Category: Class 12 Chemistry Objective Questions

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

1. Which of the following methods cannot produce aldehydes?
a) Oxidation of primary alcohols
b) Dehydrogenation of secondary alcohols
c) Ozonolysis of alkenes
d) Hydration of ethyne with acid
Answer: b
Clarification: The dehydrogenation of secondary alcohols give ketones. Aldehydes are obtained by the dehydrogenation of primary alcohols.

2. Which of the following reactions can produce ketones?
a) Oxidation of primary alcohols
b) Dehydrogenation of primary alcohols
c) Dehydrogenation of tertiary alcohols
d) Oxidation of secondary alcohols
Answer: d
Clarification: Oxidation and dehydrogenation of secondary alcohols results in ketones. The same reactions with primary alcohols give aldehydes.

3. Conversion of propyne to acetone requires three important reagents. Identify which of the following is not one of the three?
a) Water
b) Zinc dust
c) H₂SO₄
d) HgSO₄
Answer: b
Clarification: Addition of water to propyne in the presence of H₂SO₄ and HgSO₄ gives acetone. Zinc dust is an important reagent in the ozonolysis of alkenes.

4. What is the catalyst used in the hydrogenation of acetyl chloride to produce ethanal?
a) Pt over BaSO₄
b) Pt over CuSO₄
c) Pd over BaSO₄
d) Pd over CuSO₄
Answer: c
Clarification: Acyl chlorides are hydrogenated in the presence of catalyst palladium over barium sulphate. This is known as Rosenmund reaction.

5. Which of the following carbonyl compounds can be prepared from Rosenmund reaction?
a) Methanal
b) Acetone
c) Butanone
d) Benzaldehyde
Answer: d
Clarification: Rosenmund reaction is exclusively used for the preparation of aldehydes by the substitution of chloride by hydrogen. Given this, methanal cannot be formed from this reaction because its corresponding acyl chloride, i.e., formyl chloride, is unstable at room temperature. Benzaldehyde is formed from benzoyl chloride.

6. Which of the following is required in Stephen reaction?
a) LiCl
b) NiCl₂
c) SnCl₂
d) TiCl₄
Answer: c
Clarification: Nitriles are converted to respective imines with SnCl₂ in the presence of HCl, which on hydrolysis gives corresponding aldehyde.

7. Which of the following compounds helps in reducing esters to aldehydes?
a) BINAL-H
b) DIBAL-H
c) DIPT
d) TBAF
Answer: b
Clarification: DIBAL-H or diisobutylaluminium hydride is a reducing agent which are used to reduce nitriles to imines or esters to aldehydes. These are important in the preparation of aldehydes.

8. Identify ‘X’ in the reaction given below.

a) CrO₃
b) CrO₂Cl₂
c) Alkaline KMnO₄
d) Anhydrous AlCl₃
Answer: b
Clarification: This is known as Etard reaction. The chromyl chloride oxidises the methyl group to a chromium complex in CS₂. This complex on hydrolysis gives benzaldehyde.

9. Benzaldehyde can be obtained from the hydrolysis of the product formed during side chain chlorination of toluene.
a) True
b) False
Answer: a
Clarification: Side chain chlorination of toluene in the presence of light (represented by hv) gives benzal chloride which on hydrolysis gives benzaldehyde. This is an important commercial method for the production of benzaldehyde.

10. Identify the reagent(s) for the conversion of chlorobenzene to 3-chlorobenzaldehyde.
a) CrO₃ and (CH₃CO)₂O
b) CrO₂Cl₂; H₂O
c) Cl₂/hv; H₂O
d) CO, HCl and CuCl
Answer: d
Clarification: When benzene or its derivative is treated with CO and HCl in the presence of anhydrous AlCl₃ or CuCl, it gives a benzaldehyde or substituted benzaldehyde. This is known as Gatterman-Koch reaction.

11. Acetyl chloride reacts with _______ to give butan-2-one.
a) cadmium chloride
b) methyl magnesium chloride
c) dimethyl cadmium
d) diethyl cadmium
Answer: d
Clarification: For butan-2-one to form from acetyl chloride, the Cl group needs to be replaced by an ethyl group. This ethyl group is obtained from diethyl cadmium which is produced from the reaction between ethyl magnesium bromide and cadmium chloride.

12. What is formed after the hydrolysis of the product of the reaction between benzonitrile and methyl magnesium bromide in dry ether?
a) Phenyl acetaldehyde
b) Acetophenone
c) 1-Phenylpropanone
d) Benzaldehyde
Answer: b
Clarification: When a nitrile is treated with a Grignard reagent in the presence of dry ether, an addition product is formed which on hydrolysis gives a ketone.

13. Friedel-Crafts benzoylation of benzene gives ________
a) acetophenone
b) propiophenone
c) benzophenone
d) no reaction
Answer: c
Clarification: When benzene is treated with benzoyl chloride in the presence of anhydrous aluminium chloride, it forms benzophenone which is an aromatic ketone.

14. Identify the reagent for the conversion of but-2-ene to ethanal.
a) O₃/H₂O-Zn dust
b) H₂O, H₂SO₄, HgSO₄
c) PCC
d) DIBAL-H

Answer: a
Clarification: But-2-ene on ozonolysis followed by reaction with water and zinc dust gives ethanal or sometimes a mixture of an aldehyde and ketone depending on the substitution pattern.

15. Which of the following reactions does not give benzaldehyde?
a) Rosenmund reaction
b) Etard reaction
c) Gatterman-Koch reaction
d) Friedel-Craft acylation reaction
Answer: d
Clarification: Friedel-Crafts acylation of benzene produces an aromatic ketone and not benzaldehyde. Benzene on Gatterman-Koch reaction gives benzaldehyde, so does toluene on Etard reaction. Rosenmund reaction also gives benzaldehyde but from benzoyl chloride.

Chemistry Multiple Choice Questions on “Preparation of Amines – 1”.

1. Which of the following is the most preferred reagent for reducing nitroethane to ethylamine?
a) H₂/Pt
b) Sn/HCl
c) Fe/HCl
d) Zn/HCl
Answer: c
Clarification: Reduction of nitroalkanes with iron scrap and HCl is preferred because of the formation of FeCl₂ which gets hydrolysed to release HCl during the reaction. Thus, only a small amount of HCl is required to initiate the reaction.

2. Which of the following reagents cannot be used to convert nitrobenzene to aniline?
a) LiAlH₄-dry ether
b) SnCl₂/HCl
c) H₂/Pd-ethanol
d) Zn in HCl
Answer: a
Clarification: LiAlH₄ may be used for the reduction of aliphatic nitro compounds to respective amines. But with aromatic nitro compounds (nitrobenzene), it gives azobenzene and not primary amines (aniline).

3. What compound is formed when hydrogen gas is passed through nitrobenzene in the presence of finely divided nickel?
a) Aniline
b) 2-Nitroaniline
c) 3-Nitroaniline
d) 4-Nitroaniline
Answer: a
Clarification: The nitro (NO₂) group of nitrobenzene is reduced to amino (NH₂) group by the replacement of two oxygens by two hydrogen to give aniline or benzenamine.

4. How many water molecules are formed as the by product of reduction of one molecule of nitropropane to one molecule of propanamine, with hydrogen gas in Pt catalyst?
a) 1
b) 2
c) 3
d) 4
Answer: b
Clarification: One molecule of nitropropane reacts with 3 molecules of hydrogen gas (H₂) in the presence of Ni/Pt/Pd catalyst to form one molecule of propanamine along with two molecules of water.

5. Ammonolysis is a type of _________ reaction.
a) electrophilic addition
b) electrophilic substitution
c) nucleophilic addition
d) nucleophilic substitution
Answer: d
Clarification: Alkyl and benzyl halides on reaction with an alcoholic solution of ammonia (nucleophile) undergoes nucleophilic substitution of the halogen atom by the amino group (-NH₂). This process is known as ammonolysis.

6. What is the type of amine obtained from the ammonolysis of alkyl halides?
a) Primary
b) Primary and secondary
c) Secondary and tertiary
d) Primary, secondary and tertiary
Answer: c
Clarification: When an alcoholic solution of ammonia is heated with alkyl halides, a mixture of all three types of amines, i.e., primary, secondary and tertiary amines are formed along with a quaternary ammonium salt.

7. What is the most suitable condition for the ammonolysis of an alkyl halide?
a) 273K, open tube
b) 273K, sealed tube
c) 373K, open tube
d) 373K, sealed tube
Answer: d
Clarification: The cleavage of the carbon-halogen bond (of the alkyl halide) by a nucleophile takes place under high pressure and temperature. Thus, the ammonolysis is carried out in a sealed tube at about 100°C.

8. What is obtained when the following compound undergoes ammonolysis in a sealed tube at 373K with iodomethane?

a) Dimethylamine
b) N-Iodo-N-methylmethanamine
c) Tetramethyl ammonium iodide
d) No reaction
Answer: c
Clarification: Tertiary amines are formed as the ammonolysis of alkyl halides proceeds to form 1° and 2° amines, which further reacts with the alkyl halide. The compound shown is trimethylamine which reacts with iodomethane (alkyl halide) to form a quaternary ammonium salt [(CH₃)4N⁺]I^–.

9. Aniline can be formed from the ammonolysis of chlorobenzene.
a) True
b) False
Answer: b
Clarification: Ammonolysis is not suitable for the production of aryl amines, because aryl halides are stable and relatively less reactive toward nucleophilic substitution than alkyl halides. Since chlorobenzene is an aryl halide, it does not undergo ammonolysis to form aniline.

10. The main product formed when the ammonolysis is carried out using excess of iodoethane is _______
a) 1° amine
b) 2° amine
c) 3° amine
d) quaternary ammonium salt
Answer: d
Clarification: When an excess of alkyl halide is used, the presence of a base will allow the excess hydrogen halide formed to be consumed, resulting in the formation of quaternary ammonium salt as a major product.

11. What is the correct order of reactivity of the following alkyl halides towards ammonolysis reaction?
a) CH₃I > CH₃Br > CH₃Cl
b) CH₃I > CH₃Cl > CH₃Br
c) CH₃Cl > CH₃Br > CH₃I
d) CH₃Br > CH₃Cl > CH₃I
Answer: a
Clarification: Iodine is the most electropositive atom of chlorine and bromine halogens. This makes it more susceptible to a nucleophilic attack than towards Cl or Br. Hence, CH₃I is more reactive towards ammonolysis.

12. Ammonolysis is a reaction between an alkyl halide and most preferably an ______ solution of NH₃.
a) alkaline
b) acidic
c) alcoholic
d) aqueous
Answer: c
Clarification: An alcoholic medium promotes the occurrence of a nucleophilic attack of NH₃ molecule on the alkyl/benzyl halide. Generally, ethanol is used for this purpose.

13. Identify the product B in the shown reaction.

a) Benzylamine
b) N-Methylphenylmethanamine
c) N,N-Dimethylphenylmethanamine
d) N,N-Dichlorophenylmethanamine
Answer: c
Clarification: Benzyl chloride reacts with ethanolic ammonia to form benzylamine (A) which reacts with one molecule of CH₃Cl to give N-methyphenylmethanamine, which further reacts with one more molecule of CH₃Cl to give N, N-dimethylphenylmethanamine.

14. Only primary amine is formed as the main product when ammonolysis of an alkyl halide takes place with large excess of ammonia.
a) True
b) False
Answer: a
Clarification: When NH₃ is taken in excess in the reaction mixture, the alkyl halide is more likely to react with an NH₃ molecule rather than with a molecule of amine which is present in smaller amount. Therefore, only primary amine is formed as the main product.

15. Which of the following would not be a good choice for reducing an aryl nitro compound to an amine?
a) H₂/Pt
b) LiAlH₄-ether
c) Fe and HCl
d) Sn and HCl
Answer: b
Clarification: Lithium aluminium hydride is a reducing agent but does not react with nitro compounds. Instead it reduces nitriles to give corresponding primary amines.

Chemistry Multiple Choice Questions on “Biomolecules Carbohydrates – 4”.

1. Disaccharides on _______ with dilute HCl yield two same or different monosaccharides.
a) hydration
b) hydrolysis
c) oxidation
d) carbonation
Answer: b
Clarification: Disaccharides react with water in the presence of dilute acids or enzymes to give two monosaccharide units. These units may be same of different.

2. Which of the following disaccharides on hydrolysis with invertase gives two different monosaccharides?
a) Sucrose
b) Lactulose
c) Lactose
d) Maltose
Answer: a
Clarification: Invertase is an enzyme that specifically acts on sucrose. It is named so because it forms invert sugar which is a mixture of glucose and fructose.

3. Two monosaccharides are joined through a ______ bond to form a disaccharide.
a) ionic
b) peptide
c) glycosidic
d) phosphodiester
Answer: c
Clarification: Two monosaccharide units together lose a molecule of water and combine by an oxide linkage. Such a bond between two monosaccharide units through oxygen atom is called glycosidic linkage.

4. Sucrose is a _______ compound and the product mixture obtained from its hydrolysis is _______ in nature.
a) dextrorotatory; dextrorotatory
b) dextrorotatory; laevorotatory
c) laevorotatory; dextrorotatory
d) laevorotatory; laevorotatory
Answer: b
Clarification: Sucrose is dextrorotatory but on hydrolysis gives a mixture dextrorotatory glucose and laevorotatory fructose. Since, the specific rotation of fructose is greater than that of glucose, the resultant mixture is laevorotatory.

5. Which of the following is false regarding the reaction of sucrose to give fructose and glucose?
a) It can take place in the presence of enzyme sucrase
b) It results in the formation of a glycosidic bond.
c) It is an inversion reaction
d) It is a hydrolysis reaction
Answer: b
Clarification: Hydrolysis of sucrose (in the presence of either dilute acid, invertase or sucrase) results in the cleavage of glycosidic bond to give glucose and fructose. Sucrase results in the cleavage of O-C(glucose) bond, whereas invertase results in the cleavage of O-C(fructose bond).

6. Sucrose is made of which of the following monosaccharides?
a) α-D-glucose, α-D-fructose
b) α-D-glucose, β-D-fructose
c) β-D-glucose, α-D-fructose
d) β-D-glucose, β-D-fructose
Answer: b
Clarification: Sucrose is composed of α-D-glucose and β-D-fructose. These two are held together by an α,β-glycosidic linkage between C₁ of glucose (pyranose) and C₂ of fructose (furanose).

7. Identify the enzyme X in the following reaction.

a) Amylase
b) Maltase
c) Zymase
d) Diastase
Answer: d
Clarification: Diastase is an enzyme present in malt sugar that particularly catalyses the hydrolysis of starch to maltose. Amylase, maltase and zymase help in the breakdown of certain saccharides into other compounds.

8. Which of the following is incorrect regarding maltose?
a) It consists of two glucopyranose units
b) It is a non-reducing sugar
c) Glycosidic bond between C₁ of one unit and C₄ of the other unit
d) It is a disaccharide
Answer: b
Clarification: The free aldehyde group can be produced at the C₁ carbon of the second α-D-glucose unit in solution and it shows reducing properties. Hence, it is a reducing sugar.

9. Which of the following is also known as beet sugar?
a) Fructose
b) Sucrose
c) Maltose
d) Lactose
Answer: b
Clarification: Sucrose is widely distributed among plants, particularly sugarcane and sugar beet. The sugar obtained from sugar beet is known as beet sugar, whereas that obtained from sugarcane is cane sugar.

10. Identify the saccharide from the Haworth projection shown below.

a) Lactose
b) Maltose
c) Sucrose
d) Trehalose
Answer: a
Clarification: It is composed of β-D-galactose (left) and β-D-glucose (right). There is a β-glycosidic linkage between the C₁ of galactose and C₄ of glucose unit. It is a non-reducing sugar.

11. Which of the following statements is incorrect with respect to starch?
a) It is a reducing carbohydrate
b) It is a polymer of α-D-glucose
c) It gives blue colour with iodine
d) It consists of branched chains
Answer: a
Clarification: Starch is a non-reducing saccharide as it does not reduce Fehling’s solution or Tollen’s reagent. This indicates that all hemiacetal hydroxyl groups of glucose units are not free nut are linked by glycosidic bonds.

12. Amylum is one of the components of starch.
a) True
b) False
Answer: b
Clarification: Starch is a polymer of α-glucose consisting of two components namely, amylose and amylopectin. Amylum is just another common name for starch.

13. The branching in the ________ component of starch occurs by the glycosidic linkage between _______ carbons.
a) amylose; C₁-C₄
b) amylose; C₁-C₆
c) amylopectin; C₁-C₄
d) amylopectin; C₁-C₆
Answer: d
Clarification: Amylopectin is the water insoluble branched constituent of starch which makes up 80-85% of its structure. It consists of straight chains of glucose through C₁-C₄ linkage, with the C₁ of the terminal glucose of each chain linking with the C₆ of another glucose unit in the next chain.

14. Cellulose is a _______ saccharide.
a) reducing
b) branched chain
c) β-glucose straight chain
d) oligo
Answer: c
Clarification: Cellulose is a straight chain polysaccharide composed of only β-D-glucose units which are joined by glycosidic links between C₁ of one glucose unit and C₄ of the next glucose unit.

15. Starch : Plants : : X : Animals. Identify X.
a) Starch
b) Glucose
c) Cellulose
d) Glycogen
Answer: d
Clarification: Starch is the main storage polysaccharide of plants. On the other hand, carbohydrates are stored in animals in the form of glycogen. It is also known as animal starch because of its similar structure to amylopectin.

Chemistry Online Test for IIT JEE Exam on “Therapeutic Action of Different Classes of Drugs – 2”.

1. Which of the following is a bactericidal antibiotic?
a) Erythromycin
b) Ofloxacin
c) Tetracycline
d) Chloramphenicol
Answer: b
Clarification: Antibiotics can either kill the target microbes (cidal effect) or can prevent it from pathogenic action (static effect). The former are known as bactericidal and the latter is known as bacteriostatic.

2. Identify the incorrect statement with respect to antibiotics.
a) Arsphenamine is toxic to humans
b) Ampicillin is a broad-spectrum antibiotic
c) Penicillin is bacteriostatic antibiotic
d) Sulphanilamide has antibacterial properties
Answer: c
Clarification: Penicillin is an antibacterial compound derived from Penicillium. It has a bactericidal effect on microbes. Ampicillin is a modified version of penicillin with a wider spectrum. Arsphenamine is an -As=As- linkage compound that resists bacterial action and is also toxic to human beings. Sulphanilamide is an active antibiotic compound formed in the body.

3. A certain antibiotic X is effective only against a few types of harmful microbes and cells. X is a _______ antibiotic.
a) broad-spectrum
b) narrow spectrum
c) limited spectrum
d) antibacterial
Answer: b
Clarification: Antibiotics are classified based on the range of microbes that are affected by it. Broad spectrum antibiotics affect Gram-positive as well as Gram-negative bacteria. Narrow spectrum are used only against either Gram-positive or Gram-negative bacteria, whereas limited spectrum antibiotics kill or inhibit only one particular organism.

4. Which of the following is a narrow spectrum antibiotic?
a) Amoxycillin
b) Chloramphenicol
c) Vancomycin
d) Dysidazirine
Answer: d
Clarification: Dysidazirine is effective in treatment of only certain strains of tumour cells, hence they are narrow spectrum. Chloramphenicol, vancomycin and amoxycillin are examples of broad-spectrum antibiotics, that are used for treating different types of problems.

5. Which of the following are not supposed to treat humans directly?
a) Disinfectants
b) Antimalarials
c) Antiseptics
d) Antibiotics
Answer: a
Clarification: Antimalarials and antibiotics are ingested in the form of pills. Antiseptics are applied on the skin. Disinfectants are used for cleaning objects like floors, toilets, drains, etc. and protect them from pathogenic activity.

6. Which of the following is used as an antiseptic for eyes?
a) Iodoform
b) Tincture of iodine
c) Dettol
d) Boric acid
Answer: d
Clarification: Dettol, tincture of iodine and iodoform are common antiseptics that are directly applied on wounds to treat them. Boric acid is prepared as a dilute aqueous solution and acts as an antiseptic for the eyes.

7. Which of the following compounds is added in soaps as an antiseptic?
a) Chloroxylenol
b) Terpineol
c) Bithionol
d) Thymol
Answer: c
Clarification: Bithionol consists of two dichloro and dihydroxy substituted aromatic rings connected by a sulphur bridge. It is added to soaps as an antiseptic. It also helps to reduce the smell produced by microbial decomposition on the skin.

8. Phenol acts as an antiseptic.
a) True
b) False
Answer: a
Clarification: Phenol acts as a disinfectant or as an antiseptic based on its concentration in solution. A 0.2% phenol solution acts as an antiseptic, while a 1% phenolic solution is a disinfectant.

9. Which of the following antifertility drugs consists of Estrogen derivatives?
a) Norethindrone
b) Novestrol
c) Mifepristone
d) Ormeloxifene

Answer: b
Clarification: Progesterone is a hormone that is known to suppress ovulation. Antifertility drugs are made of synthetic Estrogen and progesterone derivatives. Ethynylestradiol is a drug with a combination of these two derivatives.

Chemistry Online Test for IIT JEE Exam,

Chemistry Multiple Choice Questions on “Solid State Characteristics”.

1. Solid state is denser than the liquid and gaseous states of the same substance. Which of the following is an exception to this rule?
a) Mercury
b) Carbon dioxide (dry ice)
c) Ice
d) NaCl
Answer: c
Clarification: The density of ice is about 0.92 g/cm³ while that of water is 1 g/cm³. Mercury has density 14.184 g/cm³ as solid and 13.69 g/cm³ as liquid. Carbon dioxide has density 1.56 g/cm³ as solid and 1.10 g/cm³ as liquid. NaCl has density 2.71 g/cm³ as solid and 1.556 g/cm³ in molten state. Therefore only ice has lesser density as a solid than as a liquid.

2. Which of the following can be used to describe a crystalline solid?
a) Heterogeneous, anisotropic
b) Homogeneous, anisotropic
c) Heterogeneous, isotropic
d) Homogeneous, isotropic
Answer: b
Clarification: Homogeneity refers to uniformity in composition, which is a characteristic property of crystalline solids. Isotropy is when the values of physical properties do not change with direction throughout the body of the solid. Crystalline solids are anisotropic because the composition of the solid changes with direction, hence the physical properties also change with direction.

3. When a single substance can crystallize in two or more forms under different conditions provided, it is called as _________
a) Polymorphous
b) Isomorphous
c) Semimorphous
d) Multimorphous
Answer: a
Clarification: Isomorphous is when two or more substances have the same crystal structure. Polymorphous is when a single substance can crystallize in two or more forms depending upon the conditions.

4. Which of the following is an amorphous solid?
a) Quartz
b) Quartz glass
c) Graphite
d) Salt (NaCl)
Answer: b
Clarification: Quartz glass does not have a perfectly ordered structure, hence it is classified as an amorphous solid. The rest are crystalline solids due to ordered structures.

5. Amorphous solids are actually super cooled liquids.
a) True
b) False
Answer: a
Clarification: Amorphous solids behave like fluids and flow very slowly under the influence of gravity. Hence, they are said to be super cooled liquids.

6. Which type of solid structure melts at a definite, sharp melting point?
a) All types of solids
b) No type of solid
c) Amorphous solids
d) Crystalline solids
Answer: d
Clarification: Crystalline solids have a perfectly ordered structure which collapses immediately at a specific temperature. Amorphous solids melt over a range of temperatures, not one specific value.

7. Which of the following describes a general solid?
a) Compressible
b) Incompressible
c) Fluid
d) Semi-compressible
Answer: b
Clarification: The intermolecular forces of attraction in a solid are very strong, making it incompressible. Gases are highly compressible, while liquids are semi-compressible. Fluid is a property of a substance that can ‘flow’.

8. _________ is the basic repeated structural unit of a crystalline solid.
a) Monomer
b) Molecule
c) Unit cell
d) Atom
Answer: c
Clarification: Crystalline solids are composed of many small crystals, each of which is called a unit cell. It is a specific term. Monomer is the basic unit for a polymer, and atoms make up molecules, which can further arrange themselves to form solids, liquids or gases.

9. Which of the following statements is true for an amorphous solid?
a) Long range order is present
b) Short range order is present
c) There is no orderly arrangement
d) Complete order is present at lower temperatures
Answer: b
Clarification: For an amorphous solid there is short range order present which is independent of the temperature. Long range order is present in crystalline solids.

10. Sulfur exists in two polymorphic forms ____________ and ______________
a) rhombic and monoclinic
b) rhombic and triclinic
c) hexagonal and triclinic
d) hexagonal and monoclinic

Answer: a
Clarification: There are two polymorphous structures of sulfur, rhombic and monoclinic. Polymorphous structures occur when a single substance can crystallize in two or more forms depending upon the conditions.

Chemistry Multiple Choice Questions on “Colligative Properties and Determination of Molar Mass – 1”.

1. What are the properties arising due to varying concentrations of solute in a given solvent, irrespective of the nature of solute with respect to the solvent?
a) Colligative properties
b) Intensive properties
c) Extensive properties
d) Solute properties
Answer: a
Clarification: Colligative properties are set of four properties. properties purely depends on the number of solute particles present in the solution/solvent, independent of the nature of the particles with respect to the solution/solvent. The properties are 1.) Relative lowering of vapor pressure, 2.) Elevation in boiling point, 3.) Depression in freezing point and 4.) Osmotic pressure.

2. Which of the following is a colligative property?
a) Relative lowering of fluid pressure
b) Decrease in boiling point
c) Decrease in freezing point
d) Change in volume after mixing

Answer: c
Clarification: Decrease in freezing point is the correct colligative property, known as ‘Depression in freezing point’. This is caused by solute particles present on the surface which lowers the equilibrium solid-vapor pressure. Therefore, a lower freezing temperature is required to match the pressure outside. The other correct colligative properties are 1.) Relative lowering of vapor pressure, 2.) Elevation in boiling point, 3.) Depression in freezing point and 4.) Osmotic pressure.

3. Which law specifically governs the relative lowering of vapor pressures in solutions?
a) van’t Hoff law
b) Boyle’s law
c) Raoult’s law
d) Amagat’s law
Answer: c
Clarification: Raoult’s law quantifies the relative lowering in vapor pressure. From the law, it follows that p₁ = X₁ x p⁰₁. If p⁰₁ was the original pressure before X₂ mole fraction of solute was added to the solvent then reduction in vapor pressure is given as:
Δp₁ = p⁰₁ – p₁
= p⁰₁ – p⁰₁ X₁
= p⁰₁(1 – X₁)
= p⁰₁X₂
Therefore, the relative reduction in vapor pressure (∆p₁/p⁰₁) = X₂ i.e. mole fraction of solute in the solution.

4. If 1.5 grams of a non-volatile solute (M_W = 100) is added to 200 ml pure CS₂ (ρ = 1.3 g/cc) whose vapor pressure is 400 mm of Hg at 28.0°C, what is the resulting vapor pressure of the dilute solution?
a) 401.246 mm Hg
b) 398.754 mm Hg
c) 401.754 mm Hg
d) 398.246 mm Hg
Answer: d
Clarification: Given,
Mass of solute, m₂ = 1.5 grams
Molar mass of solute, M₂ = 100
Vapor pressure of pure CS₂, p⁰₁ = 400 mm Hg
Volume of solvent, ρ = 200 ml
Density of solvent, ρ = 1.3 g/cc
Number of moles of solute, n₂ = m₂/M_W = 1.5/100 = 0.015 mole
From the law of relative lowering of vapor pressure, Δp₁ = p⁰₁X₂, where X₂ is the mole fraction of solute and Δp₁ is the difference in pressure.
Mass of solvent, m₁ = ρ x V = 1.3 x 200 = 260 grams
Number of moles of solvent, n₂ = 260g/[(12 + 32 + 32)g/mole] = 3.421
Since the solution is dilute we can approximate X₂ = n₂/(n₁ + n₂) ≈ n₂/n₁ (since n₂1)
Using ∆p₁ = p⁰₁X₂,
∆p₁ = (400 mg of Hg) (0.015/3.421) = 1.754 mm of Hg
Using Δp₁ = p⁰₁ – p₁, we get p₁ = p⁰₁ – Δp₁
Hence, resulting lowered vapor pressure, p₁ = 400 mm Hg – 1.754 mm Hg = 398.246 mm Hg.

5. If the relative lowering of pressure of o-xylene is 0.005 due to addition of 0.5 grams of non-volatile solute in 500 grams of solvent, what is the molecular weight of the solute?
a) 21.3 g/mole
b) 23.1 g/mole
c) 32.1 g/mole
d) 1.23 g/mole
Answer: a
Clarification: Given,
∆p₁/p⁰₁ = 0.005
Mass of solute, m_W = 0.5 grams
Mass of solvent, m_S = 500 grams
Number of moles of solute, n_S = 0.5/M_W, where M_W is the molecular weight of the solute.
Number of moles of solvent, n_Solvent = 500 grams/(106 g/mole) = 4.7 mole
Since m_SSolvent , the solution can be considered to be dilute. Hence, equation for relative lowering for vapor pressure becomes:
Δp₁/p⁰₁ = n₂/(n₁ + n₂) ≈ n₂/n₁
On substituting values,
0.005 = (0.5/M_W)/4.7
On solving, M_W = 21.3 g/mole.

6. On addition of non-volatile potassium iodide in water at 298K it is noticed that vapor pressure reduces from 23.8 mm Hg to 2.0 cm Hg. What is the mole fraction of solute in the solution?
a) 0.916
b) 0.160
c) 0.084
d) 0.092
Answer: b
Clarification: Given,
P⁰_water =23.8 mm Hg
P_water= 2.0 cm Hg = 20.0 mm Hg (after addition of solute)
From the law of relative lowering of vapor pressure, ∆p₁ = X₂ x p⁰₁ (where X₂ is the mole fraction of solute)
On rearranging, Δp₁/p⁰₁ =X₂
Δp₁ = 23.8 – 20.0 = 3.8 mm Hg
X₂ = 3.8/23.8 = 0.160.

7. When a non-volatile solute is added to a solvent what is the difference in vapor pressure expressed as a faction of original vapor pressure equal to?
a) Mole fraction of solute in vapor phase
b) Mole fraction of solvent in vapor phase
c) Mole fraction of solute in liquid phase
d) Mole fraction of solvent in liquid phase
Answer: c
Clarification: Relative lowering of vapor pressure is the difference in vapor pressure expressed as a fraction of original vapor pressure. Using Raoult’s law:
ΔP₁ = X₂ x P₁
Therefore, ΔP₁/P₁ = X₂ which implies the relative lowering is equal to mole fraction of solute in liquid phase. Since the solute is non-volatile, it cannot be present in vapor phase.

8. When 2.0 grams of copper (II) nitrate is added to 1000 ml of pure water, by how much is the vapor pressure of water decreased, given that at 20°C the vapor pressure of pure water is 17.535 mm Hg?
a) 3.1 x 10^-4
b) 0.303
c) 0.0333
d) 0.0033
Answer: d
Clarification: Given,
Mass of solute, m_solute = 2.0 grams
Volume of solvent, V_solvent = 1000 ml
P⁰_water = 17.535 mm Hg
Molecular mass of Cu(NO₃)₂ = 188 g/mole
We know that ΔP₁/P⁰₁ = X₂, where X₂ is the mole fraction of the solute.
We are required to find out the difference in vapor pressure i.e. ΔP₁.
Therefore, ΔP_water= P⁰_water x X₂
Number of moles of copper (II) nitrate = 2.0/188 = 0.0106
Number of moles of water = 1000g/(18g/mole) = 55.56 mole (since density of water is 1g/ml)
Mole fraction of solute, X₂ = 0.0106/(0.0106 + 55.56) = 1.9 x 10^-4
ΔP_water = 17.535 x 1.9 x 10^-4 = 0.0033.

9. Two components A and B have their pure vapor pressures in the ratio 1 ∶ 4 and respective mole fractions in solution in ratio 1 ∶ 2. What is the mole fraction of component B in vapor phase?
a) 0.8889
b) 0.1250
c) 0.8000
d) 0.2000
Answer: a
Clarification: Given,
4P⁰_A = P⁰_A ———- (1)
2X_A = X_B ———— (2)
Using law of relative lowering of vapor pressure:
P_A = P⁰_A x X_A ——- (3)
P_B = P⁰_B x X_B ——- (4)
Using values of (1) and (2) in (4)
P_A = P⁰_A x X_A
P_B = 4P⁰_Ax 2X_A =8(P⁰_A x X_A)
Therefore, P_B = 8P_A
Mole fraction of component B in vapor phase, Y_B = P_B/(P_A + P_B)
On simplifying, Y_B = 8P_A/(P_A + 8P_A) = 8/9 = 0.8889.

10. Which of the following is Raoult’s law applicable to, in order to determine molar masses correctly?
a) Ionic solute in liquid
b) Non-ionic solute in dilute solution
c) Non-ionic solute in concentrated solution
d) Ionic solid in insoluble form in solvent
Answer: b
Clarification: In order to determine molar masses correctly, it is essential that the solute is non-volatile, non-ionic and present in dilute form only. If it is ionic one has to account for its van’t Hoff factor, i, for association in concentrated solution and dissociation in dilute solution. Molar masses can only be calculated from dilute solutions containing non-dissociable non-ionic solutes.