Category: Class 12 Chemistry Objective Questions

Chemistry Written Test Questions for IIT JEE Exam on “Alcohols, Phenols and Ethers Nomenclature – 2”.

1. What is the correct name of the following compound?

a) But-3-enol
b) But-3-en-2-ol
c) But-1-en-3-ol
d) But-2-en-3-ol
Answer: b
Clarification: The numbering preference is given to the hydroxyl group over the double bond and hence, the OH lies on the second carbon and double bond on the third.

2. What is the correct IUPAC name of the following phenol?

a) Benzene-1,2,4-triol
b) 1,2,4-Benzenetriol
c) Benzene-1,2,4-triphenol
d) 1,2,4-Triphenol
Answer: a
Clarification: Polyhydric phenols are named as hydroxy derivatives of benzene in the IUPAC system. The positions 1,2,4 comply with the lowest set of locants rule.

3. What is the correct common name of 2-Bromophenol?
a) 2-Bromophenol
b) m-Bromophenol
c) o-Bromophenol
d) 2-Hydroxybromobenzene
Answer: c
Clarification: Phenol is used as the common as well as the IUPAC name and haloarenes with hydroxy groups are named as halo derivatives of phenols.

4. Which of the following is given highest preference while naming when present in the same structure?
a) Hydroxyl group
b) Nitro group
c) Triple bond
d) Halogen group
Answer: a
Clarification: Functional groups are given the highest preference while naming compounds according to the IUPAC system.

5. What is the correct common name for CH₃OC₂H₅?
a) Methyl methyl ether
b) Methyl ethyl ether
c) Ethyl ethyl ether
d) Ethyl methyl ether
Answer: d
Clarification: It is an unsymmetrical ether with one methyl group and one ethyl group. It is named such that the alkyl group are in alphabetical order.

6. Which of the following compounds is ethoxyethane?
a) CH₃OCH₃
b) CH₃OC₂H₅
c) C2H₅OC₂H₅
d) C2H₅OC₃H₇
Answer: c
Clarification: Ethoxyethane is an ether which is a derivative of ethane with one hydrogen atom replaced by ethoxy group (OC₂H₅).

7. Identify the incorrect name for CH₃-O-C₆H₅.
a) Methyl phenyl ether
b) Methoxybenzene
c) Anisole
d) Phenetole
Answer: d
Clarification: This is an aromatic ether with a methoxy substituent and has the common name anisole or methyl phenyl ether.

8. What is the IUPAC name of CH₃-O-CH₂-CH₂-OCH₃?
a) 1,2-Dimethoxyethane
b) 1,3-Dimethoxybutane
c) 2,3-Dimethoxyethane
d) 1,4-Dimethoxypropane
Answer: a
Clarification: The base hydrocarbon is ethane, in which one hydrogen of each carbon atom is replaced by a methoxy group, making it a dimethoxy ether.

9. What is the IUPAC name of CH₃(CH₂)₆-OC₆H₅?
a) Heptyl phenyl ether
b) 1-Phenoxyheptane
c) 1-Hepoxybenzene
d) Phenyl heptyl ether
Answer: b
Clarification: Since heptane is a larger group than benzene, it is chosen as the parent hydrocarbon and the compound is named as an aryloxy derivative of heptane.

10. The name 1-Ethoxy-2, 2-dimethylcyclohexane is correct according to IUPAC nomenclature.
a) True
b) False
Answer: b
Clarification: The methyl group should be given higher preference over the alkoxy group in ethers. The correct naming will be 2-Ethoxy-1, 1-dimethylcyclohexane.

11. Phenetole is an aromatic ether.
a) True
b) False
Answer: a
Clarification: Phenetole is also known as ethoxybenzene or ethyl phenyl ether and has an aromatic ring on one side.

Chemistry Written Test Questions for IIT JEE Exam,

Chemistry Exam Questions for IIT JEE Exam on “Methods of Carboxylic Acids Preparation”.

1. Which of the following is known as Jones reagent?
a) KMnO₄ in alkaline medium
b) CrO₃ in H₂SO₄
c) K₂Cr₂O₇ in acidic medium
d) KMnO₄ in H₂SO₄
View Answer

Answer: b
Clarification: Chromium trioxide in an aqueous solution with sulphuric acid in known as Jones reagent. It Is an important compound in the preparation of carboxylic acids from alcohols.

2. Benzoic acid is obtained from the oxidation of _______ with alkaline KMnO₄ followed by treatment with mineral acid.
a) phenol
b) benzaldehyde
c) acetophenone
d) benzyl alcohol
Answer: d
Clarification: Only primary alcohols undergo oxidation in the presence of common oxidising agents followed by reaction with H₃O⁺ to give the respective carboxylic acids. This reaction is proceeded by the removal of both hydrogen atoms from the alpha carbon and formation of a double bond between C and O.

3. Identify the most suitable reagent for the conversion of ethanal to acetic acid.
a) Alkaline KMnO₄; H₃O⁺
b) Jones reagent
c) Tollen’s reagent
d) LiAlH⁴
Answer: c
Clarification: Aldehydes are easily oxidized to carboxylic acids having the same number of carbon atoms as the parent aldehyde, when reacted even with mild oxidising agents like Tollen’s reagent.

4. Identify the product B in the reaction chain shown.

a) Ethanoic acid
b) Propanoic acid
c) Butanoic acid
d) 2-Methylpropanoic acid
Answer: b
Clarification: Ethyl bromide on reaction with alc. KCN gives ethyl cyanide by nucleophilic substitution. Nitriles can be hydrolysed to give amides and on further heating give a carboxylic acid which contains one more carbon than the original ethyl bromide. Hence, the acid contains three C atoms, making it propanoic acid.

5. Identify X in the following conversion.

a) Alkaline KMnO₄; H₃O+
b) KOH; heat
c) H₃O⁺; heat
d) KMnO₄-KOH; heat
Answer: c
Clarification: Amides undergo hydrolysis in the presence of heat to give corresponding carboxylic acids along with ammonia gas. In the reaction shown, benzamide undergoes oxidation to form benzoic acid.

6. Benzoic acid can be prepared by oxidation of tert-Butyl benzene.
a) True
b) False
Answer: b
Clarification: tert-Butyl benzene consists of a tertiary group with no benzylic H. It is highly stable and does not undergo oxidation even under drastic conditions, to give aromatic carboxylic acids.

7. Which of the following pairs do not give the same compound on heating with alkaline potassium permanganate?
a) Toluene and propyl benzene
b) Toluene and n-Butyl benzene
c) Propyl benzene and isopropyl benzene
d) o-Xylene and n-Butyl benzene
Answer: d
Clarification: All mono-substituted alkyl benzenes, with primary or secondary alkyl groups, on vigorous oxidation give benzoic acid. The entire side chain is oxidised irrespective of the length. In case of o-xylene, there are two methyl groups on the benzene ring and both of them are oxidised to carboxyl group, resulting in phthalic acid.

8. p-Xylene on reaction with acidified potassium dichromate at high temperature gives ________
a) benzoic acid
b) phthalic acid
c) terephthalic acid
d) no reaction
Answer: c
Clarification: p-Xylene is a dimethyl substituted benzene which on vigorous oxidation, gets oxidised to an aromatic dicarboxylic acid, with the two carboxyl groups at para positions with respect to each other. This compound is called terephthalic acid.

9. Which of the following cannot be converted to benzoic acid on reaction with KMnO₄-KOH followed by H₃O⁺?
a) Ethyl benzene
b) Acetophenone
c) 4-Methylacetophenone
d) Styrene
Answer: c
Clarification: The ethyl group, acetyl group and ethene group are all oxidised to potassium carboxylate groups which are further oxidised to carboxyl groups. In the case of 4-Methylacetophenone, there are two groups that will be oxidised to carboxyl groups, hence forming terephthalic acid instead of benzoic acid.

10. How can methyl magnesium bromide be converted to propanoic acid?
a) Jones reagent
b) KMnO₄-KOH; heat
c) H₃O⁺; heat
d) CO₂-dry ether; H₃o⁺
Answer: d
Clarification: Methyl magnesium bromide (Grignard reagent) on reaction with CO₂ forms an addition product containing an additional C carbon atom. This is decomposed in the presence of mineral acid to form propanoic acid.

11. 3-Chlorophenyl magnesium bromide on reaction with dry ice followed by acidification in mineral acid gives _________
a) 3-Chlorophenol
b) 3-Chlorophenylethanoic acid
c) 3-Chlorobenzaldehyde
d) 3-Chlorobenzoic acid
Answer: d
Clarification: The MgBr group of 3-Chlorophenyl magnesium bromide (Grignard reagent) will be substituted by COOH group in the above reaction, to give a halogen substituted aromatic carboxylic acid.

12. Benzoic ethanoic anhydride on hydrolysis gives __________
a) benzoic acid and methanoic acid
b) benzoic acid and ethanoic acid
c) phenylethanoic acid and methanoic acid
d) no products
Answer: b
Clarification: Benzoic ethanoic anhydride (C₆H₅COOCOCH₃) is easily hydrolysed with water to give its corresponding acids, benzoic acid (by adding H to C₆H₅COO) and ethanoic acid (by adding OH to COCH₃).

13. Ethanoyl chloride on hydrolysis with aqueous NaOH gives _______
a) acetate ion
b) acetic acid
c) propanoic acid
d) no reaction
Answer: a
Clarification: Acid chlorides are readily hydrolysed with aqueous base to give carboxylate ions which further give corresponding carboxylic acids on acidification. It can be directly obtained by direct hydrolysis with water.

14. The final product(s) of basic hydrolysis followed by acidification of ethyl butanoate is _______
a) ethanoic acid
b) butanoic acid
c) ethanoic acid and butanoic acid
d) butanoic acid and ethanol
Answer: b
Clarification: Ethyl butanoate (CH₃CH₂CH₂COOC₂H₅) on basic hydrolysis forms CH₃CH₂CH₂COONa and ethanol. Then this sodium carboxylate compound gets acidified to give butanoic acid.

15. Acidic hydrolysis of ethyl benzoate directly gives benzoic acid.
a) True
b) False
Answer: a
Clarification: Acidic hydrolysis of esters directly gives carboxylic acids. In this case, the ethyl group of the ester is separated out as ethanol along with the main product, benzoic acid.

Chemistry Assessment Questions for Class 12 on “Amines Chemical Reactions – 4”.

1. Which gas is produced when ethanamine reacts with nitrous acid?
a) N₂
b) H₂
c) HCl
d) O₂
Answer: a
Clarification: Nitrogen gas is evolved due to the decomposition of diazonium salts which are formed from the reaction of 1° aliphatic amines with HNO₂. This is used in the estimation of amino acids and proteins.

2. What is the main product of the following reaction?

a) Benzene diazonium chloride
b) Phenol
c) o-Nitrosoaniline
d) p-Nitrosoaniline
Answer: b
Clarification: Primary aromatic amines react with nitrous acid at cold temperatures (273-278K) to form diazonium salts. However, if the temperature is more than 278K, phenol is formed along with the evolution of nitrogen gas.

3. Tertiary amines dissolve in nitrous acid to form corresponding salts.
a) True
b) False
Answer: a
Clarification: Tertiary amines on reaction with cold HNO₂ remain dissolved, forming amine nitrite salts, which decompose to nitrosoamines and alcohol on warming.

4. Which of the following is not a final product of the reaction between propylamine and nitrous acid?
a) CH₃CH₂CH₂N₂Cl
b) CH₃CH₂CH₂OH
c) N₂ gas
d) H₂O
Answer: a
Clarification: CH₃CH₂CH₂N₂Cl is a diazonium salt derivative of propanamine and is first formed when it reacts with HNO₂. This is a very unstable compound and immediately decomposes to give propanol and evolves nitrogen gas.

5. Diethylamine reacts with nitrous acid in the cold to form _______
a) diazonium salt
b) alcohol
c) imine
d) nitrosoamine
Answer: d
Clarification: Secondary amines react slowly with nitrous acid at low temperatures to give yellow oily nitrosoamines. Diethylamine specifically produces N,N-diethylnitrosoamine.

6. Hinsberg’s reagent is _______
a) benzenesulfonic acid
b) benzenesulphonyl chloride
c) p-toluenesulphonyl chloride
d) chlorosulphuric acid
Answer: b
Clarification: Benzenesulphonyl chloride (C₆H₅SO₂Cl), or arylsulphonyl chloride, is known as Hinsberg’s reagent. It is an important compound used in the distinction of different classes of amines.

7. Which of the following amines, on reaction with benzenesulphonyl chloride, will give a sulphonamide that is insoluble in alkali?
a) Ethylamine
b) Ethylmethylamine
c) Trimethylamine
d) Aniline
Answer: b
Clarification: Secondary amines on reaction with Hinsberg’s reagent produce sulphonamides without any hydrogen atom attached to the nitrogen atom. Hence, it is not acidic and therefore insoluble in alkali. The amide produced in this case is N-ethyl-N-methylbenzenesulphonamide.

8. Which of the following amines will form a product that is soluble in KOH, on reaction with Hinsberg’s reagent?
a) Isopropylamine
b) Diethylamine
c) N,N-Dimethylpropylamine
d) N,N-Dimethylaniline
Answer: a
Clarification: The reaction of Hinsberg’s reagent with Isopropylamine gives N-isopropylbenzene sulphonamide. The lone hydrogen attached to the N atom is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, is it soluble in alkali like KOH.

9. Which of the following reactions/tests does not help in the distinction between ethylamine and diethylamine?
a) Carbylamine test
b) Hinsberg’s test
c) Reaction with HNO₂
d) Reaction with CH₃CH₂Br
Answer: d
Clarification: Both ethylamine and diethylamine on reaction with CH₃CH₂Br eventually gives quaternary ammonium salt. Hence, the alkylation of primary amines cannot be used as a distinction method.

10. Aniline is a _______ directing compound.
a) ortho
b) meta
c) ortho and para
d) ortho, meta and para
Answer: c
Clarification: The ortho and para positions with respect to NH₂ group in aniline are centres of high electron density. This is due to the resonance structures of aniline. Thus, NH₂ is ortho and para directing and a powerful activating group.

11. The activating effect of -NHCOCH₃ group is ______ as compared to -NH₂ group.
a) less
b) same
c) more
d) very high
Answer: a
Clarification: The -NHCOCH₃ group forms two resonance structures, where the lone pair of nitrogen interacts with the lone pairs on oxygen atom. This makes the lone electron pair on N less available for donation to benzene ring.

12. What is the major product formed when aniline reacts with bromine water at room temperature?
a) 2-Bromoaniline
b) 4-Bromoanline
c) 2,6-Dibromoaniline
d) 2,4,6-Tribromoaniline
Answer: d
Clarification: Since amino group is highly ortho and para directing, it substitutes the Br group at the para as well as both ortho positions (2,4,6) to give a white precipitate of a tribromo substituted aniline.

13. What is the order of quantities of all isomers of nitroaniline formed on the reaction of aniline with nitric acid and sulphuric acid at 288K?
a) ortho > meta > para
b) para > ortho > meta
c) para > meta > ortho
d) meta > para > ortho
Answer: c
Clarification: Nitration of aniline produces nitro derivatives. Under controlled conditions, aniline is protonated (by strong acidic medium) to form anilinium ion, which is meta directing. Therefore, apart from p-nitroaniline (51%) and o-nitroaniline (2%), a significant amount of m-nitroaniline (47%) is also formed.

14. It is possible to obtain para isomers of anilines as the major product in electrophilic ring substitution of aniline.
a) True
b) False
Answer: a
Clarification: This is possible by protecting the amino group of aniline by reacting it with acetic anhydride (acetylation) to give acetanilide. This compound on desired substitution reaction followed by hydrolysis gives para substituted aniline as major product.

15. A compound ‘P’ on treating with concentrated H₂SO₄ forms ‘R’. The product ‘R’ on heating at 460K forms a zwitter ionic compound. Identify P, R respectively.
a) Aniline; anilinium hydrogensulphate
b) Aniline; sulphanilic acid
c) Anilinium hydrogensulphate; sulphanilic acid
d) Sulphanilic acid; anilinium hydrogensulphate
Answer: a
Clarification: Aniline (P) reacts with conc. H₂SO₄ to form anilinium hydrogensulphate (R) which on heating at 453-473K produces sulphanilic acid (zwitter ionic compound).

Chemistry Assessment Questions for Class 12,

Chemistry Multiple Choice Questions on “Biomolecules – Hormones”.

1. Which of the following hormone is a polypeptide?
a) Estrogen
b) Insulin
c) Androgen
d) Epinephrine
Answer: b
Clarification: Insulin consists of a chain of 51 amino acids and hence is a polypeptide. Estrogen and androgen are steroids, whereas epinephrine is an amine.

2. Hormones are ______
a) messengers
b) catalysts
c) enzymes
d) inhibitors
Answer: a
Clarification: Hormones are chemical substances produced by the endocrine glands in the human body. They are carried to different parts of the body through the blood stream. Because of the action of hormones as intercellular communicators, they are called chemical messengers.

3. Which of the following is not an amine hormone?
a) Norepinephrine
b) Adrenaline
c) Thyroxine
d) Oxytocin
Answer: d
Clarification: Amine hormones are water-soluble compounds which have amino groups and are structurally derived from amino acids. Oxytocin is a peptide hormone.

4. Identify the hormone that increases the glucose level in blood.
a) Insulin
b) Glucagon
c) Oxytocin
d) Vasopressin
Answer: b
Clarification: Insulin is released in response to the rapid rise in blood glucose level. On the other hand, the hormone glucagon tends to increase the glucose level in blood. These two hormones together regulate the blood glucose level.

5. Which of the following is known as fight or flight hormone?
a) Epinephrine
b) Norepinephrine
c) Insulin
d) Thyroxine
View Answer

Answer: a
Clarification: Epinephrine and norepinephrine are amine hormones that mediate response to external stimuli. Adrenaline plays an important role in fight or flight situations by increasing blood flow to muscles, blood sugar level and pulse rate.

6. Which hormone plays an important role during child birth and post it?
a) Estrogen
b) Progesterone
c) Cortisone
d) Oxytocin
Answer: d
Clarification: Oxytocin controls the contraction of the uterus during child birth and helps in the release of milk from mammary glands. Progesterone is responsible for the preparation of uterus for implantation of fertilised egg.

7. The condition goitre is associated with which hormone?
a) Insulin
b) Thyroxine
c) Adrenaline
d) Cortisone
Answer: b
Clarification: Thyroxine is an iodinated derivative of tyrosine. Abnormally low levels of thyroxine leads to hypothyroidism, which causes enlargement of thyroid gland (goitre). Increases level of thyroxine causes hyperthyroidism.

8. Lack of which component in diet causes hypothyroidism?
a) Potassium
b) Vitamin C
c) Iodine
d) Water
Answer: c
Clarification: Low levels of iodine in the diet may lead to hypothyroidism and enlargement of the thyroid gland. This can be controlled by adding sodium iodide to table salt to form iodised salt.

9. Which of the following does not release steroid hormones?
a) Testes
b) Ovary
c) Adrenal cortex
d) Pancreas
Answer: d
Clarification: Steroid hormones are produced by adrenal cortex and gonads. Hormones released by the adrenal cortex play a very important role in the functions of the bod.

10. Which hormone controls the balance of water and minerals in the body?
a) Vasopressin
b) Mineralocorticoids
c) Testosterone
d) Thyroxine
Answer: b
Clarification: Mineralocorticoids are steroid hormones which control the level of excretion of water and salt by the kidney, thus balancing the water and mineral levels as required.

11. Lack of which hormone causes Addison’s disease?
a) Glucocorticoids
b) Oxytocin
c) Insulin
d) Norepinephrine
Answer: a
Clarification: If adrenal cortex does not function properly then one of the results may be Addison’s disease characterized by hypoglycaemia, weakness and increased susceptibility to stress. This may be fatal unless treated by glucocorticoids and mineralocorticoids.

12. All hormones are proteins.
a) True
b) False
Answer: b
Clarification: Hormones are compounds having varied chemical structures. They may be polypeptide chains or amino acids or contain a steroid nucleus.

13. Estradiol is the main sex hormone in females.
a) True
b) False
Answer: a
Clarification: Estradiol is a steroid hormone which is majorly responsible for the development of secondary sex characteristics in females and participates in the control of menstrual cycle.

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Chemistry MCQs for Schools on “Solid State – Packing Efficiency”.

1. What does the ratio ‘space occupied/total space’ denote?
a) Packing factor
b) Packing efficiency
c) Particle fraction
d) Packing unit
Answer: a
Clarification: Packing factor is a fraction of total space of the unit cell occupied by the constituent particles.

2. What is the dimensional formula of packing fraction?
a) M⁰L³T⁰
b) M⁰L⁰T⁰
c) ML⁰T⁰
d) M⁰L²T⁰
Answer: b
Clarification: Packing fraction is a dimensionless quantity which is the ratio of space occupied to total crystal space available. Since both the quantities have the same units the ratio renders dimensionless.

3. Arrange the types of arrangement in terms of decreasing packing efficiency.
a) BCC b) HCP c) HCP d) CCP Answer: c
Clarification: HCP and CCP have the highest packing efficiency of 74% followed by BCC which is 68%. The simple cubic structure has a packing efficiency of 54%.

4. If the body-centered unit cell is assumed to be a cube of edge length ‘a’ with spherical particles of radius ‘r’ then how is the diameter, d of particle and surface area, S of the cell related?
a) S = 32d⁴/3
b) S = 2d²
c) S = 4d²
d) S = 8d²
Answer: d
Clarification: For BCC unit cell the relation between radius of a particle ‘r’ and edge length of unit cell, a, is r = (frac{sqrt{3}}{4})a.
We know that diameter, d = 2r = (frac{sqrt{3}}{2})a
Implying d² = (frac{3}{4})a²
Therefore, 4d²/3=a²
Multiplying by 6 on both sides gives S = 6a² = 8d², where S is the surface area of the cube = 6a².

5. Which of the following metals would have the highest packing efficiency?
a) Copper
b) Potassium
c) Chromium
d) Polonium
Answer: a
Clarification: Copper metal bears face-centered unit cells in its crystal structure. Potassium and chromium both have body-centered unit cells whereas polonium is the only known metal to bear a simple cubic structure. FCC structure has the highest efficiency.

6. “The packing efficiency can never be 100%”. Is this true or false?
a) False
b) True
Answer: b
Clarification: Packing efficiency can never be 100% because in packing calculations all constituent particles filling up the cubical unit cell are assumed to be spheres.

7. What are the percentages of free space in a CCP and simple cubic lattice?
a) 52% and 74%
b) 48% and 26%
c) 26% and 48%
d) 74% and 52%
Answer: c
Clarification: The packing efficiency in CCP and simple cubic lattice are 74% and 52%, respectively. Hence the corresponding free spaces will be 100% – 74% = 26% and 100% – 52% = 48%.

8. How many atoms surround the central atom present in a unit cell with the least free space available?
a) 4
b) 6
c) 8
d) 12
Answer: d
Clarification: FCC, CCP and HCP are unit cells with least free space available i.e. highest packing efficiency. The coordination number of given cells are 12.

9. If metallic atoms of mass 197 and radius 166 pm are arranged in ABCABC fashion then what is the surface area of each unit cell?
a) 1.32 × 10⁶ pm²
b) 1.32 × 10^-18pm²
c) 2.20 × 10⁵ pm²
d) 2.20 × 10^-19 pm²
Answer: a
Clarification: ABCABC arrangement is found in CCP.
In closed cubic packing, relation between edge length of unit cell, a, and radius of particle, r, is given as a=2(sqrt{2})r.
Surface area (S.A.) = 6a²
From the relationship,
a² = 8r²
S.A. = 6a² = 48r²
When r = 166 pm, S.A. = 48(166pm) = 1.32 x 10⁶ pm².

10. If copper, density = 9.0 g/cm³ and atomic mass 63.5, bears face-centered unit cells then what is the ratio of surface area to volume of each copper atom?
a) 0.0028
b) 0.0235
c) 0.0011
d) 0.0323
Answer: b
Clarification: Density, d of unit cell is given by d = (frac{zM}{a^3N_A})
Given,
Density, d = 9.0 g/cm³
Atomic mass, M = 63.5 g/mole
Edge length = a
N_A = Avogadro’s number = 6.022 x 10²³
z = 4 atoms/cell
On rearranging the equation for density we get a³ = (frac{zM}{dN_A})
Substituting the given values:
a³ = (frac{4 times 63.5}{9 times 6.022 times 10^{23}})
Therefore, a = 360.5 pm
The relation of edge length ‘a’ and radius of particle ‘r’ for FCC packing i.e. a = 2(sqrt{2})r.
On substituting the value of ‘a’ in the given relation, r = (frac{360.5}{2sqrt{2}})=127.46 pm
Now, for spherical particles volume, V = 4πr³/3 and surface area, S = 4πr²
Required ratio = S/V=4πr²/(4πr³/3) = 3/r (after simplifying)
Thus, S/V = 3/127.46 = 0.0235.

Chemistry MCQs for Schools,

Chemistry Assessment Questions for Schools on “Conductance of Electrolytic Solutions”.

1. The voltameter is an instrument in which electrical energy is converted to chemical energy.
a) True
b) False
Answer: a
Clarification: The voltameter is a scientific instrument used to measure the quantity of electricity through electrolytic action. Since it is a type of electrolytic cell, it uses electrical energy in order to perform electrolysis. Therefore, it converts electrical energy to chemical energy.

2. Which of the following solutions cannot conduct electricity?
a) Sugar in water
b) NaCl in water
c) MgCl₂ in water
d) KCl in water
Answer: a
Clarification: For a solution to conduct electricity, it requires the solution to have movable ions. A solution of sugar in water does not conduct electricity as the sugar molecules do not dissociate to form ions. Whereas the salts in the other options dissociate to form ions and help conduct electricity.

3. Which of the following scientists stated that the equivalent conductivity of an electrolyte at infinite dilution is equal to the sum of the conductances of the ions and cations?
a) Svante Arrhenius
b) Friedrich Kohlrausch
c) Hermann Kolbe
d) Antoine Lavoisier
Answer: b
Clarification: Friedrich Kohlrausch was a German scientist who conducted research on the conductive properties of electrolytes. He stated that “at infinite dilution, each ion of an electrolyte contributes a characteristic ionic conductance towards equivalent conductance of electrolyte which is independent of the nature of the other ion present in the solution”.

4. Which of the following statements is correct with respect to electrolytic solutions?
a) Its conductance increases with dilution
b) Its conductance decreases with dilution
c) Its conductivity increases with dilution
d) Its equivalent conductance decreases with dilution
Answer: a
Clarification: Conductance is dependent on the concentration of the electrolytic solution. It is also inversely proportional to the conductivity of the solution. On dilution, the number of ions per unit volume reduces, decreasing conductivity and increasing the conductance. The equivalent conductance increases on dilution ionic mobility increases on dilution.

5. The limiting equivalent conductance for weak electrolytes can be calculated using Kohlrausch’s law.
a) True
b) False
Answer: a
Clarification: According to Kohlrausch’s law, the equivalent conductivity of an electrolyte at infinite dilution is the sum of two values one depending upon the cation and the other upon the anion. Therefore, it can be used to calculate the limiting equivalent conductance for weak electrolytes.

6. Which of the following statements is correct regarding the conductivity of solutions of electrolytes?
a) It is independent of the size of the ions
b) It is independent of the viscosity of the solution
c) It depends on the solvation of ions present in solution
d) It decreases with temperature
Answer: c
Clarification: Conductivity of a solution is inversely proportional to the size of the ions present in it, the viscosity of the solution and the solvation of the ions in the solution. Conductivity increases with temperature as it can cause an increase in the number of ions in solution.

7. Which of the following solutions has theleast value of conductivity?
a) 0.01M Na₂SO₄
b) 0.01M NaCl
c) 0.01M CH₃COOH
d) 0.01M HCl
Answer: c
Clarification: The conductivity of a solution depends on the strength of the ions that make up the solution because strong electrolytes dissociate easily and increase the number of ions, increasing the conductivity of the solution. Since CH₃COOH is not a strong electrolyte, and because all of the solutions have the same concentration, it has the least value of conductivity.

8. What is the value of the cell constant when the conductance of an electrolytic solution is equal to its conductivity?
a) 0
b) 1
c)10
d) 100
Answer: b
Clarification: Given,
Conductance = conductivity
We know that, conductivity = conductance x cell constant
Since the value of conductivity is equal to conductance, the value of the cell constant is equal to 1.

9. Which of the following conditions are satisfied when the electrolytic solutions are infinitely dilute?
a) Electrolyte is 100% dissociated
b) Interionic effects increase
c) Conductance is infinite at infinite dilution
d) Molecules continue to exist in solution
Answer: a
Clarification: At infinite dilution, even weak electrolytes behaves like strong electrolytes and undergo complete ionization. Hence, the electrolyte is 100% dissociated. Also, at infinite dilution, the ions are far apart and interionic effects disappear.

10. Which of the following salts show maximum value of equivalent conductance in their fused state?
a) NaCl
b) KCl
c) RbCl
d) CsCl
Answer: d
Clarification: On moving down a group, the atomic size of the elements increases and so their covalent character also decreases. Thus, CsCl, being the least covalent compound will readily give its ions in its fused state. Therefore, CsCl has the maximum value of equivalent conductance in its fused state.

11. Which of the following is an additive property?
a) Surface tension
b) Viscosity
c) Conductance
d) Volume
Answer: c
Clarification: An additive property is a property which is equal to the sum of corresponding properties of its constituent atoms. In the given list, viscosity, surface tension and volume are not additive properties since it does not depend on the interactions of molecules whereas conductance does. Therefore, conductance is an additive property.

12. Which of the given solutions have an equal value of molar conductivity and equivalent conductivity?
a) 1M BaSO₄
b) 1M KCl
c) 1M BCl₃
d) 1M CaSO₄
Answer: b
Clarification: We know that for electrolytic solutions, ⋀_M = ⋀_E × valency factor
Where ⋀_M = molar conductivity and ⋀_E = equivalent conductivity
For KCl, the valency factor is 1. Therefore, for KCl, the value of molar conductivity is equal to its equivalent conductivity. For other ions, ⋀_M > ⋀_E as their valency factors are > 1.

13. Which of the following is ionic mobility independent of?
a) Size of ion
b) Charge on ion
c) Distance of separation between the electrodes
d) Concentration of electrolyte
Answer: c
Clarification: Ionic mobility is an inherent property of the electrolytic solution and is hence, independent of the distance of separation between the two electrodes. As the charge to size ratio increases, the ionic mobility decreases.

14. Which of the following statements is true regarding ionic speed?
a) It is independent of the size of the ion
b) It depends on the potential difference between the two electrodes
c) It is independent of the concentration of the electrolyte
d) It is independent of the charge of the ion
Answer: b
Clarification: Ionic speed is the speed with which the ion moves in the electrolyte during current passage. It depends on the nature of the ions, potential difference, separation between the electrodes and the concentration of the electrolyte. The ionic speed is directly proportional to the potential difference between the electrodes.

15. Which of the following complex compounds will have minimum conductance in solution?
a) [Co(NH₃)₃Cl₃]
b) [Co(NH₃)₄Cl₂]Cl
c) [Co(NH₃)₆]Cl₃
d) [Co(NH₃)₅Cl]Cl₂
Answer: a
Clarification: Electrolytes that ionise in solution have a higher value of conductance. Among the given options, [Co(NH₃)₃Cl₃] does not ionise, whereas the other complexes ionise. Therefore, [Co(NH₃)₃Cl₃] will have minimum conductance in solution.

Chemistry Assessment Questions for Schools,