
Clarification: The compounds Hg[Co(SCN)4], [Cr(H2O)5(NO2)]Cl2 and K[Cr(NH3)2(ONO)4] have ambidentate ligands thiocyanate or nitrate in them and these ligands can bind through different donor atoms to produce different structures or isomers.

7. Linkage isomers exhibit different physical properties.
a) True
b) False

Answer: a
Clarification: It was observed that [Co(NH3)5(NO2)]Cl2 was either red in colour or yellow depending on whether the nitrite ligand bonded through O atom or N atom respectively, thus showing different physical properties.

8. Which of the following compounds does not have a coordination isomer?
a) [Ag(NH3)2][Ag(CN)2]
b) [Cr(NH3)6][Co(CN)6]
c) [Zn(NH3)4][PtCl4]
d) [Cu(NH3)4][FeCl4]

Answer: a
Clarification: Coordination isomerism takes place when exchange of ligands happens between cationic and anionic complex ions having different metal ions. In the compound [Ag(NH3)2][Ag(CN)2], both the complexes have the same metal ion Ag+, and hence does not have a coordination isomer.

9. Identify the coordination isomer of [Fe(CO)4][Zn(CN)4].
a) Tetracyanidozinc(II) tetracarbonylferrate(II)
b) Tetracarbonylzinc(II) tetracyanidoferrate(II)
c) Tetracyanidoiron(II) tetracarbonylzincate(II)
d) Tetracarbonyliron(II) tetracyanidozincate(II)

Answer: b
Clarification: The coordination isomer will be [Zn(CO)4][Fe(CN)4] by interchanging the central atoms between the two complexes with [Zn(CO)4]2+ being the cationic complex and [Fe(CN)4]2- being anionic. The naming will be tetracarbonylzinc(II) tetracyanidoferrate(II).

10. The compounds [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl are ________ isomers.
a) linkage
b) coordination
c) ionisation
d) solvate

Answer: c
Clarification: The sulphate counter ion is a potential ligand and can displace the Cl atom to make the latter the counter ion.

11. [Co(NH3)5Cl]SO4 + Ag+ = _________
a) AgCl
b) BaSO4
c) white precipitate
d) no reaction

Answer: d
Clarification: The compound [Co(NH3)5Cl]SO4 dissociates into a cobalt complex ion and a sulphate ion when dissolved ion water. The sulphate ion does not react with silver ion. On the other hand, the ionisation isomer of [Co(NH3)5Cl]SO4, that is [Co(NH3)5(SO4)]Cl dissociates into cobalt complex ion and Cl ion which reacts with silver ion to precipitate AgCl.

12. The compound [Co(NH3)5(NO2)](NO3)2 does not show _________ isomerism.
a) coordination
b) optical
c) ionisation
d) linkage

Answer: a
Clarification: There is only one complex ion and there is no possibility of a coordination isomer. The ionisation isomer is [Co(NH3)5(NO3)](NO3)(NO2) and the linkage isomer is [Co(NH3)5(ONO)](NO3)2.

13. Hydrate isomerism is a form of ________ isomerism.
a) coordination
b) linkage
c) ionisation
d) solvate

Answer: d
Clarification: When the solvate involved in solvate isomerism is water molecules; it is called as hydrate isomerism.

14. What is the number of water molecules that are present as ligands in the solvate isomer of hexaaquachromium(III) chloride which is grey-green in colour?
a) 1
b) 4
c) 5
d) 6

Answer: c
Clarification: The solvate isomer of [Cr(H2O)6]Cl3 that is grey-green in colour is [Cr(H2O)5Cl]Cl2.H2O. This compound has 5 water molecules directly bonded to metal ion and one water molecule as a free solvent in the crystal lattice.

250+ TOP MCQs on Alcohols, Phenols and Ethers Classification and Answers

Chemistry Multiple Choice Questions on “Alcohols, Phenols and Ethers Classification”.

1. What is the general formula for an aliphatic alcohol? (R=alkyl group)
a) R-H
b) R-OH
c) R-CHO
d) R-COOH
View Answer

Answer: b
Clarification: Alcohols are compounds having OH group attached to an alkyl group and are hence hydroxy derivatives of hydrocarbons.

2. Which of the following is true regarding polyhydric alcohols?
a) It should have one or more OH groups
b) It should have two or more OH groups
c) It should have three or more OH groups
d) It should have more than four OH groups
Answer: b
Clarification: Alcohols may be classified as mono-, di- tri- or polyhydric depending on whether it has one, two, three or more OH groups in its structure. Di- and trihydric alcohols are also classified as polyhydric.

3. Which of the following types of alcohol contain a bond between sp2 hybridised carbon and OH group?
a) Primary allylic alcohols
b) Secondary allylic alcohols
c) Tertiary allylic alcohols
d) Vinylic alcohols
Answer: d
Clarification: Allylic alcohols are those in which OH group is attached to a sp3 hybridised carbon adjacent to a C-C double bond, i.e., an allylic carbon. Whereas, vinylic alcohols contain OH group attached directly to a C-C double bond.

4. Which of the following terms does not describe CH2=CH-CH2OH?
a) Primary
b) Monohydric
c) Allylic
d) Vinylic
Answer: d
Clarification: The given compound contains a sp3 hybridised C-OH bond and is allylic. Ince it has only one alkyl group attached to the C bonded to the OH group, it is primary and the presence of only one OH group makes it monohydric.

5. Choose the most suitable classification for the shown compound?

a) Secondary alcohol
b) Allylic alcohol
c) Dihydric alcohol
d) Benzylic alcohol
Answer: b
Clarification: The compound shown is a monohydric, tertiary alcohol in which the OH group is attached to the C next to a C-C double bond, hence make it allylic.

6. Primary alcohols van be benzylic in nature.
a) True
b) False
Answer: a
Clarification: Benzylic alcohols contain a sp3 C-OH bond next to an aromatic ring, which means that the C atom can have zero or two alkyl groups attached to it along with the ring and OH group.

7. Which of the following is not a vinylic alcohol?
a) CH2=CH-OH
b) HO-CH=CH-CH3
c) CH2=CH-CH2-OH
d) CH3-CH2-CH=CH-OH

Answer: c
Clarification: Vinylic alcohols contain an OH group bonded to a sp2 hybridised carbon of a double bond. CH2=CH-CH2-OH is an allylic alcohol.

8. Which of the following compounds contain an aryl carbon?
a) Ethanol
b) Benzyl alcohol
c) Vinyl alcohol
d) Phenol

Answer: d
Clarification: The sp2 hybridised carbon of an aromatic ring to which the hydroxyl group is attached is called an aryl carbon. In phenol, the OH group is attached to an aryl carbon.

9. R-O-R, where R represent an alkyl or aryl group is the general formula of which compound?
a) Ether
b) Ester
c) Aldehyde
d) Ketone
View Answer

Answer: a
Clarification: Ethers are compounds where the hydrogen of a hydrocarbon is replaced by an alkoxy or aryloxy group. It may also be thought of as compounds formed when the hydrogen of the OH group of an alcohol is replaced by an alkyl or aryl group.

10. When the alkyl groups attached to either side of the oxygen atom of an ether id different, it is known as _______ ether.
a) simple
b) symmetrical
c) mixed
d) diethyl
Answer: c
Clarification: If the alkyl or aryl groups on either side of O atom is different, it is known as a mixed or unsymmetrical ether.

11. Identify the simple ether from the following.
a) CH3OC2H5
b) C2H5OC6H5
c) C6H5OCH3
d) C6H5OC6H5
Answer: d
Clarification: Since C6H5OC6H5 has the same aromatic ring on both sides of the oxygen atom, it a simple or symmetrical ether.

12. Which of the following is not an aromatic ether?
a) CH3-O-C6H5
b) CH3-O-CH2CH3
c) C2H5-O-C6H5
d) C6H5-O-C6H5
Answer: b
Clarification: Aromatic ethers are those in which at least one of the groups on either side of O atom is an aryl group. In CH3-O-CH2CH3, both the groups are alkyl and is hence an aliphatic ether.

13. Diethyl ether is a mixed ether.
a) True
b) False
Answer: b
Clarification: Diethyl ether consists of ethyl (C2H5) groups on either side of O, hence it is a simple symmetrical ether.

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250+ TOP MCQs on Uses of Aldehydes and Ketones and Answers

Chemistry Multiple Choice Questions on “Uses of Aldehydes and Ketones”.

1. Which of the following is used in the preservation of biological specimens?
a) Methanal
b) Ethanal
c) Acetone
d) Benzaldehyde
View Answer

Answer: a
Clarification: Methanal is easily soluble in water and is commercially sold as formalin (40% methanal, 8% methanol, 52% water) and is used for preserving biological specimens.

2. Which of the following is not a use of formaldehyde?
a) Preservation of biological specimens
b) Manufacturing of bakelite
c) Silvering of mirrors
d) Preparation of acetic acid.
Answer: d
Clarification: Formaldehyde consists of only a single carbon atom and cannot be used as a starting compound for the preparation of acetic acid. However, it is used in the manufacture of bakelite, glues and polymeric products. It also acts as a reducing agent in the silvering of mirrors.

3. Which of the following carbonyl compounds are known for their pleasant odour?
a) Vanillin
b) Acetophenone
c) Acetone
d) Camphor
Answer: c
Clarification: Acetone belongs to the lower sizes of ketones and is less fragrant than the higher compounds like acetophenone, camphor and vanillin which are used as flavouring agents in different industries.