Explore topic-wise InterviewSolutions in Current Affairs.

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1.

N_(2) + 3H_(2) to 2NH_(3) molecular weight and equivalent weight of NH_(3) and N_(2) are 17.03 g,14 g and Y_(1),Y_(2)respectively. Find the value of (Y_(1)-Y_(2)).

Answer»


Solution :`6e^(-) + N_(2) to 2N^(3-)`
`THEREFORE E_(N_(2)) = 14/6 =Y_(2)`
`therefore E_(NH_(3)) = (17.03)/3 = Y_(1)`
`therefore Y_(1)-Y_(2) = (17.03)/3- X_(2)/6= 5.67 - 2.33 = 3.34`
2.

N_(2)+3H_(2)hArr2NH_(3)+ heat . What is the effect of the increase of temperature on theequilibrium of the reaction ?

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equilibrium is unaltered
reaction rate does not change
equilibrium is SHIFTED to the left.
equilibrium is shifted to the right

Solution :`N_(2)+3H_(2)hArr2NH_(3)+`heat FORMATION of ammonia is exothermic reaction so increase in temperature favours backward reaction and equilibrium is shifted to the left.
3.

N_(2) + 3H_(2) hArr 2NH_(3)""K=4xx 10^(6)"at"298 ""K=41 "at" 400 k Which statements is correct?

Answer»

If `N_(2)` is added at equlibrium condition, the equilibrium will shift to the FORWARD direction because according to `II^(nd)` law of thermodynamics the entropy must increases in the direction of spontaneous reaction .
The condition for equlibrium is `2DeltaG_(NH_(3)) = 3DeltaG_(N_(2)) + DeltaG_(N_(2))` where G is Gibbs FREE energy per mole of the gaseous speciesmeasured at that partial pressure.
ADDITION of catalyst does not change `K_(p)` but changes`DELTAH`.
At 400 K addition of catalyst will increase forward reaction by `2` times while reverse reaction rate will be changed by 1.7 times.

Solution :When nitrogen is added at equilibrium condition, the equilibrium will shift according to Le- chatelier principle at equilibrium `DeltaG^(@)=0` and catalyst changes the rate of forward and backward reactions by equal extent. `K_(P)` of reaction isa function of temperture only.
4.

n_(1)Cr_(2)O_(7)^(-2) + n_(2)FeC_(2)O_(4) underset("Medium")overset("Acidic")rarr n_(3)Cr^(+3) + n_(4)Fe^(+3) + n_(5)CO_(2) + n_(6)H_(2)O here n_(1), n_(2), n_(3), n_(4), n_(5) and n_(6) are simplest non fractional coefficients of the balanced redox reaction. Then value of [n_(1) + n_(2) + n_(3) + n_(4) + n_(5) + n_(6)] is :

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SOLUTION :`14H^(+) + Cr_(2)O_(7)^(-2) + 2FeC_(2)O_(4) RARR 2Cr^(+3) + 2Fe^(+3) + 4CO_(2) + 7H_(2)O`
`therefore n_(1)= 1, n_(2)=2, n_(3)= 2, n_(4)= 2, n_(5)= 4, n_(6)= 7`
And `(n_(1) + n_(2) + n_(3) + n_(4) + n_(5) + n_(6))= 18`
5.

n-valeric acid is functional isomer of

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METHYL propionate
n-butyl FORMATE
propyl isobutyrate
ethyl isobutyrate

Answer :B
6.

n the completecombustionof c_(n) H_(2n+2), thenumberof oxygenmolesrequiredis :

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`(N) /(2O_(2))`
`((n+1)/(2))O_(2)`
`((3n+1)/(2))O_(2)`
`((n+2)/(2))O_(2)`

ANSWER :C
7.

n-Propylmagnesium bromide on treatment with carbon dioxide and further hydrolysis gives

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ACETIC acid
propanoic acid
butanoic acid
formic acid.

Answer :C
8.

n-propylbromide on treatment with ethanolic potassium hydroxide produced

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Propane
PROPENE
Propyne
Propanol

Solution :`underset("n-propyl BRO MIDE")(CH_(3)-CH_(2)-CH_(2)Br) overset("ethanolic KOH")toCH_(3)underset("propene")(-CH)=CH_(2)+KBr+H_(2)O`
9.

n-Propylamine yields a volatile compound X on warning with alc alkali and chloroform X has an offensive odour. The structure of X is

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`CH_(3)CH_(2)CH_(2)CN`
`(CH_(3))_(2)CHCN`
`CH_(3)CH_(2)CH_(2)NC`
`(CH_(3))_(2)CHNC`

Answer :C
10.

n-propyle bromide on treatment with ethanolic potassium hybroxide produces :

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Propane
Propene
Propyne
Propanol

Answer :B
11.

n-Propyl chloride reacts with sodium metal in dry ether to give

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`CH_(3)-CH_(2)-CH_(2)-CH_(2)-CH_(2)-CH_(3)`
`CH_(3)-CH_(2)-CH_(3)`
`CH_(3)-CH_(2)-CH_(2)-CH_(3)`
`CH_(3)-CH_(2)-CH_(2)-CH_(2)-CH_(2)-CH_(2)-CH_(3)`

Solution :`underset("n-propylchloride")(CH_(3)CH_(2)CH_(2)-Cl+2Na)+Cl -CH_(2)CH_(2)CH_(3) underset("ETHER")OVERSET("dry")to underset("n-Hexane")(CH_(3)CH_(2)CH_(2))-CH_(2)CH_(2)CH_(3)`
12.

n-Propyl chloride and benzene react in the presence of anhydrous AlCl_(3) to form :

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ETHYL BENZENE
METHYL benzene
n-propyl benzene
iso-propyl benzene

SOLUTION :
13.

n-Propyl bromide on treatment with ethanolic potassium hydroxide produces :

Answer»

Propane
PROPENE
Propyne
Propanol

Solution :`UNDERSET("n-Propyl bromide")(CH_(3)-CH_(2)-CH_(2)-CH_(2))-Br+KOH(alc.)rarr underset("Propene")(CH_(3)CH)=CH_(2)+CH_(2)KBr+H_(2)O`
14.

n-propyl bromide on treatment with alcoholic KOH produces

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PROPANE
propane
PROPYNE
hypnotic

Answer :B
15.

n-propyl bromide reacts with ethanolic KOH to

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propane
propene
Propyne and propene
propyl alcohol

Answer :B
16.

n-propyl bromide on treatment with ethanolic potassium hydroxide produces

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Propane
PROPENE
Propyne
Propanol

Solution :`UNDERSET"n-Propyl BROMIDE"(CH_3-CH_2-CH_2-Br)+KOHoverset(C_2H_5OH)tounderset"Propene"(CH_3-CH=CH_2+KBr+H_2O)`
17.

n-propyl bromide on treating with alcoholic KOH produces

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PROPYNE
Propanol
Propane
Propene

Solution :`underset"Propylbromide"(CH_3CH_2CH_2Br)+underset"(alc.)"(KOH)to underset"propene"(CH_3CH)=CH_2+KBr+H_2O`
This reaction REMOVES a molecule of HX and therefore, the reaction is called dehydrohalogenation. The hydrogen atom is ELIMINATED from `beta`-carbon atom (carbon atom next to the carbon to which halogen is ATTACHED). Therefore the reaction is ALSO called `beta`-elimination reaction
18.

n-propyl amine reactswith HNO_(2)gives

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ISO - propylalcohol
N -propylalcohol
n - propylnitrosoamine
NONE of these

ANSWER :B
19.

n- propyl bromide on treating with alcoholic KOH produces

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propyne
propanol
propane
propene

Solution :`underset("Propyl bromide")(CH_(3)CH_(2)CH_(2)Br)+underset(("alc"))(KHO)tounderset("propene""")(CH_(3)CH=CH_(2)+KBr+H_(2)O)`
This reaction REMOVES a molecule of HX and therefor, the reaction is called dehydrohalogenation.The hydrogen atom is eliminated from `beta` CARBON atom (carbon atom NEXT to the carbon to which HALOGEN is attached). Therefore the reaction is also called `beta` - ELIMINATION reaction.
20.

n-Propyl alcohol and isopropyl alcohol can be chemically 5· distinguished by which reagent?

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`PCl_(5)`
Reduction
Oxidation with potassium dichromate
Ozonolysis

Solution :Primary ALCOHOL on oxidation give aldehyde which on further oxidation give CARBOXYLIC acid whereas secondary alcohols give ketone.
`underset("n-Propylalcohal")(CH_(3)CH_(2)CH_(2)OH)overset([O])(to) CH_(3)CH_(2)CHO overset([O])(to)CH_(3)CH_(2)COOH`
21.

n-Propyl alcohol and iso-propyl alcohol can be distinguished by

Answer»

using PCI
oxidation with potassium dichromate
`I_(2)//NaOH`
reduction

Solution :On oxidation PRIMARY alcohol GIVES carboxylic acid.
On oxidation secondary alcohol gives KETONE.
Iso-propyl alcohol will give a +ve IODOFORM test whereas n-propylalcohol will not.
22.

n-propyl alcohol and isopropyl alcohol are:

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POSITION isomers
chain isomers
tautomers
geometrical isomers

Answer :A
23.

n- Prophylbenzene can be obtained in quantitative yield by (i)treating benzene with n-propyl chloride in presence of AICI_(3) (ii)treating excess of benzene with n - prophyl chloride in presence of AICI_(3) (iii) treating benzene with allyl chloride in presence of AICI_(3) followed by reduction (iv)treating benzene with propionyl chloride in presence of AICI_(3) followed by Clemmensen reduction

Answer»

by (II), (III) and (IV)
by (i), (iii) and (iv)
by (iii)
by (iv)

ANSWER :C, D
24.

N-phenyl hydroxylamine is obtained when nitro benzene is reacted with

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`Sn//HCl`
`Zn//NH_(4)Cl//H_(2)O`
`As_(2)O_(3)//NAOH`
`Zn//NaOH`

ANSWER :2
25.

n-prepyl alcohol and isopropyl alcohol can be chemically distinguished by which reagent

Answer»

`PCl_5`
reduction
oxidation with POTASSIUM DICHROMATE
`PCl_3`

ANSWER :C
26.

N-phenyl ethanmideis obtainedform CH_(3)COCOl and what

Answer»




SOLUTION :
27.

n-pentane and neopentane are:

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FUNCTIONAL isomers
Geometrical isomers
Chain isomers
Position isomers

Answer :C
28.

n-pentane and 2-methylbutane are a pair of

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Enantiomers
Stereoisomers
Diastereomers
Constitutional isomers

Solution :Constitutional isomerism or strcuture isomerism arises DUE to the difference in the manner in which the CONSITUENT ATOMS or groups are linked to one another within the molecule, without any reference to SPACE.
The type of consitutional isomerism that exists between n-pentane and 2-methyl BUTANE is chain isomerism.
29.

N, P, As, Sb, and Bi belong to

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VA GROUP
IV A group
VIII group
VB group

ANSWER :A
30.

n-pentane and iospentane can be distinguished by

Answer»

`Br_2`
`O_3`
Conc. `H_2SO_4`
`KMnO_4`

SOLUTION :On OXIDATION, with `KMnO_4` , they give DIFFERENT alcohols
`underset"n-pentane"(CH_3CH_2CH_2CH_2CH_3)underset([O])OVERSET(KMnO_4)to underset(1^@"alcohol")(CH_3CH_2CH_2CH_2CH_2OH)`
`CH_3CH_2-underset(CH_3)underset(|)CH-CH_3underset"[O]"overset(KMnO_4)to underset(3^@"alcohol")(CH_3CH_2-undersetunderset(CH_3)(|)oversetoverset(OH)(|)C-CH_3)`
31.

N, P, As, Sb and Bi are in group 15, So, Bi in its external most shell contains:

Answer»

THREE electrons
One electrons
Five electrons
Seven electrons

Answer :C
32.

'n' number of alkenes yield 2,2,3,4,4-pentamethyl-pentane on catalytic hydrogenation and 'm' number of monochloro structural isomers are possible for this compound. Report your answer as (n+m).

Answer»


Solution :Only one alkene
`H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C )-underset(CH_(2))underset(||)(C )-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C )-CH_(3) overset(H_(2)//"catalyst") rarr`H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C )-underset(CH_(3))underset(|)(CH)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C )-CH_(3)`
THREE monochloro isomers are possilbe as it has three different TYPES of `'H'` atoms.
33.

N-N- dimethyl phenylaminereact withnitrous acids gives

Answer»




SOLUTION :
34.

N, N-dimethyl butanamine -2 is the functional isomer of

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N-butanamine -2
N-methyl-2-ethyl butanamine-2
trimethyl amine
triethyl amine

Answer :D
35.

N-N bond is not present in

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`N_(2)O_(3)`
`N_2O_(4)`
`N_(2)O_(5)`
`N_2`

Answer :C
36.

Nessler's reagent is :

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`HG^+`
`Hg^(2+)`
`HgI_2^(2-)`
`HgI_4^(2-)`

ANSWER :D
37.

N, N - dimethyl benzamide cannot be prepared by

Answer»

`C_(6)H_(5)COO C_(2)H_(5)+(CH_(3))_(2)NH to`
`C_(6)H_(5)COCl+(CH_(3))_(2)NH to`
`C_(6)H_(5)COO.CO C_(6)H_(5)+(CH_(3))_(2)NH`
`C_(6)H_(5)CO NH_(2)+CH_(3)MgCl`

Solution :`C_(6)H_(5)-overset(O)overset(||)C-NH_(2)+CH_(3)MgCl` not give benzomide
38.

N-methylaniline reacts with nitrous acid to form____while N,N-dimethylaniline gives_____

Answer»

SOLUTION :N-nitroso-N-methylaniline, p-nitroso-N , N-dimethylaniline
39.

N-Methyl aniline into N-Methyl benzanilide ?

Answer»

SOLUTION :
40.

(N is number of isomeric products) underset("distillation")overset("Fractional")to (M is the number of isomeric products). N and M are respectively.

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6,6
6,4
4,4
7,3

Solution :
41.

N-isopropyl hydroxyl amineis obtained by reductionof

Answer»

`CH_(3)-overset(NO_(2))overset(|)(CH)-CH_(3) andZn+ NH_(4)Cl`
`CH_(3)-overset(NO_(2))overset(|)(CH)-CH_(3) andSnCl_(2) + HCl`
`CH_(3)-overset(NO_(2))overset(|)(CH)-CH_(3) ANDFE + CONC. HCl`
`CH_(3)-overset(NO_(2))overset(|)(CH)-CH_(3) and H_(2)//Ni`

Answer :A
42.

N kg^(-1)is the unit of:

Answer»

Momentum
Velocity
Pressure
Acceleration

Solution :`N kg^(-1)= kg MS^(-2) xx kg^(-1) = ms^(-2)`
These are the UNITS of acceleration
43.

N- Hexane can beconverertedto benzene by thetreatmentwith :

Answer»

ALK.`KMnO_(4)`
`alc.KOH`
`Cr_(2)OAT 770K`
`LiAiH_(4)`

ANSWER :C
44.

n-hexadecane (or cetane) has cetane number:

Answer»

100
Zero
90
110

Answer :A
45.

n-Heptane passed over chromium trioxide supported over alumina at 873 K finally gives

Answer»

Toluene
Cycloheptane
Carboxycyclohexane
Ethylcyclopentane

Solution :This REFERS to the AROMATISATION PROCESS.
46.

N- Heptaneon heatingto 773K at10-20atmpresurein thepresenceof V_(2)O_( 5) a

Answer»

BENZENE
2- heptene
toluene
cycloheptane

ANSWER :C
47.

N- ethyl N- methylpropan -1- amineis _____ amine

Answer»

PRIMARY amine
Secondaryamine
Tertiaryamine
Quanternary

Solution :`H_(3) C- H_(2) C - CH_(2)- overset( C_(2) H_(5)) overset(|)underset(CH_(3))underset(|) ( N)`Tertiaryamine
48.

N-ethyl formamide on treatment with POCI_(3) in pyridine gives

Answer»

Ethyl isocyanate
Ethyl isocyanide
Acetaldoxime
Ethyl CYANIDE

Solution :`H-OVERSET(O)overset(||)C-NH-CH_(2)CH_(3)overset(POCl_(3))toH_(3)C-CH_(2)-NC`
49.

N- ethyl benzene sulphonyl amide is strongly acidic and soluble in alkali due to presence of

Answer»

STRONG ELECTRON donating sulphonyl group
strong electron withdrawing sulphonyl group.
weak electron donating sulphonyl group.
weak electron withdrawing sulphonyl group.

Solution :The hydrogen attached to nitrogen in sulphonamide is strongly acidic DUE to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali.
50.

n-Butyl chloride and iso butyl chloride are

Answer»

POSITION isomers
Functional GROUP isomers
Chain isomers
Metamers

ANSWER :C