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.

The number ofisomerspossiblefor disubstitutedborazine, B_(3)N_(3)H_(4)X_(2) is

Answer»

3
4
5
2

Solution :FOUR ISOMERS are possible.These are

2.

The total number of isomers (containing benzene ring) of molecular formula C_(7)H_(8)O is :

Answer»

2
3
4
5

Answer :D
3.

The number of isomers of the aromatic compound C_(8)H_(10) are :

Answer»

3
4
2
5

Solution :
4.

The number of isomers of octanc with five carbons in their principal chain is

Answer»


SOLUTION :CHAIN ISOMERISM
5.

The number of isomers of C_(6)H_(14) is (MLNR)

Answer»

4
5
6
7

Answer :B
6.

The number of isomers including stereoisomers possible for the compound having molecular formula C_4 H_8 is

Answer»

one
two
three
four

Answer :D
7.

The number of isomers of C_(6)H_(14) are……..

Answer»

4
5
6
7

Answer :B
8.

The number of isomers for the compound with molecular formula C_2BrClFI is

Answer»

3
4
5
6

Solution :SIX ISOMERS are :
9.

The number of isomeric primary amines possible for the formula C_(3)H_(9)N

Answer»

2
3
5
6

Solution :
10.

The number of isomeric pentyl alcohols possible are

Answer»

Two
Four
Six
Eight

Answer :C
11.

The number of isomeric compounds having a benzene ring and molecular formula C_7 H_9 N are_____

Answer»


SOLUTION :
12.

The number of isomeric alkyl group possible for C_4H_5- is

Answer»

FOUR
Three
Two
Five

SOLUTION :Four.
13.

The number of ions present in 1 ml of 0.1M CaCl_(2) solution is

Answer»

`1.8 XX 10^(20)`
`6.0 xx 10^(20)`
`1.8 xx 10^(19)`
`1.8 xx 10^(21)`

Answer :A
14.

The number of ion pairs that constitute one unit cell of CsCI

Answer»

4
2
8
1

Answer :D
15.

The number of intensivepropertiesamong the following isTemperature, Pressure, Volume , Heatcapacity ,Density ,pH of a solution , EMF of a cell,Entropy, Free energy,Enthalpy , Surface busion, Viscosity , Boiling point

Answer»


ANSWER :8
16.

The number of hyperconjugable hydrogen atoms of followingspecies are respectively :

Answer»

3, 5, 9, 8
3, 5, 9, 5
5, 5, 3, 5
5, 2, 6, 5

Answer :B
17.

The number of hydroxide ions produced by one molecule of Na_(2)CO_(3) on hydrolysis

Answer»

4
2
3
0

Answer :B
18.

The number of hydrogen bonds that can be formed by each borax acid molecule.

Answer»


Solution :each OXYGEN forms 2 hydrogen BONDS. TOTAL hydrogen bonds .6..
19.

The number of hydrogen - boded and coordinated water molecule present in hydrated copper sulphate salt are respectively

Answer»

1, 4
2, 3
4, 1
0, 5

Answer :A
20.

The number of hydrogen atoms present in 25.6 g of sucrose (C_(12)H_(22)O_(11))which has a molar mass of 342.3 g is (N_(A)=6.023xx10^(23))

Answer»

`22xx10^(23)`
`9.91xx10^(23)`
`11xx10^(23)`
`44xx10^(23)` H atoms

Solution :No. of moles of sucrose `=(25.6)/(342.3)=0.0748~~0.075`
1 MOLE sucrose CONTAINS `=22xx6.023xx10^(23)` of H-ATOM
0.075 moles of sucrose contain `= 0.075xx6.023xx10^(23)xx22`
`=9.91xx10^(23)` of H -atom
21.

The number of P-OH bonds present in pyrophosphoric acid and hypophosphoric acid is respectively.

Answer»

120
24
36
12

Answer :D
22.

The number of hybrid orbitals involved in the formation of B_(2)H_(6), B_(3)N_(3)H_(6), BCl_(3),H_(3)BO_(3) are p, q, r & s, then the sum of (p+q+r+s) is 8y, then y = ?

Answer»


Solution :`B_(2)H_(6)to 8 ["each "B-4sp^(3)]`
`B_(3)N_(3)H_(6)to 18 ["each "B, N-3sp^(2)]B_(3)N_(3)H_(6)rarr 18 ["each "B, N-3sp^(2)]`
`BCl_(3)to 3 ["each "B-3sp^(2)]`
`H_(3)BO_(3)to 3["each "B-3sp^(2)]`
`P+Q+R+S rArr 8+18+3+3=32(8y)`
`therefore y=4`.
23.

The number of hexagonal faces that are present in a truncated octahedron is

Answer»


Solution :TRUNCATED octahedra ( cubooctahedra) are the structures shown by zeolites. A truncated octahedron is an Archimedean solid. It has 14 FACES ( 8 regular hexagons and 6 square) , 36edges and 24 vertices.
24.

The number of water molecule(s) derectly bonded to the metal centre in CuSO_4.5H_2O is

Answer»

`H_3PO_2`
`H_(3)PO_(2)`
`H_(3)PO_(4)`
`PH_(3)`

ANSWER :D
25.

The number of grams of oxygen in 0.10 mol of Na_(2)CO_(3) .10H_(2)O is

Answer»


Solution :`Na_(2)CO_(3).10H_(2)O`=1 mole
1 mole of `Na_(2)CO_(3) .10H_(2)O` CONTAINS 13 OXYGEN ATOMS.
Mass of 13 oxygen atoms = `13 xx 16 = 208`
1 mole of `Na_(2)CO_(3) .10H_(2)O` contains 208 G of oxygen.
`:.` 0.10 mole of `Na_(2)CO_(3).10H_(2)O` contains `208/1xx0.1=20.8g`
26.

The number of grams of H_2SO_4 required to dissolve 5 g of CaCO_(3) is:

Answer»

10.24
4.9
5.12
2.56

Answer :B
27.

The number of gram - atoms of sulphur in 2 moles of peroxydisulphuric acid is

Answer»

2
3
1
4

Answer :D
28.

The number of gram atoms of hydrogen present in 1.5 mole of hydrogen sulphide is

Answer»

3
2
1
0.5

Answer :A
29.

The number of Glucose molecules present in 10 ml of decimolar solution is

Answer»

`6.0 XX 10^(20)`
`6.0 xx 10^(19)`
`6.0 xx 10^(21)`
`6.0 xx 10^(22)`

ANSWER :A
30.

The number of isomers (geomctrical and optical) possible for the compound with the structure CH_3 CH=CH-CH=CH-CH_2 CHOHCH_3 is

Answer»

2
4
6
8

Answer :B
31.

The number of geometrical isomers of CH_3CH=CH-CH-CH=CH-Cl

Answer»

2
4
6
8

Solution :An organic COMPOUND has .n. number of non-equivalent double BONDS with different groups are present then the total number of geometrical ISOMERS ` = 2^n "" n = 3 rArr 2^3 = 8`
32.

The total number of gaseous elements are

Answer»


11 
12 
15 

Answer :B
33.

The number of esters possible for the molecular formula C_5H_10O_2is ____

Answer»


SOLUTION :
34.

The number of equivalents of Na_(2)S_(2)O_(3) required for the volumetric estimation of one equivalent of Cu^(2+) is :

Answer»

`1//3`
`1`
`3//2`
`2//3`

SOLUTION :NUMBER of EQUIVALENTS of REACTING species in a chemical reaction are same.
35.

The number of enantiomers of the compound CH_(3)CHBrCHBrCOOH is

Answer»

0
1
3
4

Solution :
NUMBER of ENANTIOMERS ` = 2^n` n = number of CHIRAL centres `n = 2 rArr 2^2 = 4`
36.

The number ofelements present in 2nd, 3rd, 4th and 5th IJeriods of modern periodic table respectively are

Answer»

2, 8, 8 & 18
8, 8, 18 & 32
8, 8, 18 & 18
 8, 18, 18 & 32 

ANSWER :C
37.

The number of elliptical orbits excluding circular orbits in the N-shell of an atom is:

Answer»

3
4
2
1

Solution :For, N-shell, n=4. this shell will have ONE CIRCULAR and three ELLIPTICAL ORBITS.
38.

The number of elements in the first period is only 2. Give reason.

Answer»

Solution :The first period n=1 ,it is the first energy level and has only 1 s orbital which can accommodate a maximum of 2 electrons HENCE the NUMBER of ELEMENTS in the first period is only 2.
39.

The number of elements classified by Newland was .......

Answer»

72
61
66
56

Solution :The English chemist, John Newland PRESENTED a RESEARCH PAPER in 1863, in which he classified 56 ELEMENT in 11 groups based on their physical properties.
40.

The number of electrons in M-shell of an element with atomic number 24 is

Answer»


Solution :CONFIGURATION (Z=33) `= 1s^2 2s^2 2p^6 3s^2 3p^64s^2 3d^(10) 4p^3`
Each subshell has one ORBITAL with m = 0
`therefore ` No. of electrons `=2xx 7 +1 = 15`
41.

The number of electrons to balance the equation NO_(3)^(-)+4H^(+)+e^(-)to2H_(2)O+NO

Answer»

5
4
3
2

Answer :C
42.

The number of electrons that are involved in the reduction of permanganate to managanes (II) salt managanate and managanese dioxide respecitvely are

Answer»

5,1,3
5,3,1
2,7,1
5,2,3

Solution :`overset(+7)MnO_(4)^(-)+e^(-)RARROVERSET(2+)Mn`
`overset(+7)MnO_(4)^(-)+e^(-)rarroverset(+6)MnO_(4)^(2-)`
`overset(+7)MnO_(4)^(-)+3e^(-)rarroverset(+4)MnO_(2)`
Thus option (a) is CORRECT
43.

The number of electrons shared between the two Boron atoms directly in the formation of bonds in diborane molecule

Answer»

4
2
0
8

Answer :C
44.

The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 respectively. Assign the proper symbol to the species.

Answer»

Solution :ELECTRONS = 18, PROTONS = 16, Neutrons = 16
ATOMIC number = No. of protons = 16
Mass number = No. of prtons + No. of neutrons `= 16 + 16 = 32`
As the species has two electrons more than protons, it has two UNITS of -ve charge. Hence, the species is `._(16)^(32)S^(2-)`
45.

The number of electrons, protons and neutrons in a species are equal to 18,16 and 16 respectively. Assign the proper symbol to the species.

Answer»

SOLUTION :ATOMIC number = 16. HENCE, the element is sulphur (S). Mass number = 16+16 = 32 CHARGE = 16 - 18 = -2 THEREFORE, the symbol is `""_16^33""s^(2-)`
46.

Thenumberof electronsprotons and neutronsin aspeciesare equalto 18,16 and16 respectively . Assigh the proper symbolto the species.

Answer»

Solution :Numberof electron=18
Atomicnumber= numberof protons( z) = 16
Numberofneutron (n ) in sulphur = 16
A= massnumber= z+ n
Numberof electron(18) `lt`numberproton(16)
(18-16)=12electronisexcess
chargeis -2so SYMBOL of NEGATIVE ION : `s^(2)`
Symbolof species`._(16) ^(18) S^(2)`
47.

The number of electrons, protons and neutrons in a species are equal to 18, 16 and 16 res­pectively .Assign the proper symbol to the species.

Answer»

Solution :The atomic number is EQUAL to number of protons = 16. The ELEMENT is sulphur (S). Mass number = number of protons + number of neutrons = 16+ 16 = 32 Species is not NEUTRAL as the number of protons is not equal to electrons It is anion, (negatively charged) with charge equal to excess electrons =18-16 = 2. SYMBOL is `""_16^32""s^(-2)`
48.

The number of electrons per second which pass through a cross section of a copper wire carrying 10^(-16) A is

Answer»

`1.6xx10^(-3)` e/s
60 e/s
625 e/s
`16xx10^(-2)` e/s

Solution :Since 1A=1 C/s
CHARGE on one electron`=1.6xx10^(-19) C`
Rate `=(10^(-16)C//s)/(1.6xx10^(-19))=625 C//s`
49.

The number of electrons lost or gained during the change, Fe + H_(2)O rarr Fe_(3)O_4 + H_2

Answer»

2
4
6
8

Solution :`3overset(0)(FE)RARROVERSET(+8)(Fe_3O_4)`
50.

The number of electrons lost during the oxidation of Cr^(3+) " to " Cr_(2)O_(7)^(2-)are

Answer»


SOLUTION :`7H_2O+2Cr^(3+)rarrCr_(2)O_(7)^(2-)+14H^(+)+6e^(-)`
1 MOL of `CR^(3+)`will LOOSE 3 mol electrons while oxidised to `Cr_(2)O^(2-)`