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.

If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element, are in whole number ratio. (a) Is this statement true ? (b) If yes, according to which law ? (C) Give one example related to this law.

Answer»

Solution :(a) Yes, the given statement is TRUE.
(b) According to the law of multiple proportions
(c) `{:(H_(2),+,O,rarr,H_(2)O,,,H_(2),+,O_(2),rarr,H_(2)O_(2)),(2g,,16g,,18g,,,2g,,32G,,34g):}`
Here, the masses of OXYGEN which combine with fixed MASS of hydrogen are in the simple ratio i.e., `16:32` or `1:2`.
Discussion
2.

If two compounds have same empirical formula but different molecular formula then.....

Answer»

percentage PROPORTION of elements is DIFF.
MOLECULAR MASS is different.
density is same
vapour density is same.

Answer :A::B::C::D
Discussion
3.

If travelling at same speeds, which of the following matter waves have the shortest wave­length ?

Answer»

Electron
Alpha particle `(He^(2+))`
Neutron
Proton

Solution :According to de-Broglie equation, wavelength, `lambda=(h)/(MV)`For same speed of different PARTICLES, i.e., electron, proton, neutron and `alpha` particle `lambda` particle `lambda prop (1)/(m)`
h is CONSTANT. So, GREATER the mass of matter waves, lesser is wavelength and vice-versa. In these matter waves, alpha particle `(He^(2+))` has higher mass, therefore, shortest wavelength.
Discussion
4.

If travelling at same speeds, which of the following matter waves have the shortest wavelength?

Answer»

Electron
Alpha PARTICLE `(He^(2+))`
Neutron
Proton

Solution :`lamda = (h)/(MV)`. For same value of v, larger the value of mass m, shorter is the wavelength, `lamda`. HENCE `alpha`-particles have the largest mass
Discussion
5.

If three elements P, Q and R crystallize in a cubic solid lattice with P atoms at the corners, Q atoms at the cube centre and R atoms at the centre of the faces of the cube, then write the formula of the compound.

Answer»


Solution :Atoms P PER unit cell = 8 x 1/8=1, Atoms Q per unit cell =1, Atoms R per unit cell = 6 x 1/2=3
Hence, the formula is `PQR_3`
Discussion
6.

If three element, P , Q and R crystallize in a cubic solid lattice with P atoms at he corners, Q atoms at the cube centre and R atoms of the faces of the cube, then write the formula of the compound.

Answer»

SOLUTION :Atoms P PER formula CELL =` 8 xx 1//8 =1 ` , Atoms Q per unit cell =1 , Atoms R per UNITCELL =1 , Atoms R per unit cell = ` 6 xx 1//2 =3`
Hence the formula is ` PQR_(3)`
Discussion
7.

If three electrons are lost by Mn^(3+), its final oxidation state would be

Answer»

ZERO
`+6`
`+2`
`+4`

Answer :B
Discussion
8.

If this is passed in solution of 0.1 M Zn^(2+) and 0.01 M Cu^(2+) and concentration of S^(2-) made 8.1xx10^(-31) M . Precipitation of ZnS and CuS will take place ? K_(sp) of ZnS=3.0xx10^(-23) & K_(sp) of CuS=8.0xx10^(-34) .

Answer»

SOLUTION :PRECIPITATION of CUS TAKE PLACE
Discussion
9.

If thermal energy predominates over intermolecular forces, then the substance changes from……..to……..

Answer»

GAS to LIQUID
liquid to SOLID
gas to solid
liquid to gas

ANSWER :D
Discussion
10.

If thermal energy predominates over intermolecular forces, then the substance changes from __to__

Answer»

GAS to LIQUID 
liquid to SOLID 
gas to solid 
liquid to gas 

ANSWER :D
Discussion
11.

If there is no change in concentration, why is the equilibrium state considered dynamic?

Answer»

Solution :Chemical reactions which are reversible do not cease, when equilibrium is ATTAINED. At equilibrium the FORWARD and the BACKWARD reactions are proceeding at the same RATE and no macroscopic change is observed. So chemical equilibrium is in a STATE of dynamic equilibrium.
Discussion
12.

If there is no change in concentration why is the equilibrium state considered dynamic?

Answer»

Solution :At chemical equilibrium the RATE of two opposing reactions are EQUAL and the concentration of reactants and products do not CHANGE with time. This condition is not STATIC and is DYNAMICS because both the forward and reverse reactions are still occurring with the same rate and macroscopic change is observed. So chemical equilibrium is in a state of dynamic equilibrium.
Discussion
13.

If there is no change in concentration, why is the equilibrium state considered dynamic ?

Answer»

Solution :At chemical equilibrium the rate of two OPPOSING reactions are equal and the concentrationof reactants and PRODUCTS do not change with time. This CONDITION is not static and is dynamic, because both the forward and REVERSE reactions are still OCCURRING with the same rate and no macroscopic change is observed. So chemical equilibrium is in a state of dynamic equilibrium.
Discussion
14.

If there is a large energy gap between the filled valence band and empty conduction band, the substance acts as _______

Answer»


ANSWER :INSULATOR
Discussion
15.

If there be a compoud of the formula CH_(3)C(OH)_(3) which one of the following compounds would be obtained from it without reaction with any reagent

Answer»

`CH_(3)OH`
`CH_(3)COOH`
`CH_(3)COCH_(3)`
`CH_(3)CHO`

Solution :
Discussion
16.

If there are three possible values (-1//2,0,+1//2) for the spin quantum, then electronic configuration of K (19) will be:

Answer»

`1S^(3),3s^(3)2p^(9),3s^(2)3p^(1)`
`1s^(2),2s^(2)2p^(6),3s^(2)3p^(6),4s^(1)`
`1s^(2),2s^(2)2p^(9),3s^(2)3p^(4)`
none of these

Answer :A
Discussion
17.

If there are only two H-atoms each is in 3rd excited state then:

Answer»

maximum number of DIFFERENT PHOTONS emitted is 4
maximum number of different photons emitted is 3
minimum number of different photons emitted is 1
minimum number of different photons emitted is 2

Answer :A::C
Discussion
18.

If there are 13 electrons in M shell and 1 electron in N shell th en the num ber of unpaired electrons in such atom will be.....

Answer»

5
4
3
6

Solution :K shell `therefore`n = 1 L shell `therefore`n = 2 M shell, n = 3 an d 13 electrons in it `therefore 3S^(2) 3p^(6) 3D^(5)`configuration N shell `therefore`n = 4 and 1 electron in it`therefore``4S^(1)` configuration This configuration `3s^(2) 3p^(6) 3d^(5) 4s^(1)`is of CR as below

`therefore`Six unpaired electrons are p RESEN t in this
Discussion
19.

If the weight of 5.6 lit of a gas at NTP is 11g, the gas may be (at.wt of P = 31, N = 14,0 = 16 and Cl = 35.5)

Answer»

`PH_3`
`COCl_2`
`NO`
`N_2 O`

ANSWER :D
Discussion
20.

If the wavelength of an electromagnetic radiation is 2000 A. what is its energy in ergs?

Answer»

`9.94xx10^(-12)`
`9.94xx10^(-10)`
`4.97xx10^(-12)`
`4.97xx10^(-19)`

ANSWER :A
Discussion
21.

If the wavelength of the electron is numerically equal to the distance travelled by it in one second, then

Answer»


ANSWER :`LAMBDA = SQRT((H)/(m))`
Discussion
22.

If thewavelengthanddistancetravelbyelectronin 1 sec . Issame thancalculatethe itsvelocity .

Answer»

SOLUTION :`2.7 XX 10^(2) MS^(-1)`
Discussion
23.

If the volume of the racion flask is reduced to half of its initial value and temperature is kept constant the in which of the following cases the position of equilibrium will shift?

Answer»

`CO(g)+H_(2)O(g)hArrCO(2)(g)+H_(2)(g)`
`I_(2)(g)hArr2I(g)`
`NH_(4)HS_(s)hArrNH_(3)(g)+H_(2)S(g)`
`2NOCI(g)hArr2NO(g)+CI_(2)(g)`

SOLUTION :N//A
Discussion
24.

IF the volume of each molecule is 3.4 xx 10^(-24) cc, calculate the vacant space of 8 g of oxygen gas at 27^(@)C and one atmosphere pressure.

Answer»

SOLUTION :`5.65 LIT
Discussion
25.

If the volume of a fixed mass of a gas is reduced to half at constant temperature, the gas pressure _________

Answer»

REMAINS CONSTANT
DOUBLES
REDUCES to half
becomes zero

Answer :B
Discussion
26.

If the velocity of the electron in Bohr's first orbit is 2.19 xx 10^(6) m s^(-1), calculate the de Broglie wavelength associated with it.

Answer»

SOLUTION :`lamda = (H)/(MV) = (6.626 XX 10^(-34) kg m^(2) s^(-1))/((9.11 xx 10^(-31) kg) (2.19 xx 10^(6) ms^(-1))) = 3.32 xx 10^(-10) m = 332` pm
Discussion
27.

If the values of Delta_(f) H^(@) for H_(2) O_(2) and H_(2) O are -188 and -286 "kJ mol"^(-1), then the value of Delta H^(@) = …. kJ for the following reaction - 2H_(2) O_(2(l)) to 2H_(2)O_((l)) + O_(2(g))

Answer»

`-494`
`-196`
`-98`
`+196`

ANSWER :B
Discussion
28.

If the value of principal quantum number is 3, the total possible values for magnetic quantum number will be

Answer»

5
9
8
10

Answer :A
Discussion
29.

If the value of equilibrium constant for a particular reaction is 1.6 xx 10^(12) , then at equilibrium , the system willcontain

Answer»

mostly products
similar amounts of reactants and products
all reactants
mostly reactants

SOLUTION :As the value of K is very high, this SHOWS that the reaction PROCEEDS almost to completion , i.e., the system will contain mostly products .
Discussion
30.

if the value of azimuthal quantum number is 3, the possible values of the magnetic quantum number would be

Answer»

0,1,2,3
0,-1,-2,-3
`0,pm1, pm2,pm3`
`pm1, pm2, pm3`

Solution :When l=3 then m=-3,-2,-1,0,+1,+2,+3. m=-1 to +1 INCLUDING ZERO
Discussion
31.

If the value of Avogadro's number is changed to 1.0 xx 10^20, what would be the molecular mass of nitrogen gas ?

Answer»


ANSWER :`4.6 XX 10^(-1)` AMU
Discussion
32.

If the value of Avogadro number is 6.023 xx 10^23 "mol"^(-1)and the value of Boltzmann constant is 1.380 xx 10^(-23) JK^(-1), then the number of significant digits in the calculated value of the universal gas constant is [Hint: Since k and N, both have four significant figures, the value of R is also rounded off up to four significant figures.]

Answer»


ANSWER :4
Discussion
33.

If the universal gas constant is 8.3 joule mol^(-1) K^(-1) and the Avogadro's number is 6 xx 10^(23). The mean kinetic energy of the oxygen molecules at 327^(@)C will be :

Answer»

`415 XX 10^(23)` joule
`2490 xx 10^(-22)` joule
`1245 xx 10^(-23)` joule
`830 xx 10^(-22)` joule

Answer :C
Discussion
34.

If the value of an equilibrium constant for a particular reaction is 1.6 xx 10^12, then at equilibrium the system will contain:

Answer»

all reactants.
mostly reactants.
mostly products.
similar AMOUNTS of reactants and products

Solution :The value of equilibrium CONSTANT for reaction
`K=1.6xx10^12`
The value of K is very high so the SYSTEM will contain mostly products at equilibrium.
Discussion
35.

If the unit cell of a mineral has a cubic close packed (ccp) array of oxygen atoms with m fraction of octahedral holes occupied by magnesium ions , m and n, respectively are

Answer»

`1/2,1/8`
`1,1/4`
`1/2,1/2`
`1/4,1/8`

SOLUTION :In ccp lattice, Z=4 . `therefore` No. of O-atoms per unit cell= `4(O^(2-))` .
`therefore` No. of octahedral voids = 4 and No. of tetrahedral voids=8
As m fraction of octahedral voids is occupied by `Al^(3+)` ions, therefore , `Al^(3+)` ions present =4 m . Similarly, `Mg^(2+)` ions =8 n. Hence, formula of the MINERAL is `Al_(4m) Mg_(8n)O_4`. As TOTAL charge on the COMPOUND is zero, hence
4 m (+3)+8 n(+2)+4 (-2)=0
or 12 m +16 n -8 =0
Substituting the GIVEN values of m and n , equation is satisfied only when `m=1/2` and `n=1/8`
(as `12 xx1/2+16xx1/8-8=0`)
Discussion
36.

If the unit cell of a minearl has a cubic close packed (ccp) array of oxygen atoms with m fraction of octahderal holes occupied by aluminium ions and n fraction of tetrahdral holes occupied by magnesium ions, m and n, respectively are

Answer»

`1/2 , 1/8`
`1,1/4`
`1/2, 1/2`
`1/4,1/8`

Solution :In ccp LATTICE , Z =4No. of O-atoms per unit cell =`4 ( O^(2-))`
No.of octahedral voids = 4 and No. of tetrahedral voids = 8.
As m fraction of octahedral voids is occupied by `Al^(3+)`ions. Therefore, `Al^(3+)`ions present = 4 m. similarly, ` Mg^(2+)`ions= 8 N. Hence, formula of the MINERAL is ` Al_(4m) Mg_(8n)O_(4)` . As total charge on the compound is zero, hence.
4 m ( +3)+ 8 n ( +2) + 4( -2) =0
or 12 m+ 16 n -8 =0
Substituting the given values of m and n m, EQUATION is satisfied only when ` m = 1/2 and n = 1/8`
` ( as 12 xx 1/2 + 16 xx 1/8 -8 =0 )`
Discussion
37.

If the uncertainity in velocity of a moving object is 1.0 xx 10^(-6) ms^(-1) and the uncertainty in its position is 58m, the mass of this object is approximately equal to that of (h = 6.626 xx10^(-34) Js)

Answer»

Helium
Deuterium
Lithium
Electron

Answer :D
Discussion
38.

If the system does not lose heat or does not receive heat, then the process is called ......... .

Answer»

ISOBARIC process
Adiabatic process
lsothermic process
reversible process

Answer :B
Discussion
39.

If the subsidiary quantum number of a subenergy level is 4, the maximum and minimum values of the spin multiplicities are :

Answer»

9,1
10,1
10,2
8,1

Answer :B
Discussion
40.

If the starting material for the manufactureof silicons is RSiCl_(3), write the structure of the productformed .

Answer»

SOLUTION :HYDROLYSIS of alkyltrichlorosilanes GIVES cross-linked SILICONS.
Discussion
41.

if the starting material is labelled with deuterium as indicated, predict how many total deuterium atoms will be present in the major elimination product ?

Answer»
Discussion
42.

If the starting material for the manufacture of silicones is RSiCl_(3), write the structure of the product formed.

Answer»

Solution : If the STARTING MATERIAL is `RSiCl_3`, a cross-linked silicone is OBTAINED as SHOWN below.
Discussion
43.

If the starting material for the manufacture of silicones is RSiCl_3, write the structure of the product formed.

Answer»

SOLUTION :
Discussion
44.

If the standard molar enthalpy of formation of SO _(2 (g)) and SO _(3(g)) is - 296 . 82 kJ mol ^(-1) and -395 . 72 kJ mol ^(-1) respectively, then enthalpy change for thereaction in kJ mol ^(-1) ,SO _(2 (g)) + (1)/(2) O _(2 (g)) to SO _(3(g)) , is

Answer»

`-296.82`
`+98.9`
`-395.72`
`-98.9`

ANSWER :D
Discussion
45.

If the standard molar enthalpy of formation of CaO_((s)), CO_(2(s)) and CaCO_(3(s)) is -635, -393 and -1207kJ "mol"^(-1) respectively, the Delta_r H^(o.) in kJ mol^(-1) for the reaction CaCO_(3(s)) rarr CaO_((s)) + CO_(2(g)) is

Answer»

`-179`
`265`
`223.5`
`+179`

ANSWER :D
Discussion
46.

If the speed of light is 3.0xx10^(8) ms^(-1), calculate the distance covered by light in 2.00 ns.

Answer»



Solution :`2 ns = 2.00xx10^(-8)s`
Distance `= "speed " xx "TIME"`
`=3XX10^(8) ms^(-1) xx 2.0 xx 10^(-9)s`.
`=6.0xx10^(-1) m = 0.6 m`
Discussion
47.

If the speed of light is 3.0 xx 108 m s^(–1), calculate the distance covered by light in 2.00 ns.

Answer»

Solution :Distance COVERED = speed `XX` time
`=3.0 xx 10^(8) MS^(-1)xx 2.00 s`
`=3.0 xx 10^(80) ms^(-1) xx 2.00 ns xx (10^(-9) s)/(1 ns)=6.00 xx 10^(-1) m = 0.600 m`
Discussion
48.

If the solubility product of silver chloride is 1.8xx10^(-10). What is the solubility of silver ion if concentration of Cl^(-) is 0.01 molar.

Answer»

Solution :`K_(sp)=[Ag^(+)][CL^(-)]""1.8xx10^(-10)=[Ag^(+)]0.01`
`[Ag^(+)]=(1.8xx10^(-10))/(0.01)=180xx10^(-10)=1.8xx10^(-8)"MOL"//dm^(3)`
Discussion
49.

If the solubilityproduct of MOH is 1xx 10^(-10)mol^(2)dm ^(-2)Then the P^(H)of its aqueous solutions will be

Answer»

12
9
6
3

Solution :` K_(sp) = 1 XX 10 ^(_10 )M, [OH]= 10 ^(-5)M, POH = 5`
` therefore pH = 9`
Discussion
50.

If the solubility of zirconium phosphate is x, what is its solubility product? What are the units of solubility product of Zr_3 (PO_4 )_4?

Answer»

Solution :Zirconium phosphate DISSOCIATES in aqueous solutions as `Zr_3 (PO_4)_4 hArr+3Zr^(4+)+ 4PO_(4)^(3-)`
If x is the solubility, `[Zr^(4+)] =3S and [PO_(4)^(3-) ] = 4S`
The solubility product `K_(sp)` is `Kp_(sp)= [Zr^(4+)] [PO_(4)^(3-) ]^4 =(3s)^2 (4s)^4 = 6912 s^7`
The units of solubility product are : `mol^7L^(-7)`
Discussion