Category: Class 12 Chemistry Objective Questions

Chemistry Multiple Choice Questions on “Thermodynamic Principles of Metallurgy”.

1. Ellingham Diagram does not give any information regarding the kinetics of the reduction process.
a) True
b) False
Answer: a
Clarification: Ellingham diagrams are based on thermodynamic concepts, therefore, these diagrams simply suggest whether the reduction of a given metal oxide with a particular reducing agent is possible or not. It, however, does not tell us anything about the kinetics of the reduction process.

2. Which of the following is the correct Gibbs equation?
a) ΔG = ΔH + TΔS
b) ΔG = ΔH – TΔS
c) ΔG = ΔH – 2TΔS
d) ΔG = ΔH – 3TΔS
Answer: b
Clarification: For any process, the change in Gibbs free energy (ΔG) is given by the equation,
ΔG = ΔH – TΔS
Where ΔH is the enthalpy change, ΔS is the entropy change and T is the absolute temperature.

3. What is the criterion of the feasibility of a reaction at any temperature?
a) ΔG of the reaction must be positive
b) ΔG of the reaction must be negative
c) ΔG of the reaction must be equal to zero
d) Does not depend on ΔG of the reaction
Answer: b
Clarification: The criterion of the feasibility of a reaction at any temperature is that the ΔG of the reaction must be negative. For a reaction to be spontaneous, the value of the change in Gibbs energy of the reaction must always be negative.

4. How can a reaction with positive ΔG be made to occur?
a) By increasing the temperature
b) By decreasing the temperature
c) By coupling it with another reaction
d) It is not possible for the reaction to occur
View Answer

Answer: c
Clarification: A reaction with ΔG positive can still be made to occur by coupling it with another reaction having large negative ΔG so that the net ΔG of the two reactions is negative (ΔG = change in the Gibbs energy of the reaction).

5. What does the Ellingham diagram consist of?
a) Plots of Δ_fG° vs T
b) Plots of T vs Δ_fG°
c) Plots of Δ_fG° vs ΔS
d) Plots of ΔS vs Δ_fG°
Answer: a
Clarification: Ellingham diagram normally consists of plots of Δ_fG° vs T for formation of oxides of elements. Similar diagrams can also be constructed for sulphides and halides of elements. These diagrams help us in predicting the feasibility of thermal reduction of an ore.

6. Under what condition can Al reduce MgO?
a) Above 1000°C
b) Below 1000°C
c) Above 1350°C
d) Below 1350°C
Answer: c
Clarification: For a reaction to be feasible, the change in the Gibbs free energy of the reaction (Δ G) must be negative. Lesser the ΔG, greater is the feasibility of the reaction. Above 1350°C, the standard Gibbs free energy of formation of Al₂O₃ from Al is less than that of MgO from Mg. Therefore, Al can reduce MgO above 1350°C.

7. Which of the following is false regarding the Ellingham diagram?
a) It consists of plots of Δ_fG° vs T
b) Ellingham diagrams are based on thermodynamic concepts
c) They do not tell us anything about the kinetics of the reduction process
d) They do not assume reactant-product equilibrium
Answer: d
Clarification: The interpretation of ΔG° is based upon K (i.e., ΔG° = -RT ln K). Thus, it is presumed that the reactants and the products are in equilibrium. But, this is not always true because the reactant/product may be solid.

8. Any metal will reduce the oxide of other metals which lie below it in the Ellingham diagram.
a) True
b) False
Answer: b
Clarification: No, any element (metal) will reduce the oxide of other metals which lie above it in the Ellingham diagram because the free energy change (Δ_rG°) for the combined redox reaction will be negative by an amount equal to the difference between the free energy of formation (Δ_fG°) of the two oxides at that temperature.

9. Which of the following statements is true?
a) Al can reduce ZnO more readily than Mg
b) Zn can reduce Cu₂O more readily than Al
c) Mg can reduce ZnO more readily than Al
d) Zn can reduce Cu₂O more readily than Mg
Answer: c
Clarification: The relative tendency of these metals to act as reducing agents is Mg > Al > Zn > Fe > Cu. Mg can reduce Al₂O₃, ZnO, FeO and Cu₂O more readily than Al can reduce ZnO, FeO and Cu₂O. Similarly, Al can reduce ZnO, FeO and Cu₂O more readily than Zn reduces FeO and Cu₂O.

10. How are the oxides of less reactive metals like silver and mercury reduced?
a) By thermal decomposition
b) By electrolysis
c) By using reducing agents
d) Cannot be reduced
Answer: a
Clarification: In case of less reactive metals like silver and mercury, Δ_fG° becomes positive at higher temperatures. This suggests that both silver oxide (Ag₂O) and mercury oxide (HgO) are unstable and hence decompose at high temperatures to liberate the corresponding metal.

Chemistry Multiple Choice Questions on “P-Block Elements – Simple Oxides”.

1. What type of oxides is formed by metals?
a) Oxides with pH = 7
b) Oxides with pH > -log₁₀[1 x 10^-7]
c) Oxides with pH 10[1 x 10^-7]
d) Amphoteric oxides
Answer: b
Clarification: Metals usually form basic oxides. E.g. sodium, potassium, calcium. Meaning, the pH of the salts is greater than 7 on the pH scale. pH is calculated as the negative logarithm of concentration of hydronium ions. Metals form oxides because upon dissolution, they release oxide ions which in turn give hydroxide (basic) ions. However, a few metals form oxides which are both acidic and basic, called amphoteric oxides. E.g. zinc, lead.

2. What type of oxides is formed by non-metals?
a) Oxides with pH b) Amphoteric oxides
c) Oxides with pH > 7
d) Oxides which are acidic
Answer: d
Clarification: Non-metals form acidic oxides because upon dissolution in water they liberate free H⁺ ions. Oxides with pH 7 are basic in nature. Amphoteric oxides behave both like acid and base.

3. Which of the following can classified as an amphoteric oxide?
a) Iron (III) oxide
b) Zinc oxide
c) Mercury (II) oxide
d) Lime

Answer: b
Clarification: Zinc oxide is the only amphoteric oxide out of these four. On dissolution in acid like HCl, it acts as a base and forms ZnCl₂ salt. On dissolving in base like NaOH, it acts an acid to form sodium zincate, Na₂ZnO₂.

4. What is the starting substance fed into the electrolytic cell in Hall-Heroult process?
a) Cryolite
b) Sodium hydroxide
c) Aluminum hydroxide
d) Alumina
Answer: d
Clarification: Hall-Heroult process is used to extract aluminum by using an electrolytic cell. Alumina means aluminum oxide. It is dissolved in cryolite, Na₃AlF₆, to increase electrical conductivity, reduce melting point and achieve greater yield of aluminum metal.

5. What is the primary constituent of sand?
a) Silicon tetroxide
b) Silica
c) Silicon
d) Feldspar
Answer: b
Clarification: Sand is primarily composed of silicon dioxide, also called silica. Silicon tetroxide is not a compound but an oxo-anion of silicon. Silicon metal does not form native in nature and feldspar is a mineral ore consisting of several tectosilicate compounds.

6. Before subjecting Mg ribbons to flame, it is rubbed with sandpaper. Why?
a) To increase surface area for combustion
b) To increase roughness of surface
c) To remove oxide layer
d) To remove moisture
Answer: c
Clarification: In order to ensure effective burning of the magnesium ribbon, it is cleaned with sandpaper to remove dust particles along with the layer of magnesium oxide, which may prevent it from burning completely.

7. What product is formed when iron metal is dipped into concentrated nitric acid?
a) Iron (III) nitrate + NO₂
b) Iron (II) nitrate + NO
c) Iron (III) oxide
d) Iron (III) nitrite
Answer: c
Clarification: Iron becomes passive when treated with concentrated nitric acid. The formation of a tough oxide layer prevents it from further reacting with the acid, hence inhibiting the reaction.

8. Which of the following processes involves heating a carbonate ore to form metal oxide?
a) Leaching
b) Smelting
c) Ore Reduction
d) Calcination
Answer: d
Clarification: In general metallurgy, calcination is the process of heating a metal ore to convert it to its corresponding metal oxide. For e.g. zinc carbonate (calamine ore) is heated to form zinc oxide which is then reduced by electrolysis in the presence of sulfuric acid to obtain zinc metal.

9. Which of the following metals does not react with hot water to form an oxide?
a) Calcium
b) Magnesium
c) Iron
d) Lithium
Answer: c
Clarification: Iron is relatively unreactive with respect to other metals mentioned. Iron forms iron (III) oxide on heating vigorously. However, lithium, magnesium and calcium all form hydroxides and oxides on reaction with hot water.

10. What is formed when amphoteric oxides react with an alkali solution?
a) Salt only
b) No reaction
c) Acid
d) Salt and water
Answer: d
Clarification: Amphoteric oxides react with both acids and alkalis. On treatment with alkali solution, they behave as acids and undergo reactions analogous to neutralization, thus forming salt and water. For e.g. Al₂O₃ + NaOH → NaAlO₂ + H₂O.

Chemistry Multiple Choice Questions on “D and F-Block Elements – Lanthanoids”.

1. The lanthanides are d-block elements.
a) True
b) False
Answer: b
Clarification: The lanthanides are elements in which the differentiating electron enters (n-2)f-orbital of anti-penultimate shell and are called as f-block elements. These have three outer shells incomplete. Thus, lanthanides are f-block elements, not d-block.

2. What is the general electronic configuration of the lanthanides?
a) (n-2)f ^1-14 (n-1)d ^1-10 ns²
b) (n-2)f ^1-14 (n-1)d ^1-2 ns²
c) (n-2)f ^1-14 (n-1)d ^0-1 ns²
d) (n-2)f ^1-14 (n-1)d ⁰ ns²
Answer: c
Clarification: In lanthanides, the differentiating electron occupies 4f subshell and are rarely found on the earth’s crust, so called rare earths. The general electronic configuration of lanthanides is 6s²5d^0-14f^1-14.

3. What happens to the atomic size of the lanthanides with increase in atomic number?
a) The radius remains unchanged
b) The radius decreases
c) The radius increases
d) The radius first increases and then decreases
Answer: b
Clarification: The gradual decrease in the atomic and ionic radii of the lanthanides with an increase in atomic number is called lanthanide contraction. It occurs due to the poor shielding effect of the 4f electrons.

4. Which of the following is not a consequence of lanthanide contraction?
a) From La⁺³ to Lu⁺³, the ionic radii changes from 106 pm to 85 pm
b) As the size of the lanthanide ions decreases the basic strength increases
c) The basic character of oxides and hydroxides decreases with increase in atomic number
d) The atomic radii of 4d and 5d series is similar
Answer: b
Clarification: The small average decreases in the atomic size is responsible for a small decrease in electronegativity and S.O.P of lanthanides. As the size of the lanthanide ions decreases the covalent character of M—OH bond increases and hence basic strength decreases.

5. What is the most common oxidation state of lanthanides?
a) +2
b) +4
c) +6
d) +3
Answer: d
Clarification: The most common and stable oxidation state of lanthanides is +3. Some elements also exhibit +4 oxidation states. Some elements exhibit +2 oxidation state also due to their half-filled, fully-filled and noble gas configuration.

6. In lanthanides, Ce changes from +4 to +3 oxidation state.
a) True
b) False
Answer: a
Clarification: In lanthanides, Ce changes from +4 to +3 oxidation state. Formation of Ce⁴⁺ is favorable due to its noble gas configuration. However, Ce⁺⁴ changes to Ce⁺³ as it is a strong oxidizing agent. The E° value of Ce⁺⁴/Ce⁺³ is +1.74 V.

7. Which of the following is not a property of lanthanides?
a) They are soft metals with white silvery color
b) They tarnish rapidly by air
c) The hardness of the metals increases with increase in the atomic number
d) The melting point of the metal ranges from 500-1000K
Answer: d
Clarification: All lanthanides are soft metals with silvery white color. They tarnish rapidly by air. With increase in atomic number, the harness of these metals also increases. The melting points of the lanthanides ranges from 1000 to 1200K but samarium melts at 1623K.

8. Which of the following lanthanide ions do not exhibit color?
a) Lu⁺³ and Ln⁺³
b) Lu⁺² and Ln⁺²
c) Ce⁺³ and Ce⁺³
d) Pr⁺⁴ and Ce⁺⁴
Answer: a
Clarification: Many trivalent lanthanide ions are colored both in the solid state and in aqueous state due to f-f electronic transition. Lu⁺³ and Ln⁺³ don’t exhibit color due to absence of unpaired electrons. All lanthanide ions except La⁺³, Ce⁺³ and Lu⁺³ are paramagnetic due to presence of unpaired electrons.

9. Which of the following is the correct order of arrangement of the first five lanthanides according to atomic number?
a) La, Ce, Pr, Nd, Pm
b) La, Pr, Ce, Pm, Nd
c) La, Pr, Ce, Nd, Pm
d) La, Ce, Pr, Pm, Nd
Answer: a
Clarification: The first five elements of Lanthanides are:
Lanthanum (La) – 57
Cerium (Ce) – 58
Praseodymium (Pr) – 59
Neodymium (Nd) – 60
Promethium (Pm) – 61.

10. Which is the last element of lanthanides?
a) Ytterbium
b) Lutetium
c) Thulium
d) Erbium
Answer: b
Clarification: The atomic numbers of Erbium, Thulium, Ytterbium and Lutetium is 68, 69, 70 and 71 respectfully. So the last element of lanthanide series is Lutetium with electronic configuration [Xe]4f¹⁴5d¹6s².

Chemistry Multiple Choice Questions on “Haloalkanes Nomenclature”.

1. What is the common name of 3-Bromopropene?
a) Tert-Butyl bromide
b) Vinyl bromide
c) Allyl bromide
d) Propylidene bromide
Answer: c
Clarification: 3-Bromopropene has 3 C atoms in the parent chain with a double bond at C-1 and Br at C-3. This means that Br is attached to the C which is next to the C-C double bond, hence it is an allylic halide.

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

a) 1-Bromo-3-methylprop-2-ene
b) 3-Bromo-1-methylpropene
c) 1-Bromobut-2-ene
d) 4-Bromobut-2-ene
Answer: c
Clarification: The longest chain of C atoms is 4 (butene) and the numbering should begin from the Br side to satisfy the lowest set of locants rule. So, Br is at C-1 and double bond is at C-2.

3. Which of the following is not the correct IUPAC name for the compound shown?

a) 1-Chloro-3-methylbenzene
b) 3-Chloromethylbenzene
c) 3-Chlorotoluene
d) m-Chlorotoluene
Answer: d
Clarification: All the names given are appropriate for the shown compound, but m-Chlorotoluene is the common name and not the IUPAC name.

4. In the common naming system, the prefix sym- is used for haloarenes with _____ halogen atoms.
a) 1
b) 2
c) 3
d) 4
Answer: c
Clarification: The prefix sym- is used for trihaloarenes with same halogen atom at alternate positions in the benzene ring (1, 3, 5). The prefixes o-, m- and p- are used for dihaloarenes depending on the relative positions of the two identical halogen atoms on the aromatic ring.

5. How many carbon atoms does Isobutyl chloride have in its parent carbon chain?
a) 2
b) 3
c) 4
d) 5
Answer: b
Clarification: The IUPAC name of Isobutyl chloride is 1-Chloro-2-methylpropane, which means the parent chain consists of 3 carbon atoms with an alkyl group at C-2 and a halogen at C-1.

6. Identify the correct naming of the compound H₃C-CHCl₂ from the following?
a) Ethylidene chloride
b) Ethylene dichloride
c) 1,2-Dichloroethane
d) Vic-dichloride
Answer: a
Clarification: The given compound is a dihaloalkane with both the halogens on the same carbon atom. These are also known as gem-dihalides or alkylidene halides.

7. Identify the correct common name for CH₃CH₂CH₂CH₂Cl.
a) Isobutyl chloride
b) n-Butyl chloride
c) sec-Butyl chloride
d) tert-Butyl chloride
View Answer

Answer: b
Clarification: The compound shown is 1-Chlorobutane which has 4 C atoms in the parent chain and the only halogen compound present at the end of the chain. This type structure has the prefix n- in the common name.

8. What is the IUPAC name for the compound (CH₃)₃CCH₂Cl?
a) 1-Chloro-2,2-dimethylpropane
b) 1-Chloro-2,2,2-trimethylethane
c) 2-Chloromethyl-2-methylpropane
d) 2,2-Dimethly-1-Bromopropane
Answer: a
Clarification: There are two different possible chains that consists of maximum 3 C atoms. One of them has two methyl and one chloro group and the other has one methyl and one complex substituent. The preference is given to the former with the chloro at C-1 and the 2 methyl groups at C-2.

9. What is the IUPAC name of the compound shown?

a) 1-Cholor-2-fluorocyclobut-1-ene
b) 2-Chloro-1-fluorocyclobut-1-ene
c) 1-Chloro-2-fluorocyclobut-2-ene
d) 1-Fluoro-2-chlorocyclobut-1-ene
Answer: b
Clarification: The alkyl group is alicyclic with 4 carbon atoms at the corners of a rectangle and a double bond; hence the base is cyclobutene. In the case of cyclic alkenes, the position of double bond is always given the first number.

10. What is the correct IUPAC naming of CHF₂CBrClF?
a) 1-Bromo-1-chloro-1,2,2-trifluoroethane
b) 1-Bromo-1-chloro-1,2,2-trifluoroethene
c) 2-Bromo-2-chloro-1,1,2-trifluoroethane
d) 2-Bromo-2-chloro-1,1,2-trifluoroethene
Answer: a
Clarification: There are only two C atoms in the compounds and only single bonds with two F atoms at C-2 and one of each F, Cl and Br at C-1.

11. The compound 2,2-Dichlorobutane is a geminal dihalide.
a) True
b) False
Answer: a
Clarification: The compound is dihaloalkane with 2 chlorine atoms one the same carbon atom, hence it is a gem dihalide.

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

a) 2-Propylprop-1-ene
b) 2,3-Dimethylbut-1-ene
c) 2,4-Dimethylbut-1-ene
d) 1,3-Dimethylbut-2-ene
Answer: b
Clarification: The parent carbon chain must include the double bond, so there are 4 C atoms including the C-C double bond and the double bond is present at C-1 with two methyl groups at positions 2 and 3.

13. The correct naming of the compound CH≡C-CH=CH-CH₃ is __________
a) Pent-3-en-1-yne
b) Pent-2-en-4-yne
c) Pent-3-ene-1-yne
d) Pent-2-ene-4-yne
Answer: a
Clarification: The lower set of locants is 1, 3 rather than 2, 4. This compound is named as a derivative of alkyne and not alkenes. Also, the terminal ‘e’ in ene is dropped because it comes before ‘y’ of yne.

14. The prefix bis is used when two complex substituents are present on the same of different C atoms on a hydrocarbon.
a) True
b) False
Answer: a
Clarification: The prefix bis is used instead of di for complex substituents in hydrocarbons before the name of the substituent which is enclosed in brackets.

15. How many isomers does C₅H₁₁Br have?
a) 4
b) 6
c) 8
d) 10
Answer: c
Clarification: It has a total of 8 isomers, four are primary and two each are secondary and tertiary haloalkanes.

Chemistry Multiple Choice Questions on “Alcohols and Phenols – 4”.

1. The oxidation of alcohol involves the cleavage of _____ bonds.
a) O-H
b) C-H
c) O-H and C-H
d) Hydrogen
Answer: c
Clarification: Oxidation of alcohols involves the cleavage of both O-H and C-H bonds so as to form a carbon-oxygen double bond. This results in the release of dihydrogen and as a result is also known as dehydrogenation reactions.

2. Which of the following oxidising agents is used to obtain carboxylic acids directly from alcohols?
a) Acidified KMnO₄
b) Aqueous KMnO₄
c) Alkaline KMnO₄
d) Anhydrous CrO₃
Answer: a
Clarification: Strong oxidising agents like acidified potassium permanganate or acidified potassium dichromate convert alcohol directly to carboxylic acid. However, only aldehyde can be obtained by using CrO₃ as the oxidising agent in an anhydrous medium.

3. Which of the following compounds gives a ketone when its vapours are passes over heated copper at 573K?
a) Propan-1-ol
b) Propan-2-ol
c) 2-Methylpropan-1-ol
d) 2-Methylpropan-2-ol
Answer: b
Clarification: Secondary alcohols undergo dehydrogenation when its vapours are passed over Cu at 573K to give ketone. Primary and tertiary alcohols under the same conditions give aldehydes and alkenes respectively.

4. Ortho and para isomers of nitrophenol can be separated by ________ distillation.
a) fractional
b) steam
c) vacuum
d) zone
Answer: b
Clarification: o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding between O of nitro group and H of hydroxyl group. On the other hand, p-Nitrophenol is present as associated molecules due to intermolecular hydrogen bonding, thus making it less volatile. Hence, a mixture of both can be separated by steam distillation.

5. Identify the product of the following reaction.

a) o- and p-Nitrophenol
b) 2- and 4-Phenolsulphonic acid
c) p-Nitrosophenol
d) 2,4,6-Trinitrophenol
Answer: d
Clarification: Picric acid is commonly formed by treating phenol first with concentrated sulphuric acid to form phenol-2,4-disulphinc acid, and then with concentrated nitric acid. It can also be formed treating phenol with only concentrated nitric acid, but the yield of the product in this case in very poor.

6. What is formed when phenol is reacted with bromine water?
a) White precipitate
b) Colourless gas
c) Brown liquid
d) No reaction
Answer: a
Clarification: Phenol on treatment with bromine water gives a white precipitate which is actually 2,4,6-Tribromophenol.

7. Which of the following is incorrect regarding Kolbe’s reaction?
a) The final product is Salicylic acid
b) CO₂ acts as an electrophile
c) Acidification of phenol leads to formation of main product
d) Phenoxide ion undergoes electrophilic attack

Answer: c
Clarification: The first step of Kolbe’s reaction is the formation of sodium phenoxide, which is formed by treating phenol with NaOH. This phenoxide undergoes substitution by a weak electrophile, CO₂, to form the main product, i.e., 2-Hydroxybenzoic acid or salicylic acid.

8. What is the final major product of Reimer-Tiemann reaction?
a) Salicylic acid
b) o-Salicylaldehyde
c) m-Salicylaldehyde
d) p-Salicylaldehyde
Answer: b
Clarification: Phenol is first treated with chloroform and NaOH to form an intermediate substituted benzal chloride (with-CHO group substituted at ortho position of aromatic ring). This is then hydrolysed in the presence of an alkali to form 2-Hydroxy benzaldehyde or o-Salicylaldehyde.

9. Which compound can covert phenol back to benzene?
a) CrO₃
b) Zinc dust
c) Pyridine
d) Strong acid
Answer: b
Clarification: When heated with zinc dust, the C-O bond in phenol is broken and it is reduced to benzene and zinc oxide.

10. In the presence of air, phenols slowly start to form dark coloured mixtures.
a) True
b) False
Answer: a
Clarification: Phenols are slowly oxidised in the presence of air to form dark coloured mixtures containing quinones.

11. Lucas test was conducted on three compounds A, B and C. Compounds A and B showed turbidity at room temperature, but compound C became turbid only after heating. Which of the compounds has a tertiary structure?
a) A
b) A and B
c) C
d) Cannot be determined
Answer: d
Clarification: It is said that A and B show turbidity at room temperature, but it is not mentioned whether the appearance of turbidity is immediate or after some time. So compounds A and B may be tertiary or secondary depending on whether turbidity appears immediately or after 5 minutes respectively. Compound C is may be primary.

Chemistry Multiple Choice Questions on “Uses of Carboxylic Acids”.

1. Which of the following is not a use of methanoic acid?
a) Leather tanning
b) Coagulant in rubber industry
c) Food preservative
d) Textile dyeing
Answer: c
Clarification: Methanoic acid is the simplest acid and is used in rubber manufacturing, leather tanning, textile dyeing and finishing and electroplating industries.

2. Which carboxylic acid is used in the manufacturing of nylon-6,6?
a) Propanedioic acid
b) Butanedioic acid
c) Pentanedioic acid
d) Hexanedioic acid
Answer: d
Clarification: Nylon-6,6 is manufactured from the condensation polymerisation of hexanedioic acid (adipic acid) with hexamethylenediamine under high pressure and temperature.

3. Esters of which acid are used in the perfume industry?
a) Ethanoic acid
b) Benzoic acid
c) Phthalic acid
d) Formic acid
Answer: b
Clarification: Esters of benzoic acids are used in the perfumery industry. For example, ethyl benzoate is a component of some fragrances and artificial flavours.

4. __________ is commonly used as a food preservative.
a) Sodium benzoate
b) Potassium benzoate
c) Terephthalic acid
d) Acetic acid
Answer: a
Clarification: When C₆H₅COOH reacts with NaOH, a salt called sodium benzoate is formed. When an aqueous salt solution is prepared from this product, it acts as a preservative. It is identified by the code E211.

5. Which of the following carboxylic acids is not used in the manufacturing of soaps?
a) Capric acid
b) Stearic acid
c) Oleic acid
d) Palmitic acid
Answer: a
Clarification: Soaps are sodium or potassium salts of long chain fatty acids like stearic, oleic, palmitic acids, etc. Capric acid is decanoic acid and contains 10 carbon atoms and is not classified as a fatty acid.

6. Ethanoic acid is used as vinegar in the cooking industry.
a) True
b) False
Answer: a
Clarification: Generally, a 1:10 solution of acetic acid in water is called vinegar. It is a commercial product that may be used in salads or other foods. It may include traces of other chemicals for the need of flavouring.