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.

(a)Which chemical substances is present as enamel on the surface of the teeth ? (b)Traces of fluoride ions (F^-) in drinking water (about 1 ppm) greatly reduce the incidence dental cavities (tooth decay ) . What is the reason for reduction in cavities ?

Answer»

Solution :(a)The chemical substance present as enamel on the surface of the TEETH is `3 Ca_3(PO_4)_2 . Ca(OH)_2`
(b)The fluoride ions `(F^-)` convert to enamel to much harder `3Ca_3(PO_4)_2. CaF_2` . Hence, the INCIDENCE cavities is GREATLY reduced.
Discussion
2.

Avogadro’s number is the number of molecules present in

Answer»


ANSWER :A::C
Discussion
3.

Avogadro's lasw states that under conditions of constant temp. and pressure equal volume of gases contain equal no. of particles. Experimental investigation shows that at one atmosphere pressure and a temperature of 273 k, one mole of any gas occupies a volume of which is very close to 22.4 lit. Therefore, the number of moles in any gas sample canbe found by comparing its volume at STP with 22.4 lit. Number of gram atoms of oxygen present in 0.3 gram mole of H_(2)C_(2)O_(4).2H_(2)O is

Answer»

`0.3`
`0.6`
`1.2`
`1.8`

Solution :NUMBER of oxygen atoms in MOLECULE of `H_(2)C_(2)O_(4).2H_(2)O=6`
1 moles of `H_(2)C_(2)O_(4).2H_(2)O` contains `6xx16=96g` oxygen
0.3 moles of `H_(2)C_(2)O_(4).2H_(2)O` contains `(0.3xx96)/(1)` = 28.8 g oxygen
`therefore` No. of GRAM atom of oxygen = `(28.8)/(16)=1.8`
Discussion
4.

Avogadro's lasw states that under conditions of constant temp. and pressure equal volume of gases contain equal no. of particles. Experimental investigation shows that at one atmosphere pressure and a temperature of 273 k, one mole of any gas occupies a volume of which is very close to 22.4 lit. Therefore, the number of moles in any gas sample canbe found by comparing its volume at STP with 22.4 lit. At STP 40 lit of CO_(2) contains

Answer»

5.6 moles
1.786 MOLE
7.635 mole
1.934 mole

Solution :22.4 LITRES `CO_(2)` - STP - 44g - 1, mole `CO_(2)`
40 litres `CO_(2)` - STP - `(40xx1)/(22.4)=1.786` moles
Discussion
5.

Avogadro's lasw states that under conditions of constant temp. and pressure equal volume of gases contain equal no. of particles. Experimental investigation shows that at one atmosphere pressure and a temperature of 273 k, one mole of any gas occupies a volume of which is very close to 22.4 lit. Therefore, the number of moles in any gas sample canbe found by comparing its volume at STP with 22.4 lit. If avagadro's number is 6xx10^(23) molecules then the mass of one atom of oxygen would be

Answer»

`(16)/(6.02)` amu
`6xx10^(-23)` amu
16 amu
`16xx6.02` amu

Solution :1 Mole of oxygen `-6.023xx10^(23)` molecules of oxygen
`2xx6.023xx10^(23)` ATOMS of oxygen MOLECULAR weight of oxygen = 32
`2xx6.023xx10^(23)` atom of oxygen - 32g
1atom of oxygen = `(1xx32)/(2xx6.023xx10^(23))g`
`=16xx1.66xx10^(-24)g=16` amu
Discussion
6.

Avogadro number of helium atoms have a mass of

Answer»

2g
4g
8g
`4xx6.02xx10^(23)g`

Solution :No ATOMS = 1 mole = 4g
Discussion
7.

Avogadro number is

Answer»

The NUMBER of ATOMS in ONE gram-atomic- weight
The number of molecules in one gram:The number of atoms in 0.012 kg of C-12
all of these

Answer :D
Discussion
8.

Average oxidation state of Osmium (Os) in OsO_(4) is _____

Answer»

`+7`
`+6`
`+4`
`+8`

SOLUTION :`OSO_(4)rArrOs+4(O)=0`
`thereforeOs+4(-2)=0`
`thereforeOs=+8`
Discussion
9.

Average oxidation number of iodine in KI_(3)

Answer»

`+1//3`
`-1//3`
`+3`
`-1`

ANSWER :B
Discussion
10.

Average oxidation state of sulphur atoms in the thiosulphate ion

Answer»

5,0
`+4,0`
`+3.5`
`-2, +6`

ANSWER :D
Discussion
11.

Average kinetic energy of a gas does not depend on_______at a given temperature

Answer»

PRESSURE 
volume 
NATURE of gas 
density 

SOLUTION :KE depneds onlt on T.
Discussion
12.

Average C-H energy is 416 KJ.mol^(-1) Which of the following is correct ?

Answer»

`CH_(4(g)) + 416KJtoC_((g)) +4H_((g))`
`CH_(4(g))toC_((g)) +4H_((g)) +416KJ`
`CH_(4(g)) +1664KJtoC_((g))+4 H_((g))`
`CH_(4(g))toC_((g)) +4H_((g)) +1664 KJ`

Answer :C
Discussion
13.

Average atomic mass of Ca and Ba is same as atomic mass of ..........

Answer»

Na <BR>Br
Sr
Mg

Solution :`Ca : A = 40, Ba : A = 137`
` :. (40 + 137)/2 = (177)/2 = 88.5`
ATOMIC MASS of Sr is `88.5`
Discussion
14.

Autoprotolysisconstant of NH_(3) is

Answer»

`[NH_(4)^(+)][NH_(3)]`
`[NH_(2)^(-)][NH_(3)]`
`[NH_(4)^(+)][NH_(2)^(-)]`
`[NH_(2)^(-)]//[NH_(2)^(-)]`

Solution :`NH_(3)+NH_(3) hArr NH_(4)^(+)+NH_(2)^(-)`
Hence, `K=([NH_(4)^(+)][NH_(2)^(-)])/([NH_(3)]^(2))`
`:. [NH_(4)^(+)][NH_(2)^(-)]=K[NH_(3)]^(2)`
= AUTOPROTOLYSISCONSTANT.
Discussion
15.

(A)underset(triangle)overset(Na[Cr(OH_(4))]//H_(2)O_(2))rarrunderset(B)overset(H_(2)SO_(4))rarrunderset("orange colour")(C )rarr (A) is

Answer»

`CrCl_(3)`
`Na_(2)CrO_(4)`
`Na_(2)Cr_(2)O_(7)`
All of these

Solution :`2NA[CR(OH)_+(4)+2NaOH+3H_(2)rarr2Na_(2)CrO_(4)+8H_(2)O`
Discussion
16.

Aufbau principle is not satisfied by

Answer»

CR and Cl
Cu and Ag
Cr and Mg
Cu and Na

Answer :B
Discussion
17.

Attractive forces between non-polar, real gas molecules are:

Answer»

short-range interactions
inversely proportional to the sixth POWER of the DISTANCE between TWO interacting particles
Important only at short distance (-500pm)
all of the above

Answer :a
Discussion
18.

Attainment of “equilibrium state” withthe helpof “constancyinintensityof colour” is noticedinthe case of ________ in a closed vessel

Answer»

DECOMPOSITION of `CaCO_3`
REACTION between `N_2` & `O_2`
Reaction between `H_2` & `I_2`
Decomposition of `PCI_5`

SOLUTION :`I_(2)` has violet COLOUR
Discussion
19.

Attainment of equilibrium can be noticed with the help of constancy of which of the following physical properties ?

Answer»

<P>INTENSITY of COLOUR
Density
Pressure
All the above

Solution :Equilibrium can be observed with CONSTANCY of P,d or intensity of colour
Discussion
20.

Atoms or group of atoms having more electrons attracting power than that of hydrogen is known as……

Answer»

`+I` EFFECT
`-I` effect
`+E` effect
`-E`effect

Answer :B
Discussion
21.

Atoms of the following group possess the lowest ionisation energies

Answer»

IA
IIA
VA
ZERO

ANSWER :A
Discussion
22.

Atomsof theelementsbelonging tothe group of periodic tablewill have

Answer»

SAMENUMBER ofprotons
samenumberof ELECTRONS in thevalenceshell
samenumberof NEUTRONS
samenumberof electrons

Solution :NA
Discussion
23.

Atoms of the following group possess the highest ionisaiton energies

Answer»

IA
IIA
VA
ZERO

ANSWER :D
Discussion
24.

Atoms of element B form hcp lattice and those of the element A occupy 2/3rd of tetrahedral voids. What is the formula of the compound formed by these elements A and B ?

Answer»

Solution :Suppose NUMBER of atoms B in hcp lattice =n
As number of tetrahedral voids is double the number of atoms in the CLOSE packing , therefore, number of tetrahedral voids =2n
As atoms A occupy 2/3rd of the tetrahedral voids, therefore, number of atoms A in the lattice =`2/3xx2n="4n"/3`
`therefore` Ratio of A:B=`"4n"/3:n=4/3:1=4:3`
HENCE, the FORMULA of the compound `A_4B_3`
Discussion
25.

Atoms of element B form hcp lattic and those of the element A occupy 2/3 of tetrahedral voids. What is the formuala fo the compound formed by these elements A and B?

Answer»

Solution :SUPPOSE number of atoms Bin hcp lattic = n
As number of tetrahedral voids is double the number of atoms in the close packing . Therefore, number of tetrachedral voids = 2n.
As atoms A occupy 2/3rd of the tetrahedral voids, therefore, number of atoms A in the lattic `= 2/3 xx 2n = ( 4n) /3`
Ratio of A : B = `(4n)/2 : n = 4/3 : 1= 4:3 `
HENCE, the FORMULA of the compound ` A_(4)B_(3) `
Discussion
26.

Atoms are too small with diameter of 10^-10 m and weigh approximately

Answer»

`10^(-27) KG`
`10^-27 g`
`10^-31 kg`
`10^-11g`

ANSWER :A::B
Discussion
27.

Atoms are complicated than hydrogen have more than one proton in their nucleus. Let z stands for the number of protons in a nucleus. Also imagine that an atom loses all but one of its electrons so that it changes into a positively charged ion with just one electron. Bohr.s formula for the energy levels of the hydrogen atom can easily be changed to apply to such ions What minimum amount of energy is required to bring an electron from ground state of Be^(3+) to infinity

Answer»

`4.358 xx 10^(-18)`
`2.179 xx 10^(-18)`
`3.4864 xx 10^(-17)`
`8.716 xx 10^(-18)`

ANSWER :C
Discussion
28.

Atoms more complicated than hydrogen have more than one proton in their nucleus. Let Z stands for the number of protons in a nucleus. Also imagine that an atom loses all but one of its electrons so that it changes into a positively charged ion with just one electron. Bohr.s formula for the energy levels of the hydrogen atom can easily be changed to apply to such ions. It becomes E_m = (-Z^2 e^4 m_e)/(8 epsi_0^2 h^2 n^2), where m= mass of electron , e= change of electron, n = orbit number. The ionization potential of Het in ground state is

Answer»

`2427 KJ MOL^(-1)`
`5249 kJ mol^(-1)`
`7116 kJ mol^(-1)`
`9811 kJ mol^(-1)`

ANSWER :A
Discussion
29.

Atoms are complicated than hydrogen have more than one proton in their nucleus. Let z stands for the number of protons in a nucleus. Also imagine that an atom loses all but one of its electrons so that it changes into a positively charged ion with just one electron. Bohr.s formula for the energy levels of the hydrogen atom can easily be changed to apply to such ions The potential energy of electron in the ground state of He^(+) ion is

Answer»

`4.358 XX 1^(-18)` J/atom
`-7.112 xx 1^(-18)` J/atom
`-1.743 xx 1^(-17)` J/atom
`-8.716 xx 1^(-18)` J/atom

Answer :C
Discussion
30.

Atomic weight of Eka Al predicted by Mendeleef's is' X' and its experimental value is 'Y' then correct order is

Answer»

`X GT Y`
`X LT Y`
`X=Y`
`X=Y/2`

ANSWER :B
Discussion
31.

Atomic weight of an element X is 39, and that of element Z is 132, atomic weight of their intermediate element Y, as per Doeberiner triad, will be

Answer»

`88.5`
`93.0`
171
`85.5`

SOLUTION :`(39+132)/(2)=85.5`
Discussion
32.

Atomic radius of Li is 1.23Å and ionic radius of Li^(+) is 0.76Å. The percentage of volume occupied by 2s electron is

Answer»

`50%`
`60%`
`76%`
`94%`

Answer :C
Discussion
33.

Atomicradiusof hydrogenatom H in Bohr.smodelis

Answer»

`0.529 xx 10^(7) CM`
`0.529 xx 1 0^(9) cm`
`0.529 xx 10^(8) cm`
`0.529 xx 10^(10) cm`

ANSWER :A
Discussion
34.

Atomic radius is measured by

Answer»

Mulliken oil drop method
Rutherford's a ray scattering EXPERIMENT
X-ray diffraction tenchnique
Electric discharge TUBE experiment

Answer :C
Discussion
35.

The atomic radius decreases in a period due to

Answer»

INCREASE in NUCLEAR ATTRACTION 
DECREASE in nuclear attraction 
Increase in number of electrons 
Decrease in number of electrons

Answer :A
Discussion
36.

Atomic radius depends upon (i) Number of bonds formed by the atom (ii) Nature of the bonding (iii) Oxidation state of the atom

Answer»

NUMBER of bonds formed by the ATOM
Nature of the bonding
Oxidation STATE of the atom
All the above

Answer :D
Discussion
37.

Atomic radius 34. Atomic radii of fluorine and neon in angstrom units are respectively

Answer»

0.72,1.62
0.72.0.72
1.2 1.2
1.62,0.72

Answer :A
Discussion
38.

Atomic radii of fluorine and neon (Å) respectively are given as

Answer»

0.72, 1.62 
0.72, 0.72 
1.2, 1.2
1.62, 0.72 

ANSWER :A
Discussion
39.

Atomic radii of fluorine and neon (in angstrom units) respectively are given as…

Answer»

`0.72, 1.60`
`1.60, 1.60`
`0.72, 0.72`
NONE of these

ANSWER :d
Discussion
40.

Atomic radii of alkaline earth metals are smaller than the corresponding members of alkali metals. Why?

Answer»

Solution :The atomic RADII of alkaline earth metals are smaller than alkali metals. This is due to the fact that group 2 elements have a higher NUCLEAR charge, allowing the electrons to move TOWARDS the nucleus. This reduces the size of atomic and IONIC radii.
Discussion
41.

Atomic orbitals are precisely distinguished by what are known as quantum numbers. The number of unpaired electrons, present in Ni is_____ (Atomic number of Ni = 28)a) 2b) 0c) 1d) 3

Answer»

2
0
1
3

Answer :B
Discussion
42.

Atomic numbers of actinides are

Answer»

57 to 71
80 to 103 
58 to 71
90 to 103 

ANSWER :D
Discussion
43.

Atomic number of three elements P, Q and R are 20, 38 and 56 elements respectively. Which option is correct?

Answer»

Its IONISATION ENTHALPY is INCREASING ORDER
Q has maximum ionisation enthalpy
R has maximum ionisation enthalpy
Ionisation enthalpy of Q is in middlebetween P and R.

Answer :d
Discussion
44.

Atomic number of the centrtal atom in MCl_2 is 50. The shape of gaseous MCl_2is given as

Answer»

`CL -UNDERSET(* * ) OVERSET( )M-Cl`
`Cl -overset(*-+- ) overset( * *)M-Cl`
`Cl -underset(* * ) overset( * *)M-Cl`
`Cl -( ) overset( * *)M-Cl`

ANSWER :D
Discussion
45.

Atomic number of the central atom in MCl_(2) is 50. The shape of gaseous MCl_(2) is given as

Answer»




ANSWER :D
Discussion
46.

Atomic number of nitrogen is 7. The atomic number of the third member in the same family is

Answer»

23
15
33
51

Answer :C
Discussion
47.

Atomic number of elementsX,Y,Z and A are 4,8,7 and 12 respectivity . Arrange them in the decreasing order of their electronegativity.

Answer»

SOLUTION :`YgtZgtXgtA`
Discussion
48.

Atomic mass of mercury is 200 and density is 13.6 g cc^(-1) . How many moles of metal are present in one litre ?

Answer»

Solution :weight of HG `= "VOLUME "xx"density" = 1000xx13.6=13600g`
Number of MOLES `("weigt")/("GAW")=(13600)/(200)=68`
ONE litre of mercury contains 68 moles
Discussion
49.

Atomic hydrogen is allowed to combine on the surface to be welded to generate the temperature of

Answer»

400K
3000K
4600K
4000K

Answer :D
Discussion
50.

Atomic hydrogen combines with atmost all elements but molecular hydrogen does not. Explain.

Answer»

Solution :Atomic hydrogen is HIGHLY UNSTABLE and hence it is very reactive. Therefore, it combines with ALMOST all the elements. In contrast, the BOND dissociation energy of H-H bond is very high (`435.88 KJ "mol"^(-1)`) . As a result , molecular hydrogen is almost insert at room temperature and hence REACTS only with a few elements.
Discussion