Category: Class 12 Chemistry Objective Questions

Chemistry MCQs for Class 12 on “D and F-Block Elements – Transition Elements Compounds”.

1. What is the colour of KMnO₄?
a) Green
b) Purple
c) Blue
d) Colourless
Answer: b
Clarification: The physical state of potassium permanganate (KMnO₄) is an odourless solid, and they look like dark purple or bronze-coloured crystals. If we dissolve these crystals in water, then the solution becomes purple in colour.

2. What is the formula of hematite?
a) Fe₃O₄
b) FeSO₄.7H₂O
c) Fe₂O₃
d) FeCl₃
Answer: c
Clarification: Ferric oxide(Fe₂O₃) occurs in nature as haematite. It is a red powder, insoluble in H₂O and not acted upon by air or H₂O. It is amphoteric in nature and reacts with acids and alkalis and used as a catalyst in the oxidation of CO to CO₂ in the Bosch process.

3. Mn has a maximum oxidation state of +6.
a) True
b) False
Answer: b
Clarification: The highest oxidation state in the oxides of any transition metal is equal to its group number, for example, 7 in Mn₂O₇. Beyond group 7, no higher oxides of iron above Fe₂O₃ are known. Some metals in higher oxidation state stabilize by forming oxocations.

4. Which of the following is amphoteric?
a) CrO
b) Cr₂O₃
c) CrO₅
d) CrO₃
Answer: b
Clarification: Chromium(atomic no.24, symbol-Cr) forms many oxides. Some of its common oxidation states are +2, +3 and +6. Chromium (III) oxide (green in colour) is amphoteric, i.e., it can react as both acid and base. Its formula is Cr₂O₃.

5. Which of the following is Baeyer’sreagent?
a) Acidified KMnO₄
b) Alkaline KMnO₄
c) Acidified K₂Cr₂O₇
d) Aqueous KMnO₄
Answer: b
Clarification: Alkaline KMnO₄ is called Baeyer’s reagent. Baeyer’s reagent is an alkaline solution of cold potassium permanganate, which is a powerful oxidant making this a redox reaction. Reaction with double or triple bonds (-C=C- or -C≡C-) in an organic material causes the colour to fade from purplish-pink to brown. It is a syn addition reaction.

6. Sc³⁺ is a paramagnetic ion.
a) True
b) False
Answer: b
Clarification: Scandium (atomic no.21, symbol-Sc) has the following electronic configuration:
[Ar] 3d¹4s²
In its Sc³⁺ state, scandium loses the 4s and 3d electrons and is colourless. Hence, Sc³⁺ is diamagnetic and not paramagnetic.

7. Which of the following metal is used as a thermometric liquid?
a) Iron
b) Copper
c) Mercury
d) Potassium
Answer: c
Clarification: Mercury (atomic no.80, symbol-Hg) is the only liquid metal at room temperature. It has a high coefficient of expansion and boiling point. This helps us to identify even the slightest change in temperature of the surroundings.

8. Which of the following tests does AgCl not answer?
a) Chromyl chloride test
b) Baeyer’s reagent test
c) Alkaline test
d) Acidic test
Answer: a
Clarification: Chromyl chloride test is done for detecting the presence of Cl^– ions. The chlorides of silver, lead, mercury and antimony are covalent in nature and thus do not generate Cl^– ions and so they do not give the chromyl chloride test also. So, heavy metal chlorides don’t give this test because they are not ionic.

9. Which compound forms double salt with sulphates of alkali metals?
a) Ferric oxide
b) Silver nitrate
c) Ferric chloride
d) Ferrous sulphate
Answer: d
Clarification: Ferrous sulphate forms double salts with sulphates of alkali metals with general formula R₂SO₄.FeSO₄.6H₂O. With ammonium sulphate, it forms a double salt known as Mohr’s salt. It ionises in solution to gives Fe²⁺, NH₄⁺ and SO₄^2- ions.

10. Which compound is used in Ultra-violet calibration?
a) Hg₂Cl₂
b) HgCl₂
c) K₂Cr₂O₇
d) KMnO₄
Answer: c
Clarification: Potassium dichromate (K₂Cr₂O₇) is especially useful in the visible range but also useful in UV. Potassium dichromate itself is stable and available in high purity. In dilute perchloric acid solution, it has a linear response with good temperature stability and also stable as solution.

Chemistry MCQs for Class 12,

Chemistry Multiple Choice Questions on “Haloalkanes Classification”.

1. Monohalo, dihalo, trihalo and tetrahalo are types of haloalkanes and haloarenes based on the ______
a) type of halogen atom
b) number of halogen atoms
c) nature of carbon atom
d) hybridisation of C atom to which halogen is bonded
Answer: b
Clarification: Haloalkanes may be classified as mono, di, tri, tetra and so on depending on the number of halogen atoms present in their structure.

2. A monohaloarene is an example of a/an __________
a) aliphatic halogen compound
b) aromatic halogen compound
c) alkyl halide
d) side chain substituted aryl halide
Answer: b
Clarification: A monohaloarene is a compound in which the halogen is directly attached to the benzene ring. Side chain substituted aryl halides are also aromatic halogen compounds with the halogen not directly attached to the benzene ring.

3. What is the general formula for haloalkanes? (X=halogen atom, n = 1, 2, 3…)
a) C_nH_2nX
b) C_nH_2n+1X
c) C_nH_2n-1X
d) C_nH_2n-3X
Answer: b
Clarification: The general formula for haloalkanes is C_nH_2n+1X where X is a halogen atom and n = 1, 2, 3… The formulae C_nH_2n-1X and C_nH_2n-3X are that of haloalkenes and haloalkynes respectively.

4. Which of the following compounds contains an allylic carbon?

a) A
b) B
c) C
d) D
Answer: a
Clarification: A sp³ hybridised C atom present adjacent to a C-C double bond is called an allylic carbon, and when the halogen atom is bonded to this carbon, it is called an allylic halide.

5. Which of the following categories does the compound shown belong to?

a) Primary haloalkane
b) Secondary haloalkane
c) Tertiary haloalkane
d) Haloarene
Answer: b
Clarification: The compound shown is cyclohexyl iodide, in which the halogen atom is bonded to an alkyl group that is alicyclic in nature. Since there are effectively two alkyl groups attached to the carbon bonded to the halogen atom, it is classified as secondary or 2° cyclo alkyl halide.

6. What is the nature of the circled C atom in the following compound?

a) sp² hybridised
b) allylic
c) benzylic
d) vinylic
Answer: c
Clarification: The circled C atom is sp³ hybridised and is attached directly to an aromatic ring, hence it is a benzylic carbon and the compound is a 1o benzylic halide.

7. In which of the following cases will the compound be a tertiary (3°) halogen compound? (X=halogen atom)

a) R’=R”=H
b) R’=CH₃, R”=H
c) R’=H, R”=CH₃
d) R’=R”=CH₃
Answer: d
Clarification: When R and R’ both are CH₃ groups, then the carbon atom bonded to the halogen will have three alkyl groups attached to it including the benzene ring.

8. Which of the following is a vinylic halide?
a) CH₂=CHCHCl₂
b) CH₃CHClCH₃
c) (CH₃)₂C=CHCH₂Cl
d) CH₃CH=CClCH₂CH₃
Answer: d
Clarification: In CH₃CH=CClCH₂CH₃, the Cl is bonded directly to the C atom of a C-C double bond, and hence it is a vinylic halide. CH₃CHClCH₃ is an alkyl halide whereas CH₂=CHCHCl₂ and (CH₃)₂C=CHCH₂Cl are allylic halides.

9. The compound C6H5F is an example of a ________ halide.
a) allylic
b) benzylic
c) vinylic
d) aryl
Answer: d
Clarification: In C₆H₅F, the F atom is directly attached to the sp² hybridised carbon atom of an aromatic ring, i.e., benzene.

10. The compound in which a CH₂Br group is attached to a benzene ring is an aryl halide.
a) True
b) False
Answer: b
Clarification: The Br atom is bonded to a carbon atom which is attached to sp² hybridised carbon of benzene. This is an example of a side chain substituted aryl halide.

Chemistry Exam Questions and Answers for Class 12 on “Alcohols and Phenols – 3”.

1. When the bond between O and H of the hydroxyl group is broken, alcohols react as _________
a) nucleophiles
b) electrophile
c) protonated molecules
d) electron seeking compounds
Answer: a
Clarification: Alcohols may react as nucleophiles or electrophiles depending upon whether the O-H bond or the C-O bond is broken respectively. Carbocations are formed in the nucleophile case and protonated alcohols are formed in the electrophile case.

2. Alcohols and phenols are ________
a) Lewis acids
b) Lewis bases
c) Bronsted acids
d) Bronsted bases
Answer: c
Clarification: Alcohols and phenols are known to be acidic in nature based on their interaction with metals. They are able to donate a proton to a stronger base. Additionally, alcohols also act as Bronsted bases due to the presence of unshared electron pairs on oxygen.

3. What is the correct relation between acidic strength of primary, secondary and tertiary alcohols?
a) 1°>2°>3°
b) 1°c) 1°=2°=3°
d) 1°>3°>2°
Answer: a
Clarification: Alkyl groups are electron releasing groups and tend to increase the electron density on oxygen and reduce the polarity of O-H bond. So, the greater the number of alkyl groups present, lesser will be the tendency to release protons, and therefore weaker will be the acidic strength.

4. Water is a ______ than alcohol.
a) better proton acceptor
b) stronger base
c) weaker acid
d) better proton donor
View Answer

Answer: d
Clarification: The reaction between water and an alkoxide illustrates that water is a better proton donor (stronger acid) than alcohol.

5. Phenols are stronger acids than alcohols.
a) True
b) False
Answer: a
Clarification: The aryl carbon in phenols is more electronegative than the sp³ hybridized carbon in alcohols due to the presence of double bond. This extra negative charge is also delocalized at positions across the benzene, resulting in resonating structure and hence stability.

6. Which of the following compounds is the best proton acceptor?
a) Ethanol
b) Phenol
c) p-Cresol
d) p-Nitrophenol
Answer: a
Clarification: Proton acceptor indicates the basic nature of the compound. Ethanol is the only alcohol in the given compounds, and it is the least acidic, because phenols are more acidic than alcohols due to their stability through resonance structures.

7. Which of the following phenols is the most acidic?
a) o-Cresol
b) m-Cresol
c) o-Nitrophenol
d) m-Nitrophenol
Answer: c
Clarification: Alkyl groups are electron releasing, and contribute towards a reduction in acidic strength of phenols. Electron withdrawing groups like nitro, enhances the acidic strength especially at ortho and para positions, due to effective delocalisation of negative charge in the phenoxide ion.

8. What is the role of concentrated H₂SO₄ in the reaction between alcohol and carboxylic acid?
a) Deprotonating agent
b) Dehydrating agent
c) Reducing agent
d) Nucleophile
Answer: b
Clarification: Concentrated H₂SO₄ is the catalyst in the esterification of alcohols, which acts a protonating and dehydrating agent. The reaction is reversible and is shifted in the forward direction by the removal of water as soon as it is formed.

9. Which of the following alcohols is the most reactive towards esterification reaction?
a) CH₃OH
b) CH₃CH₂OH
c) (CH₃)₂CHOH
d) (CH₃)₃COH
Answer: a
Clarification: Presence of bulky groups reduces the reactivity of alcohols and carboxylic acid towards esterification due to effect of stearic hinderance. Therefore, the tertiary alcohols are least reactive.

10. Identify the products of the following reaction:
Methyl alcohol + Ethyl magnesium bromide = ?
a) CH₄ and CH₃OMgBr
b) CH₄ and CH₃CH₂OMgBr
c) C₂H₆ and CH₃OMgBr
d) C₂H₆ and CH₃CH₂OMgBr
Answer: c
Clarification: Alcohols reacts with Grignard reagents to form hydrocarbons. Also, the alkane formed from this reaction corresponds to the alkyl group of the Grignard reagent and not the alcohol.

11. What is the catalyst used in the acetylation of alcohols?
a) Palladium
b) Hydrogen peroxide
c) Pyridine
d) Aluminium chloride
Answer: c
Clarification: The reaction of alcohols with acid chlorides and acid anhydride is carried out in the presence of a base so as to neutralise the acid formed during the reaction. This facilitates the shift f equilibrium in the forward direction.

12. Aspirin is produced from the reaction between salicylic acid and which compound?
a) Acetyl chloride
b) Acetic anhydride
c) Phenyl acetate
d) Benzoyl chloride
Answer: b
Clarification: Salicylic acid, which is a phenolic acid when treated with acetic anhydride produces acetyl salicylic acid, also known as aspirin.

13. What is the correct order of reactivity of alcohols toward reactions involving cleavage of C-O bond?
a) 1°>2°>3°
b) 3°>2°>1°
c) 1°>3°>2°
d) 3°>1°>2°
Answer: b
Clarification: Alkyl groups are electron releasing groups and increase the electron density towards oxygen making the C-O more polar. This makes the cleavage of the bond between C and O a lot easier.

14. What is the catalyst used in the following reaction?

a) Anhydrous AlCl₃
b) Anhydrous FeCl₃
c) Anhydrous ZnCl₂
d) No catalyst required
Answer: d
Clarification: The given alcohol is a tertiary alcohol and can easily form alkyl halides with HCl and does not require any catalyst. However, primary and secondary alcohols require ZnCl₂ as catalyst in the same reaction for it to proceed.

15. Propan-2-ol undergoes acidic dehydration relatively at a lower temperature than propan-1-ol.
a) True
b) False
Answer: a
Clarification: Acidic dehydration of alcohols occurs through the formation of a carbocation. Since tertiary alcohol produce the most stable carbocations and primary alcohols produce the least stable carbocation, the dehydration of tertiary and secondary alcohols is relatively easier.

Chemistry Exam Questions and Answers for Class 12,

Chemistry Aptitude Test for Class 12 on “Carboxylic Acids Chemical Reactions – 3”.

1. Ammonium acetate on removal of ______ forms acetamide.
a) H₂
b) NH₃
c) OH^–
d) H₂O
Answer: d
Clarification: Ammonium acetate (CH₃COO-NH₄⁺) is an ammonium salt which on heating at high temperatures, loses a water molecule to give acetamide (CH₃CONH₂).

2. Benzoic acid reacts with ______ to give ammonium benzoate salt, which on further dehydration gives benzamide.
a) N₂
b) NH₃
c) NH₄OH
d) HNO₃
Answer: b
Clarification: Benzoic acid reacts with ammonia in a reversible reaction to form ammonium benzoate by addition reaction. This salt then loses water on heating to form benzamide.

3. The conversion of phthalamide to phthalimide is brought about by the loss of ______ molecule.
a) one H₂O
b) two H₂O
c) one NH₃
d) two NH₃
Answer: c
Clarification: Phthalic acid on reaction with NH₃ forms ammonium phthalate, which on heating loses two H₂O molecules, one from each COONH₄ group, to form phthalamide. This on further strong heating loses one NH₃ molecule to form phthalimide, which is an imide derivative of phthalic anhydride.

4. Which of the following reagents does not reduce the CO group of carboxylic acids to CH₂ groups to form alcohols?
a) LiAlH₄-ether
b) B₂H₆-THF
c) H₂-CuCr₂O₄
d) NaBH₄
Answer: d
Clarification: Carboxylic acids on reduction with LiAlH₄, B₂H₆ or H₂ in the presence of suitable medium, are reduced to alcohols. The COOH group is reduced to CH₂OH group. Sodium borohydride cannot reduce the carboxyl group.

5. Soda lime consists of NaOH and CaO respectively, in the ratio of _______
a) 1:2
b) 2:1
c) 1:3
d) 3:1
Answer: d
Clarification: Soda lime is the 3:1 mixture of sodium hydroxide and calcium oxide, and is used in the decarboxylation reaction (loss of carbon dioxide) of carboxylic acids.

6. Sodium acetate on heating with NaOH and CaO (3:1 mixture) gives _______
a) methane
b) ethane
c) methanol
d) ethanal
Answer: a
Clarification: Sodium acetate undergoes decarboxylation (loss of CO₂) when heated with soda lime. This is proceeded by the loss of COONa from the sodium salt and NaO from sodium hydroxide. They combine to form CH₄ and Na₂CO₃. The hydrocarbon formed has one less carbon atom than the parent acid.

7. Which of the following acids on Kolbe’s electrolysis gives ethane?
a) HCOOK
b) HCOONa
c) CH₃COOK
d) CH₃CH₂COONa
Answer: c
Clarification: Aqueous solutions of sodium or potassium salts of carboxylic acids on electrolysis, undergo decarboxylation and form hydrocarbons having twice the number of carbon atoms as present in the alkyl group of the acid. Since ethane has two C atoms, the alkyl group of the acid should be CH₃.

8. Identify X in the following reaction.

a) i) Br₂-red P; ii) H₂O
b) i) Cl₂-red P; ii) H₂O
c) PBr₃-H₂O
d) PBr₅-H₂O
Answer: a
Clarification: Propanoic acid undergoes bromination at alpha-carbon in the presence of red phosphorus to give α-Bromopropanoic acid. This is known as Hell-Volhard-Zelinsky reaction.

9. Methanoic acid undergoes Hell-Volhard-Zelinsky reaction.
a) True
b) False
Answer: b
Clarification: Methanoic acid (HCOOH) consists of only one carbon atom and therefore does not have an alpha-hydrogen atom that can be replaced by a halogen atom.

10. What is formed when benzoic acid undergoes nitration in the presence of conc. HNO₃ and conc. H₂SO₄?
a) o-Nitrobenzoic acid
b) m-Nitrobenzoic acid
c) p-Nitrobenzoic acid
d) 3,5-Dinitrobenzoic acid
Answer: b
Clarification: Benzoic acid undergoes ring substitution in which the COOH group acts as a deactivating group and is meta-directing in nature.

11. Benzoic acid undergoes Friedel-Crafts alkylation to form o- and p-Toluic acid.
a) True
b) False
Answer: b
Clarification: The COOH group is a deactivating group and does not undergo Friedel-Crafts alkylation. The catalyst AlCl₃ being a Lewis acid, gets bonded to the carboxyl group, and the methyl group does not get substituted in the ring.

12. Benzoic acid reacts with ______ to give m-Bromobenzoic acid.
a) Br₂/light
b) Br₂-FeBr₃
c) NaBr
d) PBr₃/heat
Answer: b
Clarification: Benzoic acid undergoes electrophilic substitution with bromine at the meta position in the presence of FeBr₃ to form 3-bromobenzoic acid and hydrogen bromide as side product.

Chemistry Aptitude Test for Class 12,

Chemistry Problems for Class 12 on “Diazonium Salts Chemical Reactions – 2”.

1. The complete reaction for the conversion of aniline to benzene, involving the reduction of diazonium salt, is known as _______
a) diazotisation
b) deamination
c) sandmeyer reaction
d) gatterman reaction
Answer: b
Clarification: The complete process of diazotisation of aniline followed by the reduction of diazonium salt or replacement of the diazo group by hydrogen is called deamination.

2. When benzenediazonium chloride is treated with hypophosphorous acid, the product obtained is __________
a) phenol
b) chlorobenzene
c) benzene
d) aniline

Answer: c
Clarification: When benzenediazonium salts are treated with mild reducing agents like hypophosphorous acid, the complete diazo and anionic group is replaced by hydrogen atom to form benzene.

3. A diazonium salt, on treatment with which of the following gives benzene?
a) LiAlH₄
b) pyridine
c) CH₃CH₂OH
d) HBF₄
Answer: c
Clarification: Ethanol is a mild reducing agent which on treatment with a benzenediazonium salt, oxidises to the corresponding aldehyde, and in the process reducing the diazonium salt to give benzene.

4. Identify ‘Y’ in the following reaction.

a) PCl₃
b) H₃PO₄
c) H₃PO₃
d) H₃PO₂
Answer: d
Clarification: In the reaction, benzenediazonium chloride is reduced to benzene. This is possible in the presence of a reducing agent. Since, one of the by products is H₃PO₃, it must be formed from the oxidation of the reducing agent, and hence must either have less O atoms or more H atoms than in H₃PO₃. Therefore, it is H₃PO₂.

5. A diazonium salt ‘X’ forms sulphuric acid as one the by product when is reacts with phosphinic acid to give benzene. Identify ‘X’.
a) Benzenediazonium chloride
b) Benzenediazonium bromide
c) Benzenediazonium hydrogensulphate
d) Benzenediazonium fluoroborate
Answer: c
Clarification: The H⁺ ions from the aqueous solution of acid combine with the negative HSO₄^– ions from the benzenediazonium hydrogensulphate to form H₂SO₄ (sulphuric acid) as the by product.

6. Nitrogen gas is evolved as a product of the reaction between benzenediazonium salt and ethanol.
a) True
b) False
Answer: a
Clarification: The nitrogen from the diazo group of the diazonium salt is liberated as free nitrogen molecule (N₂) is the form of a gas, as a by-product during the reaction.

7. Which of the following pathways will result in the formation of benzene?

a) Only A
b) Only B
c) Both A and
d) Neither A nor B

Answer: b
Clarification: Hypophosphorous acid (H₃PO₂) is a mild reducing agent, and reacts with benzenediazonium chloride to form benzene. Propionaldehyde does not reduce diazonium salts to benzene.

8. Which of the following is not obtained as a product of the reaction between benzenediazonium bromide and ethanol?
a) Benzene
b) Nitrogen gas
c) Acetic acid
d) Hydrogen bromide
Answer: c
Clarification: Ethanol is a mild reducing agent that reduces benzenediazonium bromide to benzene, and itself gets oxidized to an aldehyde, ethanal (CH₃CHO).

9. Warming an aqueous solution of benzenediazonium chloride gives phenol.
a) True
b) False
Answer: a
Clarification: If the temperature of benzenediazonium salt solutions is allowed to rise upto 283K, the diazonium group is replaced by OH group, and therefore the salt get hydrolysed to phenol.

10. Which of the following is the most suitable reagent for converting benzenediazonium fluoroborate to nitrobenzene?
a) NaNO₂/HCl
b) NaNO₂/Cu powder
c) HBF₄/HNO₂
d) HNO₃/HCl
Answer: b
Clarification: Benzenediazonium fluoroborate on treatment with an aqueous solution of sodium nitrite in the presence of copper (from copper powder), gives nitrobenzene.

11. What is the most suitable temperature for the conversion of benzenediazonium fluoroborate to nitrobenzene, with sodium nitrite and copper?
a) 0°C
b) 10°C
c) 22°C
d) 40°C
Answer: d
Clarification: When diazonium fluoroborate is heated with aqueous NaNO₂ in the presence of copper metal, the diazo group is replaced by NO₂ group, to give nitrobenzene.

12. Which of the following is a by product of the reaction between fluoroborate diazonium salt and sodium nitrite in the presence of copper?
a) HBF₄
b) NaBF₄
c) Na₃N
d) NaF
Answer: b
Clarification: The diazonium salt dissociates in solution giving a free fluoroborate anion (BF₄^–). This combines with the Na⁺ ions in solution to form NaBF₄.

Chemistry Problems for Class 12,

Chemistry Multiple Choice Questions on “Molecular Mass of Polymers”.

1. Molecular mass of polymers are expressed as a/an _______
a) average
b) median
c) mode
d) percentage
Answer: a
Clarification: The length of polymer and their molecular mass depends on the number of monomers available for reaction. As a result, the chain lengths of polymers are varying and are expressed as an average.

2. Polydispersity index is defined as ______ where M_w and M_n are the weight average and number average molecular masses respectively.
a) M_w x M_n
b) M_w/M_n
c) M_n/M_w
d) M_w – M_n
Answer: b
Clarification: Polydispersity index (PDI) is defines as the ratio of weight average molecular mass to the number average molecular mass. It gives an idea of the homogeneity of the polymer.

3. The polydispersity index of natural polymers is _________
a) 0
b) c) 1
d) >1.2
Answer: c
Clarification: Natural polymers generally have chains of identical lengths and have definite molecular masses (M_w=M_n). This makes them more homogeneous and monodisperse, with a PDI of approximately 1.

4. Calculate the number average molecular mass of a polymer having four different monomers A, B, C and D present in equal number. The molecular masses of the monomers are 10000, 15000, 30000 and 50000 respectively.
a) 10050
b) 17350
c) 26250
d) 35475
View Answer

Answer: c
Clarification: Since the 4 monomers are present in equal amount, there is 25% of each monomer in the polymer. The number average molecular mass is,
(Rightarrow) M_n = (25×10000 + 25×15000 + 25×30000 + 25×50000)/(25+25+25+25)
(Rightarrow) M_n = (250000+375000+750000+1250000)/100
(Rightarrow) M_n = 2625000/100
(Rightarrow) M_n = 26250.

5. The polydispersity index can be less than 1.
a) True
b) False
Answer: b
Clarification: The weighted average molecular mass is always greater than the number average molecular mass. This is because of the 2-degree nature of M_w (because of squaring of masses). Hence PDI ≥ 1.

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