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.

Find out A in the following reaction A overset(aq.KOH) to B overset(PC C) topropanal

Answer»

n- PROPYL ALCOHOL
n-propyl HALIDE
n- propyl amine
diethyl ETHER

Answer :B
Discussion
2.

Find out A, B, C , D in following reaction (1) CH_(3) - CH = CH - CHO overset(A) to CH_(3) - CH = CH - COOH (2) CH_(3) - CH = CH - CHO overset(B) to CH_(3)- CH = CH - CH_(2) - OH (3) R - COOH overset(C) to R - CH_(2) - OH (4) R - COCl overset(D) to R - CHO

Answer»

`{:(1= h_(2) .Pf.BaSO_(4),2= "AMMONIACAL "AgNO_(3)),(3= LiAlH_(4),4=LiAlH_(4)):}`
`{:(1=LiAlH_(4),2= LiAlH_(4)),(3=H_(2).PD.BaSO_(4),4="Ammoniacal "AgNO_(3)):}`
`{:(1="Ammoniacal "AgNO_(3),2=LiAlH_(4)),(3=LiAlH_(4),4=H_(2)Pd +BaSO_(4)):}`
`{:(1=LiAlH_(4),2=LiAlH_(4)),(3="Ammoniacal" AgNO_(3),4=H_(2)Pd+BaSO_(4)):}`

ANSWER :C
Discussion
3.

Find out A and B in the following reaction respectively Ph-Cl oversetAto Ph-O Na overset(B)to Ph-OH

Answer»

NAOH and HCL
`H_2O` and HCl
HCl and NaOH
HCI and `H_2O`

ANSWER :A
Discussion
4.

Find our the ratio of coefficients of metals. Al ^ MnO_(2) rarr Al_(2)O_(3) + Mn.

Answer»


ANSWER :`4 : 3`
Discussion
5.

Find out (A) :

Answer»




ANSWER :B
Discussion
6.

Find our the entropy change in surroudings when 1 mole of H_(2)O(l) is formed under standard conditions Delta_(f)H^(ө)=-286" kJ "mol^(-1)

Answer»

959.7 `JK^(-1)mol^(-1)`
286 `JK^(-1)mol^(-1)`
`-959.7" "JK^(-1)" "mol^(-1)`
`-286JK^(-1)mol^(-1)`

Solution :`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l),Delta_(f)H^(@)=-286" kJ "mol^(-1)`
When 1 mole of heat is absorbed by the surrounding, i.e.,
`q_("surr")=+286" kJ "mol^(-1)`
`DELTAS=(q_("surr"))/(T)=(286 " kJ "mol^(-1))/(298K)=959.7JK^(-1)mol^(-1)`.
Discussion
7.

Find number of O-atoms in 1 moles O_(2)

Answer»


ANSWER : `(##ALN_NC_CHM_MC_E01_014_A01##)`
`2N_(A)`
Discussion
8.

Find number of oxygen atoms present in 100 mg of CaCO_3. (Atomic mass of Ca = 40 u, C = 12 u, O = 16 u)

Answer»

`6.02 xx 10^(23)`
`6.02 xx 10^(20)`
`1.806 xx 10^(21)`
`1.204 xx 10^(20)`

ANSWER :C
Discussion
9.

Find number of molecules in 11.35 litres SO_(2) gas at STP.

Answer»


ANSWER : `(##ALN_NC_CHM_MC_E01_016_A01##)`
`(0.5xxN_(A))`
Discussion
10.

Find number of Co-N linkage in pentaammine cobalt(III)-mu-amidodiamminetriaquacobalt(III) chloride.

Answer»


Solution :`[(H_(3)N)_(5)overset(+3)(CO)-overset(-1)(NH_(2))-overset(+3)(Co)(NH_(3))_(2)(H_(2)O)_(3)]Cl_(5)`
Number of Co-N LINKAGE =5+1+1+2
Discussion
11.

Find no. of protons in 180 ml H_(2)O . Density of water =1 gm/ml.

Answer»

SOLUTION :MASS of WATER=density `xx` volume=180g
Moles of water `=(180)/(18)=10`
1 mol water has 10 mol PROTONS
10 mol water has 100 mol protons
10 mol water has 100 `N_(A)` protons
10 mol water has `6.023xx10^(25)` protons
Discussion
12.

Find moles of Cu atom and number of Cu atoms in it's 0.635 gm

Answer»


ANSWER : `(##ALN_NC_CHM_MC_E01_015_A01##)`
`0.1"MOLE",(0.01xxN_(A))`
Discussion
13.

Find moles of electrons present in 64 g ofCH_4

Answer»

64
40
24
16

Answer :B
Discussion
14.

Find moles of O-atoms in 5.6 litres of SO_(3) at 0^(@)C ,1 atm ?

Answer»


ANSWER :`(##ALN_NC_CHM_MC_E01_010_A01##)`
0.75 MOLES
Discussion
15.

Find (i) the total number of neutrons (ii) the total mass of neutrons in 7mg of ""^(14)C (Assume the mass of a neutron = mass of a hydrogen atom)

Answer»

Solution :(i) `24.08 XX 10^(20)` (ii) 3mg
Discussion
16.

Find (I) No. of moles of cu atom In 10^(20) atoms of cu. (ii) Mass of 200atoms in amu. (iii) Mass of 100 atoms ofin amu . (iv) No. of molecules & atoms in 54 gm H_(2)O. (v) No. of atoms in 88 gm CO_(2).

Answer»


Answer :`(i)(10^(20))/(N_(A))"MOLES"""(II)3200 "amu"""(iii)14xx1.66xx10^(-24)100g""(iv)3N_(a),9N_(9)""(v)6N_(A)`
Discussion
17.

Find Delta_(r)U^(@) for the reaction 4HCl (g) + O_(2) (g) rArr 2Cl_(2) (g) + 2H_(2)O (g) at 300 K. Assume all gases are ideal {:("Given",H_(2) (g) + Cl_(2) (g) rarr 2HCl (g),Delta_(r) H_(300)^(@) - 184.5 kJ//"mole",),(,2H_(2) (g) + O_(2) (g) rarr 2H_(2)O (g),Delta_(r) H_(300)^(@) = - 483 kJ//"mole" ("Use" R = 8.3 J//"mole"),):}

Answer»

111.5 KJ/mole
`-109.01 kJ//"mole"`
`-111.5 kJ//"mole"`
NONE

Solution :`4HCl+O_(2) SQUARE hArr 2Cl_(2)+2H_(2)O""DeltaH=-114`
`DeltaU=DeltaH-Deltan_(g)RT=-114+(1xx8.314xx10^(-3) xx300)`
`=-111.5" kJ mol"^(-1)`
Discussion
18.

Find D/L configuration in the following molecules.

Answer»


ANSWER :(I) D(II) (2D, 3D)(III) Equivalent fischer projection is `(##RES_CHM_SIM_E01_011_A01##)`and configuration L.
Discussion
19.

Find correct statement

Answer»

`Cl_(2)` can react with CaO but not with `NaHCO_(3)`
Swimming pools are disinfected by passing `Cl_(2)` gas into it
Pure HF attack glass, Fe, `SiO_(2)`
TREATMENT of `Cs_(2)` with excess of `Cl_(2)` gives `C C l_(4)`

Solution :(a) `CaO+ Cl_(2) rarr CaOCl_(2)` ( b) due to its BLEACHING property
`NaHCO_(3) + Cl_(2) rarr` No REACTON
Discussion
20.

Final the total number of species having two unparired electron from the following species, Fe^(2+),Cr,Cr^(3+),Ti^(2+),V^(3+)

Answer»

SOLUTION :`{:(FE^(+2),[Ar],3d^(5)),(Cr,[Ar],3d^(5)4S^(1)),(Cr^(+3),[Ar],ed^(3)),(Ti^(+2),[Ar],3d^(2)),(Mn^(+2),[Ar],3d^(5)),(V^(+3),[Ar],3d^(2)):}`
Discussion
21.

Find basic oxides from the following: (1) MnO_(2)O_(7) (2) V_(2)O_(3) (3) V_(2)O_(5) (4) CrO (5) Cr_(2)O_(3)

Answer»

1 and 2
2 and 3
3 and 4
2 and 4

Answer :D
Discussion
22.

Find A,B,C and D. Also write equations A to B and A to C.

Answer»


ANSWER :`A=` CONC. `H_(2)SO_(4)""B=Br_(2), C=NO_(2)^(+)`
`(##MOT_CON_JEE_CHE_C34_E03_006_A01##)`
Discussion
23.

Find (a) the total number and (b) the total mass of protons in 34 mg of NH_(3) at S.T.P. (Assume the mass of proton =1.6726xx10^(-27)kg)

Answer»

Solution :a) `" 1 mol of NH"_(3) =" 17 G NH"_(3)= 6.022xx 1023` molecules of`NH_(3) = (6.022xx 10^(23))XX (7 + 3)` protons `= 6.022 xx 10^(24)` protons
In 34 mg, i.e., 0.034 g NH3, protons `=(6.022xx10^(24))/(17)xx0.034=1.2044xx10^(22)`
(b)Mass of one proton `= 1.6726 xx 10^(-27) kg`
Mass of `1.2044xx 10^(22)` protons `= (1.6726 xx 10^(-27)) xx (1.2044 xx10^(22)) kg = 2.0145 xx 10^(-5) kg`
Discussion
24.

Final product will be

Answer»




ANSWER :C
Discussion
25.

Final product of following reaction is CO_(2) + (CH_(3))_(3) C - MgBr underset((ii) H_(3)O^(+)) overset((i) "dry ether") to A underset("Red P") overset(Br_(2)) to B

Answer»

`CH_(3) - UNDERSET(BR) underset(|) OVERSET(CH_(3)) overset(|) (C) - COOH`
`CH_(3) - underset(CH_(3)) underset(|)overset(CH_(3)) overset(|) (C) -underset(Br) underset(|) (CH) - COOH`
`CH_(3) - CH_(2) - underset(Br) underset(|) (CH) - COOH`
No PRODUCT will be formed .

Answer :D
Discussion
26.

Final productof methylationof ethyl amineand ethylmethylamineis

Answer»

ETHYL TRIMETHYLAMINE
triethyl methylamine
triethylmethylammoniunhalide
ethyl trimethylammonium HALIDE

ANSWER :D
Discussion
27.

Finalproduct of methylationof ethyl amineis

Answer»

TRI ETHYL methyl AMINE
ethyl tri methyl amine
TRIETHYL methyl ammonium HALIDE
ethyl trimethylammoniumhalide

Answer :D
Discussion
28.

Final product of the oxidation of hydrocarbon is :

Answer»

acid
alcohol
`CO_2+ H_2O`
aldehyde

Answer :C
Discussion
29.

Final product is :

Answer»

`CH_(3)-OVERSET(O)overset(||)C-CH_(3)`
`HCHO+CH_(3)-overset(O)overset(||)C-UNDERSET(CH_(3))underset(|)CH-CH_(3)`

`CH_(3) CHO+HCHO`

SOLUTION :N//A
Discussion
30.

Final product formed on reduction of glycerol by huydroiodic acid is

Answer»

Propane
Propanoic acid
propene
propyne

Solution :`{:(CH_(2)-OH""CH_(2)-I""CH_(2)""CH_(3)""CH_(3)),(|""|""||""||""|""),(CH-OHunderset(-3H_(2)O)OVERSET(3HI)to CH-Iunderset(-I_(2))to CHoverset(HI)to CH-I underset(-I_(2))to CH),(|""|""|""|""||),(CH_(2)-OH""CH_(2)-I ""CH_(2)-I""CH_(2)-I""CH_(2)):}`
Discussion
31.

Final product 'B' in the following sequence of reaction C_(6)H_(5) - CH_(2) OH underset(573k) overset(Cu) to A underset("strong") overset(NaOH) to B +C_(6)H_(5)CH_(2) - OH

Answer»

benzaldehyde
acetophenone
SODIUM phenoxide
sodium BENZOATE

ANSWER :D
Discussion
32.

Final product by the treatment of isobutyl alcohol with alumina is,

Answer»

2-methyl propene
2-methyl but-1-ene
ethyl t-butyl ether
acetone and ACETIC acid

Solution :`(CH_3)_2CHCH_OH UNDERSET"425 K"OVERSET(Al_2O_3)to (CH_3)_2C=CH_2+H_2O`
Discussion
33.

Final Product :

Answer»




ANSWER :D
Discussion
34.

Finalall the structuralisomers of C , H_(14)

Answer»

SOLUTION :`C- C- C-C-C-C , C-UNDERSET(C ) underset(|)(C ) -C - C-`C`
`C- underset(C ) underset(|)OVERSET(C ) overset(|) (C ) - C- C ,,,C-C- underset(C ) underset(|)(C ) - C- C`
`C- overset(C ) overset(| ) (C ) - overset(C ) overset(|) (C ) - C`
Discussion
35.

Final by- product during preparation of Dacron fibre is

Answer»

glycerol
ETHYLENE GLYCOL
ethyl alcohol
propylene glycol

ANSWER :B
Discussion
36.

Final hydrolysis product of simple protein is

Answer»

Carboxylic acid
`alpha`-AMINO acid
Mineral acid
Acetic acid

Answer :A::B::C::D
Discussion
37.

filling of electrons in p subshell of nitrogen is on the basis of

Answer»

HUND's rule
Heisenberg UNCERTAINTY principle
Paull's EXCLUSION principle
Aufbau's principle

Answer :A
Discussion
38.

fill up the gap : 2mole of nitrogen occupies _____ ml at STP.

Answer»


ANSWER :44800
Discussion
39.

Fill up the blanks with suitablereagents to show synthesis of polyvinyl chloride . CH-=CH overset(X)toCH_(2)-=CHCl overset(Y)to(--CH_(2)-CJCl-overset(Cl)overset(|)(C)H--)_(n)

Answer»

`X=HCl,HgCl_(2)`,Y=Polymerisation,peroxide
`X=Cl_(2),FeCl_(3)`,Y=Polymerisation,heat
`X=HCl,CuCl`,`Y=H_(2)O,H^(+)`
`X=HCl,HgCl_(2)`, Y=Pt,HIGH PRESSURE

ANSWER :A
Discussion
40.

Fill the missing products in the following reactions. (i)2MnI_4^(-)+16 H^(+)+10I^(-)to2Mn^(2+)+8H_2O+"_________". (ii)2MnO_4^(-)+H-2O+I^(-)to2MnO_2+2OH^(-)+"__________" .

Answer»

(i) HI, (II) `I_2`
`(i) I_2, (ii) IO_3^(-)`
`(i)I_2,(ii) I_2`
`(i) IO_3^(-), (ii) I_2`

ANSWER :B
Discussion
41.

Fill in the reagents for the given conversion : CH_(3)COCl overset((X))to CH_(3)CHO overset((Y))to CH_(3)-overset(OH) overset(|)(CH)-CH_(2)CHO overset((Z))to CH_(3)CH=CHCHO

Answer»

`|{:(X,Y,Z),(Pd//BaSO_(4),"dil. NaOH","heat"):}|`
`|{:(X,Y,Z),("NaOH","Hydrolysis","heat"):}|`
`|{:(X,Y,Z),(I_(2)//NaOH,LiAlH_(4),H_(3)O^(+)):}|`
`|{:(X,Y,Z),(CrO_(3),"WARM",CO_(2)):}|`

Answer :A
Discussion
42.

Fill in the blanks with appropriate words : The electrolytic solution is always neutral because the total charge on ______ is equal to _____ on ____. Unlike the metallic conductor, the electrolyte conducts the electric current by virtue of movement of its _____. The property due to which a metal tends to go into solution in term of positive ions is known as _______. (i),(ii),(iii),(iv) and (v) respectively are

Answer»

cations, PARTIAL CHARGE ,anions,ELECTRONS,reduction
cations,TOTAL charge,anions,ions,oxidation
cations,ionic charge,anions,atoms,dissolution
cations,partial charge,anions,MOLECULES,electrolysis

Answer :B
Discussion
43.

Fill in the blanks with appropriate words. The electrolytic solution is always neutral because the total charge on __(i)___ is equal to __(ii)__ on ___(iii)___ .Unlike the metallic conductor, the electrolyte conducts the electric current by virtue of movement of its ___(iv)___ . The property due to which a metal tends to go into solution in term of positive ions is known as ___(v)__. (i),(ii),(iii),(iv) and (v ) respectively are

Answer»

CATIONS, partial charge , anions, ELECTRONS , reduction
cations , total charge , anions , ions , oxidation
cations , ionic charge , anions , ATOMS , dissolution
cations , partial charge , anions , molecules, ELECTROLYSIS .

Answer :B
Discussion
44.

During the transformation of ._(c )^(a)X of ._(d)^(b)Y the number of beta-particle emitted are:

Answer»


SOLUTION :`._(C)^(a)M rarr ._(d)^(B)N+x(._(2)He^(4))+y(._(-1)E^(0))`
`a=b+4x+0y`
`(a-b)/(4)=x "" ..........(1)`
`c=d+2x-y`
`c-d-(a-b)/(2)=-y`
`y=d+((a-b))/(2)=c`
Discussion
45.

The number of alphaand beta- particles emitted, when the following nuclear transformation takes place are _______ and __________ respectively.{:(238),(92):} X rarr{:(206),(82):} Y

Answer»


SOLUTION :`._(92)^(238)X rarr. _(82)^(206)Y + x alpha + y alpha`
`._(92)^(238)X rarr ._(82)^(206) Y + x(._(2)\He^(4))+y(._(-1)^(0)e)`
COMPARING mass no.
`238 rArr 206 + 4X +0y`
`x=8 i.e. 8 (alpha) `particles
Comparing no. protons.
92=82 + 2x-y
`10=2 xx 8 -y`
`y=6 i.e. 6 (beta)` particle
Discussion
46.

The half life of the nuclide Rn^(220) is 54.5 sec. What mass of radon is equivalent to 1 millicurie.

Answer»


SOLUTION :`A=Nlambda`
` 1xx10^(-3) xx3.7 xx10^(10)` d.p.s `=Nxx(0.693)/(54.5) "" ` 1curie`=3.7 xx 10^(10)` d.p.s.
`N=290.98 xx10^(7) rArr (N)/(N_(A))=(29.98 xx10^(7))/(6.023 xx10^(23))=48.31 xx 10^(-16)` moles
MASS of Rn`=220 xx48.31 xx 10^(-16)`
`rArr 10628.56 xx 10^(-16)` gm
`rArr 1.062 xx 10^(-12)gm rArr (1.06 xx 10^(-12))/(1000)KG rArr 1.06 xx 10^(-15) kg`
Discussion
47.

The nuclides with same difference of number of neutrons and number of protons are called ___________.

Answer»


ANSWER :ISODIAPHERS
Discussion
48.

Fill in the blanks : The noble gases can form compounds with ____A _____ and _____B____. The mixture of ____C____ and __D____ is used for respiration by divers.

Answer»

A-Iodine , B-Oxygen , C-Oxygen , D-Argon
A-Fluorine , B-Oxygen , C-Helium ,D-Oxygen
A-Xenon , B-Platinum ,C-Argon, D-Krypton
A-Helium ,B-Oxygen , C-Xenon , D-Argon

SOLUTION :NOBLE GASES FORMS COMPOUNDS only with `F_2` and `O_2`
Discussion
49.

Fill in the blanks in the following table which treats a reaction of a compound A with a compound B, that is the first order with respect to A and zero order with respect to B.

Answer»


Solution :The rate MAY be expressed as:
Rate = `k[A][B]^(@)=k[A]`
For expt. I: `(2.0 xx 10^(-2) MOL L^(-1))min^(-1) = k[A]=k(0.1 mol L^(-1))`
`k=([2.0 xx 10^(-2)mol L^(-1)]min^(-1))/(0.1 MOLL^(-1)) = 2.0 xx 10^(-1) min^(-1)`
For expt. II: Rate `=k[A]`
`therefore 4.2 xx 10^(-2) mol L^(-1) min^(-1)=2.0 xx 10^(-1) min^(-1)xx [A]`
`therefore [A] = ([4.0 xx 10^(-2) mol L^(-1)]min^(-1))/(2.0 xx 10^(-1) min^(-1))=0.2 mol L^(-1)`
For expt. III: Rate =k[A] `=(2.0 xx 10^(-1) min^(-1) xx (0.4 mol L^(-1))=8.0 xx 10^(-2) mol L^(-1) min^(-1)`
For expt. IV: Rate =k[A]
`(2.0 xx 10^(-2) mol L^(-1) min^(-1))=2.0 xx 10^(-1) min^(-1) xx [A]`
`[A]=(2.0 xx 10^(-2)molL^(-1)min^(-1))/(2.0 xx 10^(-1) min^(-1)) = 0.1 mol L^(-1)`
Discussion
50.

Fill in the blanks. In Hell-Volhard-Zelinsky reaction, the carboxylic acids are halogenated at ______ position by using _______ and ______.

Answer»

`alpha`, NaOH, IODINE
`alpha`, PHOSPHORUS, halogen
`BETA`, phosphorus, `H_(2)O`
`beta, PCl_(5),NaOH`

Answer :B
Discussion