Unit III MCQ Practice – Fats and Oils

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1. Fats and oils are chemically composed mainly of:

Fats and Oils · 1 Mark Practice

A. Esters of glycerol with higher fatty acids B. Ethers of glycerol C. Aromatic hydrocarbons D. Salts of fatty acids

Answer: A. Esters of glycerol with higher fatty acids

Explanation: Fats and oils consist mainly of triglycerides. A triglyceride is formed when the three hydroxyl groups of glycerol are esterified with three molecules of higher fatty acids.

Reaction:
Glycerol + 3 R–COOH ⇌ Triglyceride + 3 H₂O

2. Which of the following is a saturated fatty acid?

Fatty Acids · 1 Mark Practice

A. Oleic acid B. Linoleic acid C. Stearic acid D. Linolenic acid

Answer: C. Stearic acid

Explanation: Stearic acid contains no carbon–carbon double bond and is therefore classified as a saturated fatty acid.

Formula:
Stearic acid = C₁₇H₃₅COOH

3. Alkaline hydrolysis of fats and oils is known as:

Saponification · 1 Mark Practice

A. Hydrogenation B. Saponification C. Oxidation D. Reduction

Answer: B. Saponification

Explanation: Saponification is the alkaline hydrolysis of triglycerides. It produces glycerol and sodium or potassium salts of fatty acids, commonly called soaps.

Reaction:
Triglyceride + 3 NaOH → Glycerol + 3 R–COONa

4. Hydrogenation of an unsaturated oil involves:

Hydrogenation · 1 Mark Practice

A. Addition of hydrogen across carbon–carbon double bonds B. Removal of glycerol C. Formation of additional double bonds D. Conversion of fatty acids into amino acids

Answer: A. Addition of hydrogen across carbon–carbon double bonds

Explanation: During hydrogenation, hydrogen is added across the C=C bonds of unsaturated fatty-acid chains. The degree of unsaturation decreases and the oil generally becomes more solid.

Reaction:
R–CH=CH–R + H₂ → R–CH₂–CH₂–R
Condition: Ni catalyst, heat

5. Rancidity of fats and oils is best described as:

Rancidity · 1 Mark Practice

A. Deterioration producing unpleasant odour and taste B. Complete hydrogenation of an oil C. Formation of glycerol only D. Conversion of oil into a mineral oil

Answer: A. Deterioration producing unpleasant odour and taste

Explanation: Rancidity is the deterioration of fats and oils. Oxidative rancidity involves oxidation of unsaturated lipids, while hydrolytic rancidity involves liberation of free fatty acids.

Reaction:
Unsaturated lipid + O₂ → Peroxides → Aldehydes / Ketones / Acids

6. Drying oils form a hard film on exposure to air mainly because of:

Drying Oils · 1 Mark Practice

A. Oxidation and polymerization B. Hydrolysis only C. Saponification only D. Evaporation of glycerol

Answer: A. Oxidation and polymerization

Explanation: Highly unsaturated oils absorb oxygen from air and undergo oxidative cross-linking. This produces a tough, continuous polymeric film.

Reaction:
Unsaturated oil + O₂ → Oxidation → Cross-linked polymeric film

7. Which of the following is a typical drying oil?

Drying Oils · 1 Mark Practice

A. Coconut oil B. Linseed oil C. Liquid paraffin D. Mineral oil

Answer: B. Linseed oil

Explanation: Linseed oil contains a high proportion of polyunsaturated fatty acids and readily undergoes oxidative polymerization on exposure to air.

8. Iodine value is primarily a measure of:

Analytical Constants · Iodine Value

A. Free fatty acid content B. Hydroxyl group content C. Degree of unsaturation D. Volatile fatty acid content

Answer: C. Degree of unsaturation

Explanation: Iodine value indicates the degree of unsaturation because halogen reagent adds across carbon–carbon double bonds. More double bonds result in greater halogen uptake.

Reaction:
C=C + ICl → Addition product

9. A higher iodine value generally indicates:

Analytical Constants · Iodine Value

A. Greater unsaturation B. Lower unsaturation C. Greater free glycerol content D. Absence of double bonds

Answer: A. Greater unsaturation

Explanation: A sample containing more C=C bonds consumes more halogen reagent and therefore has a higher iodine value.

Exam Point:
Higher iodine value → greater degree of unsaturation.

10. Acid value is defined as the milligrams of KOH required to neutralize the free fatty acids present in:

Analytical Constants · Acid Value

A. 100 g of fat or oil B. 1 g of fat or oil C. 10 g of glycerol D. 1 litre of water

Answer: B. 1 g of fat or oil

Explanation: Acid value measures free fatty acid content and is expressed as the milligrams of KOH required to neutralize the free acids in one gram of sample.

Reaction:
R–COOH + KOH → R–COOK + H₂O

Formula:
Acid Value = (V × N × 56.1) / W

11. An increase in acid value during storage generally indicates:

Analytical Constants · Acid Value

A. Formation of free fatty acids B. Complete hydrogenation C. Removal of all fatty acids D. Increase in protein content

Answer: A. Formation of free fatty acids

Explanation: Hydrolytic deterioration of triglycerides releases free fatty acids. Therefore, acid value generally rises as hydrolysis progresses.

Reaction:
Triglyceride + H₂O → Glycerol + Free fatty acids

12. Saponification value is the milligrams of KOH required to saponify:

Analytical Constants · Saponification Value

A. 100 g of fat B. 10 g of oil C. 1 g of fat or oil D. 1 mole of glycerol

Answer: C. 1 g of fat or oil

Explanation: Saponification value is the number of milligrams of KOH required to neutralize free acids and saponify the esters present in one gram of sample.

13. A high saponification value generally indicates:

Analytical Constants · Saponification Value

A. Fatty acids of very high average molecular mass B. Fatty acids of lower average molecular mass C. Complete absence of esters D. Absence of fatty acids

Answer: B. Fatty acids of lower average molecular mass

Explanation: For the same mass of fat, shorter fatty-acid chains correspond to a greater number of ester groups. More KOH is therefore required for saponification.

Exam Point:
Higher SV → lower average molecular mass of constituent fatty acids.

14. Ester value is calculated by:

Analytical Constants · Ester Value

A. Acid value + Iodine value B. Acid value − Saponification value C. Iodine value − Acid value D. Saponification value − Acid value

Answer: D. Saponification value − Acid value

Explanation: Acid value represents free fatty acids, whereas saponification value includes the alkali requirement associated with free acids and esterified fatty acids. Their difference gives the ester value.

Formula:
Ester Value = Saponification Value − Acid Value

15. Acetyl value is particularly useful for estimating:

Analytical Constants · Acetyl Value

A. Free hydroxyl groups B. Carbon–carbon double bonds only C. Volatile water-soluble fatty acids D. Protein content

Answer: A. Free hydroxyl groups

Explanation: Hydroxyl-containing constituents can be acetylated. Therefore, acetyl value provides information about the amount of acetylatable –OH groups.

Reaction:
R–OH + (CH₃CO)₂O → R–OCOCH₃ + CH₃COOH

16. Which oil is characteristically associated with hydroxy fatty acids and therefore a significant acetyl value?

Analytical Constants · Acetyl Value

A. Mineral oil B. Liquid paraffin C. Castor oil D. Petroleum ether

Answer: C. Castor oil

Explanation: Castor oil contains a high proportion of ricinoleic acid, which possesses a hydroxyl group and can undergo acetylation.

17. Reichert–Meissl value is related to:

Analytical Constants · Reichert–Meissl Value

A. Degree of unsaturation B. Volatile water-soluble fatty acids C. Hydroxyl groups only D. Total glycerol content

Answer: B. Volatile water-soluble fatty acids

Explanation: RM value estimates volatile, water-soluble fatty acids obtained after saponification, acidification and distillation of the fat.

18. Which fat is known for a comparatively high Reichert–Meissl value?

Analytical Constants · Reichert–Meissl Value

A. Butter fat B. Mineral oil C. Liquid paraffin D. Petroleum jelly

Answer: A. Butter fat

Explanation: Butter fat contains appreciable quantities of short-chain volatile fatty acids, including butyric acid, and therefore shows a relatively high RM value.

19. Which catalyst is commonly used for hydrogenation of vegetable oils?

Hydrogenation · 1 Mark Practice

A. NaCl B. H₂SO₄ C. Nickel D. KMnO₄

Answer: C. Nickel

Explanation: Finely divided nickel is commonly used as a catalyst for the catalytic hydrogenation of unsaturated vegetable oils.

Reaction:
C=C + H₂ → C–C
Condition: Ni catalyst, heat

20. Partial hydrogenation of unsaturated oils may produce:

Hydrogenation · 1 Mark Practice

A. Amino acids B. Aromatic acids C. Proteins D. Trans fatty acids

Answer: D. Trans fatty acids

Explanation: During partial hydrogenation, some cis double bonds may undergo isomerization and form trans double bonds.

21. Which fatty acid contains one carbon–carbon double bond?

Fatty Acids · 1 Mark Practice

A. Stearic acid B. Oleic acid C. Linoleic acid D. Linolenic acid

Answer: B. Oleic acid

Explanation: Oleic acid is a monounsaturated fatty acid containing one C=C bond.

Formula:
Oleic acid = C₁₇H₃₃COOH

22. Linoleic acid contains:

Fatty Acids · 1 Mark Practice

A. No C=C bond B. One C=C bond C. Two C=C bonds D. Four C=C bonds

Answer: C. Two C=C bonds

Explanation: Linoleic acid is a polyunsaturated fatty acid containing two carbon–carbon double bonds.

23. Hydrolysis of a triglyceride produces:

Hydrolysis · 1 Mark Practice

A. Glycerol and fatty acids B. Only glycerol C. Only fatty acids D. Amines and alcohols

Answer: A. Glycerol and fatty acids

Explanation: Hydrolysis cleaves the three ester bonds of a triglyceride to produce glycerol and the corresponding fatty acids.

Reaction:
Triglyceride + 3 H₂O → Glycerol + 3 R–COOH

24. The sodium salts of higher fatty acids produced during saponification are commonly called:

Saponification · 1 Mark Practice

A. Waxes B. Steroids C. Proteins D. Soaps

Answer: D. Soaps

Explanation: Sodium or potassium salts of higher fatty acids are commonly known as soaps.

Reaction:
Triglyceride + 3 NaOH → Glycerol + 3 R–COONa

25. Oxidative rancidity occurs more readily in:

Rancidity · 1 Mark Practice

A. Highly saturated fats B. Highly unsaturated oils C. Mineral oils only D. Pure glycerol

Answer: B. Highly unsaturated oils

Explanation: Carbon–carbon double bonds are susceptible to oxidative attack. Therefore, increasing unsaturation generally increases susceptibility to oxidative rancidity.

26. Which of the following helps to retard oxidative rancidity?

Rancidity · 1 Mark Practice

A. Addition of antioxidants B. Increased exposure to light C. Increased contact with oxygen D. Storage at higher temperature

Answer: A. Addition of antioxidants

Explanation: Antioxidants retard lipid oxidation by interfering with free-radical chain reactions. Protection from heat, light and oxygen also improves stability.

27. Iodine value is conventionally expressed as grams of iodine absorbed by:

Analytical Constants · Iodine Value

A. 1 g of sample B. 10 g of sample C. 100 g of sample D. 1000 g of sample

Answer: C. 100 g of sample

Explanation: Iodine value is the number of grams of iodine equivalent absorbed by 100 g of a fat or oil under specified conditions.

28. Which analytical constant is most directly related to the free fatty acid content of an oil?

Analytical Constants · Revision

A. Iodine value B. Acid value C. Acetyl value D. Reichert–Meissl value

Answer: B. Acid value

Explanation: Acid value directly measures the alkali required to neutralize free fatty acids in the sample and is therefore an indicator of free fatty acid content.

29. Which analytical constant is most directly related to hydroxyl-containing fatty acids?

Analytical Constants · Revision

A. Iodine value B. Acid value C. Reichert–Meissl value D. Acetyl value

Answer: D. Acetyl value

Explanation: Acetyl value is associated with acetylatable hydroxyl groups and is particularly useful for oils containing hydroxy fatty acids.

30. Which set is correctly matched?

Unit III · Integrated Revision

A. Acid value — free fatty acids; Iodine value — unsaturation; RM value — volatile water-soluble fatty acids B. Acid value — unsaturation; Iodine value — hydroxyl groups; RM value — glycerol C. Acid value — volatile acids; Iodine value — free fatty acids; RM value — unsaturation D. Acid value — hydroxyl groups; Iodine value — ester groups; RM value — double bonds

Answer: A. Acid value — free fatty acids; Iodine value — unsaturation; RM value — volatile water-soluble fatty acids

Explanation: The major Unit III analytical constants should be distinguished clearly: acid value reflects free fatty acids, iodine value reflects unsaturation, saponification value relates to saponifiable material and average chain length, acetyl value relates to hydroxyl groups, and RM value relates to volatile water-soluble fatty acids.

Quick Revision:
AV → Free fatty acids
SV → Saponification / average fatty-acid molecular mass
EV → SV − AV
IV → Unsaturation
Acetyl value → –OH groups
RM value → Volatile water-soluble fatty acids