Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

How the following conversions can be carriedout ? Ethanol to but-1-yne

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SOLUTION :
Discussion
2.

How the following conversions can be carried out ?Ethanol to but-1-yne

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SOLUTION :
Discussion
3.

How the following conversions can be carried out ?Chloroethane to butane

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SOLUTION :`underset("Chloroethane")(2CH_(3)CH_(2)Cl)overset(" 2Na ")underset(" DRY ETHER ")rarr underset("n-butane")(CH_(3)CH_(2)CH_(2)CH_(3))`
Discussion
4.

How the following conversions can be carriedout ? Chloroethane to butane

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Solution :`underset("Choroethane") (2CH_(3)CH_(2)- Cl ) + 2Na overset("DRY ether, "Delta ) to underset("Butane") (CH_(3) CH_(2) - CH_(2)CH_(3) + 2NaCl) `
Discussion
5.

How the following conversions can be carried out ?Chlorobenzene to p - nitrophenol

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SOLUTION :
Discussion
6.

How the following conversions can be carriedout ? But-l-ene to n-butyliodide

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Solution :`underset("But -1- ene") (CH_(3)CH_(2)CH = CH_(2)) underset(("ANTI - Markovnikov addition ")) overset("HBr /Peroxide") to underset("1 - BROMOBUTANE") (CH_(3)CH_(2)CH_(2) CH_(2) BR) underset(("Finkelstein reaction ")) overset("NaI, acetone") to underset("n -Butyl iodide") (CH_(3)CH_(2)CH_(2)CH_(2)I) `
Discussion
7.

How the following conversions can be carriedout ? Chlorobenzene to p-nitrophenol

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SOLUTION :
Discussion
8.

How the following conversions can be carried out ?But -1-ene to n-butyliodide

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SOLUTION :
Discussion
9.

How the following conversions can be carried out ?Benzyl alcohol to 2-phenylethanoic acid

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SOLUTION :
Discussion
10.

How the following conversions can be carriedout ? Benzyl alcohol to 2-phenylethanoic acid

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SOLUTION :
Discussion
11.

How the following conversions can be carried out ?Benzene to diphenyl

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SOLUTION :
Discussion
12.

How the following conversions can be carriedout ? Benzene to diphenyl

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SOLUTION :BENZENE to DIPHENYL
Discussion
13.

How the following conversions can be carried out ?Benzene to 4-bromonitrobenzene

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SOLUTION :
Discussion
14.

How the following conversions can be carriedout ? Benzene to 4-bromonitrobenzene

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SOLUTION :
Discussion
15.

How the following conversions can be carried out ?Aniline to phenylisocyanide

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SOLUTION :
Discussion
16.

How the following conversions can be carriedout ? Aniline to phenylisocyanide

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SOLUTION :
Discussion
17.

How the following conversions can be carried out ?Aniline to chlorobenzene

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SOLUTION :
Discussion
18.

How the following conversions can be carriedout ? 2-methyl-1-propene to 2-chloro-2-methylpropane

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SOLUTION :`underset("2 - Methyl - 1- PROPENE") (CH_(3) - overset(CH_(3)) overset(|) C = CH_(2)) underset(("Markovnikov additon ")) overset(HCl) to underset("2- CHLORO -2- methylpropane") (CH_(3) - underset(Cl) underset(|) overset(CH_(3))overset(|) C - CH_(3)) `
Discussion
19.

How the following conversions can be carriedout ? Aniline to chlorobenzene

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SOLUTION :
Discussion
20.

How the following conversions can be carried out ?2-Methyl-1-propene to 2-chloro-2-methyl-propane

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SOLUTION :
Discussion
21.

How the following conversions can be carried out ?2-Chloropropane to 1-propanol

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SOLUTION :
Discussion
22.

How the following conversions can be carriedout ? 2-chloropropane to 1-propanol

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Solution :`underset("2-Chloropropane") (CH_(3) - underset(Cl)underset(|) (CH) - CH_(3)) underset(("Dehydrohalogenation ")) overset(KOH (alc.), Delta ) to underset("Propene") (CH_(3)- CH= CH_(2)) underset (("ANTI -Markovnikovadditon ")) overset("HBr, Peroxide") to underset("1 -Bromopropane") (CH_(3)CH_(2)CH_(2)Br) underset(("Nucleophilic SUBSTITUTION ")) overset(KOH (aq) , Delta ) to underset("1 - Propanol") (CH_(3)CH_(2)CH_(2)OH) `
Discussion
23.

How the following conversions can be carried out ?2-Chlorobutane to 3,4 - dimethylhexane

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SOLUTION :
Discussion
24.

How the following conversions can be carriedout ? 2-Chlorobutane to 3, 4-dimethylhexane

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Solution :`UNDERSET("2 - Chlorobutane") (2HC_(3) - underset(Cl) underset(|) CH - CH_(2)CH_(3)) + Na underset(("WURTZ reaction ")) OVERSET("Dry ETHER") to CH_(3)CH_(2) - underset("3,4 - Dimethylhexane")(underset(CH_(3))underset(|) CH - underset(CH_(3))underset(|) CH - ) CH_(2)CH_(3) + 2NaCl `
Discussion
25.

How the following conversions can be carried out ?1-Bromopropane to 2-bromopropane

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SOLUTION :
Discussion
26.

How the following conversions can be carried out ?2-Bromopropane to 1-bromopropane

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SOLUTION :
Discussion
27.

How the following conversions can be carriedout ? 2-Bromopropane to 1-bromopropane

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SOLUTION :`underset("2 -Bromopropane") (CH_(3) - underset(Br) underset(|) CH - CH_(3)) underset(("DEHYDROHALOGENATION ")) overset(KOH (alc.) ,Delta ) to underset("Propene") (CH_(3) - CH = CH_(2)) underset(("Anti -Markovnikov addition ")) overset("HBR, peroxide") to underset("1- Bromopropane") (CH_(3) - CH_(2) - CH_(2) - Br ) `
Discussion
28.

How the following conversions can be carriedout ? 1-Bromopropane to 2-bromopropane

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SOLUTION :`underset("1 - Bromopropane " ) (CH_(3)CH_(2)CH_(2)Br) underset("Dehydrohalogenation ") OVERSET("KOH (alc.) , D") to underset("Propene") (CH_(3) CH = CH_(2)) underset("Markovnikov ADDITON ") overset("HBr") to underset("2 - Bromopropane " ) (CH_(3)- CHBr - CH_(3)) `
Discussion
29.

Howthe followingconversions are effected ? (i) Phenol toSalicylaldehyde (ii) Phenol tophenolphthalein (iii)glycolto1,4 dioxane

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Solution :(i) Riemer Tiemann Reaction :On treatingphenol with ` CHCI_(3)/ NaOH, a - CHO_(3)`GROUP is introducedat orthoposition . This reaction proceedsthroughthe formation of substituted benzeal chlorideintermediate
` (##SUR_CHE_XII_V02_QP_E01_039_S01.png" width="80%">
(II) Phthalein reaction : On HEATING phenolwith phthalicanhydrideis presence of con. ` H_(2)SO_(4)`phenolphtaleinis obtained

(iii) When distilledwith Conc. ` H_(2)SO_(4)` glycol FORME dioxane

(##SUR_CHE_XII_V02_QP_E01_039_S03.png" width="80%">
Discussion
30.

How the density of an unit cell is calculated ?

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Solution : The density of an unit cell is calculated by,
Density `(d) = (Z xx M)/(N_A xx a^3 xx 10^(-30)) g/(cm^3)`
Where
Z = Number of atoms per unit cell
M = Mol. MASS
`N_A` = Avogadro.s number
a = Edge length of unit cell in PICOMETERS as measured by X-ray diffraction,
d = density of unit cell in `g//cm^3`
Discussion
31.

How the biodegradable polymers are decomposed ?

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By HYDROLYSIS
By MICRO ORGANISAMS
By ENZYMES
All of these

SOLUTION :All of these
Discussion
32.

How the average rate of chemical reaction is decribed?

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By increasing the CONCENTRATION of products
By decreasing the concentration of REACTANTS.
Both (A) and (B)
By CHANGING in the time of reactants and products

Answer :D
Discussion
33.

How the adsorption of a gas is related to the critical temperature

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Solution :LARGER is the critical TEMPERATURE of a gas, larger is its EASE of liquefication, and thus larger will be the adsorption
Discussion
34.

How test benzaldehyde react with a concentrated solution of sodium hydroxide? Give the equation and name the reaction.

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SOLUTION :CANNIZARO REACTION:
Discussion
35.

How solubility of gas in liquid varies with (i) temperature (ii) pressure.

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SOLUTION :Solubility of gas DECREASES with increase in temperature and increases with increase in pressure.
Discussion
36.

How solubility of a gas in liquid changes with increase in temperature?

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SOLUTION :DECREASES
Discussion
37.

How slag is formed in metal extraction ?

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Solution :FLUX combines with INFUSIBLE IMPURITY to FORM FUSIBLE slag .
Discussion
38.

How proteins are classified on the basis of structure ?

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Answer :On the basis of structure, PROTEINS are of 3 types, i.e. Primary, Secondary and TERTIARY. In primary structure ATOMS of the protein molecule are joined to one another through covalent bond and form a long CHAIN.
Discussion
39.

How proteins are classified depending on their three dimensional shape ?

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Answer :Fibrous Proteins- Fibrous proteins consist of POLYPEPTIDE chains arranged side by side in LOG filaments.
Globular Proteins- Globular proteins are USUALLY coiled and COMPACT and nearly spherical in SHAPE. Most of the proteins are globular proteins.
Discussion
40.

How potassium permanganate is prepared from MnO_2.

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SOLUTION : (a) PREPARATION of `KMnO_4`: Potassium permanganate is prepaered by the FUSION of `MnO_2`with an alkali metal hydroxidizing like `KNO_3` . It froms dark green `K_2 MnO_4` which disproportinates in a neutral of acidic solution to give peremanganate
(i) `3MnO_2 + 4KOH + O_2 OVERSET ( KNO_3) to 2K_2 MnO_4 + 2H_2O`
(ii) `3MnO_(4)^(2-) + 4H to 2MnO_4+ MnO_2 + 2H_2O`
Discussion
41.

How physical state of halogen changes with increase in atomic number ?

Answer»

Solution :Fluorinerarr GAS
Chlorine RARR gas
Bromine rarr LIQUID
Iodine rarr solid
Discussion
42.

How PCl_(3), PCl_(5) and SOCl_(2) reacts with carboxylic acids ?

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SOLUTION :`{:((i), 3RCOOH "CARBOXYLIC acid",+ PCl_(3) "Phosphorus trichloride", to RCOCl "Acid chloride", + H_(5)PO_(3),),((ii),RCOOH"Carboxylic acid", PCl_(5)"Phosphorus pentachloride", to RCOCl "Acid chloride",POCl_(3),CHL),((iii),R-COOH"Carboxylic acid",+ SOCl_(2) "Thionlyl ehloride",to R-CO-Cl"Acid chloride",+ SO_(2),+HCl):}`
Discussion
43.

How ozone is formed from atmospheric oxygen in the atmosphere/ What is the use of it in the atmosphere?

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Solution :Ozone is formed from atmospheric OXYGEN in the pressence of sun LIGHT. This ozone LAYER PROTECTS the earth.s surface from ultraviolet (UV) radiation.
Discussion
44.

How much work is required to break up 1 mole of water at 20°C into spherical droplets of radius 1 cm? Given that gamma(H_(2)O) =72.75 xx 10^(-3) Nm^(-1),density (H_(2)O)=19.015g/mole.

Answer»

Solution :SUPPOSING 1 mole of water to be spherical,
volume of 1 mole of water `=(18.015)/(0.998 xx 10^(6))`
`=1.8047 xx 10^(-5) m^(3)`
`therefore r= root(3)((3 xx V)/(4pi)) ((3 xx 1.8047 xx 10^(-5))/(4 xx 3.14))^(1//3) = 1.6272 xx 10^(-2) m`
Surface AREA `=4pir^(2) = 4 xx 3.14 xx (1.6272 xx 10^(-2))^(2)`
`=3.3273 xx 10^(-3) m^(2)`
Volume of each droplet= `4/3 xx pi xx (1 xx 10^(-3))^(3) ~~ 4 xx 10^(-6) m^(3)`
Number of droplets `=(1.8047 xx 10^(-5))/(4 xx 10^(-6)) =5`
Surface area of 5 droplets `=5 xx 4pir^(2)`
`=5 xx 4 xx 3.14 xx (1 xx 10^(2))^(2)`
`=6 xx 10^(-3) m^(2)`
When a DROP of 1 mole of water breaks into 5 droplets,
INCREASE in surface area `=(6 xx 10^(-3) -3.3273 xx 10^(-3)) m^(2)`
`=2.6727 xx 10^(-3) m^(2)`
Work DONE to break up 1 mole of water=surface tension `xx` increase in area
`=72.75 xx 10^(-3) xx (2.6727 xx 10^(-3)) J`
`=1.94 xx 10^(-4) J (1 J = 1 Nm)`
Discussion
45.

How much will the reduction potential of Zn//Zn^(2+) change if the solutionof Zn^(2+)is diluted 10 times ?

Answer»

INCREASES by 0.03 V
DECREASES by 0.03 V
INCREASE by 0.059 V
decreases by 0.059 V

Answer :B
Discussion
46.

How much will the reduction potential of a hydrogen electrode change when its solution initially at pH=0 is neutralised to Ph =7:

Answer»

Increaase by `0.059V`
DECREASE by `0.059V`
INCREASE by `0.41V`
DECREASES by `0.41V`

ANSWER :D
Discussion
47.

How much will the reduction potential of a hydrogen electrod echange when its solution initially at pH=0 is neutralized to pH=7?

Answer»

Increases by 0.059 V
Decreases by 0.59 V
Increases by 0.41 V
Decreases by 0.41 V

Solution :`H^(+)+e^(-)to(1)/(2)H_(2)`
`E_(RED)=E_(red)^(@)-(0.0591)/(2)"log"(1)/([H^(+)])`
`=E_(red)^(@)+0.0591log[H^(+)]`
`E_(red)^(@)=0`. When `pH=0,[H^(+)=]10^(@)=1M`
`thereforeE_(red)=0`
when pH=7,`[H^(+)]=10^(-7)M`
`E_(red)=0.0591" log "10^(-7)(0.0591)(-7)`
`=-0.4137`
THUS, `E_(red)` decreases by 0.41 V.
Discussion
48.

How much will the K.E. of an electron in H -atom change if the atom emits a photon of wavelength of the electron wave before emission of photon of the aforsaid wavelength .

Answer»


ANSWER :2.55 EV
Discussion
49.

How much water should be added to 2% W/V, 5 liter NaOH solution so as to make its concentration to 0.5 N ?

Answer»

5 lit
2.5 lit
1 lit
No need to add water

SOLUTION :2% W/W
i.e., 100 mL `to` 2 gram NaOH
1000 mL `to` 20 gram NaOH
To GET 0.5 N solution,
W = NORMALITY `xx` eq. weight `xx` volume
`= 0.5xx40xx1=20` gram
`therefore` There is no need to add water.
Discussion
50.

How much water should be added to 10 g of acetic acid to give a hydrogen-ion concentration equal to 1xx10^(-3)M? K_a (CH_2COOH) = 1.8 xx 10^(-5).

Answer»

SOLUTION :3.0 LITRE
Discussion