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 period in which s-block, p-block and d- block elements are present

Answer»

1
6
7
3

Answer :B
Discussion
2.

The period in which only single block elements are present by definition

Answer»

6
2
3
1

Answer :D
Discussion
3.

The percentages of all the elements present in a compound are 92. What does it indicate ?

Answer»

SOLUTION :This indicates that the compound contains in it OXYGEN ALSO and its PERCENTAGE is (100-92)=8
Discussion
4.

The percentage weight of hydrogen in H_(2)O_(2)

Answer»

5.88
6.25
25
50

Solution :`2/34 XX 100 = 5.88`
Discussion
5.

The percentage proportion of C, H and N in an organic compound are 62.07, 10.34 and 14.0 respectively. Find its empirical formula.

Answer»

`C_(2)H_(5)NO_(2)`
`C_(5)H_(10)NO_(2)`
`C_(2)H_(5)NO_(2)`
`C_(5)H_(10)NO`

Solution :
(D) `C_(5)H_(10)NO`
accurate `C_(6)H_(12)NO`..........No OPTION
Discussion
6.

The percentage to deuterium in heavy water is

Answer»

22.2
11.2
44
20

Answer :D
Discussion
7.

The percentage of silica in sodium silicate is approximately (Si=28)

Answer»

25
40
50
60

Solution :`%SiO_(2)=(60)/(122)xx100~~50%`
Discussion
8.

Thepercentage of s-character of the hybrid orbitals in methane, ethane, ethene and ethyne are respectively

Answer»

`25, 25,33.3, 50`
`50,50,33.3, 25`
`50,25,33.3, 50`
`50,25,25,50`

ANSWER :d
Discussion
9.

The percentage of s-character of the hybrid orbitals in methane, ethane, ethene and ethyne are respectively ………………….

Answer»

`25, 25, 33.3, 50`
`50, 50, 33.3, 25`
`50, 25, 33.3, 50`
`50, 25, 25, 50`

Solution :`OVERSET(sp^(3))(CH_(4)),overset(sp^(2))(CH_(3))-CH_(3),overset(sp^(2))(CH_(2))=CH_(2), overset(sp^(3))(CH)-=CH`
Discussion
10.

The percentage of pyridine (C_(5)H_(5)O) that forms pyridinium ion (C_(5)H_(5) overset(+)NH)in a 0.10 M aqueons pyridine solution is (K_(b) for C_(5)H_(5)N=1.7 xx 10^(-9))

Answer»

0.0077
0.016
6.0E-5
0.00013

Solution :Pyridine is a weak base. In AQUEOUS solution, it ionizes as
`{:(,C_(5)H_(5)N+H_(2)O hArr C_(5)H_(5)overset(+)NH+OH^(-)),("Initial conc."," "c "" 0 ""0),("At. eqm." ,C(1-alpha)"" calpha ""calpha),(,K_(b) ~=C alpha^(2) ""("By Ostwald's DILUTION law")):}`
`alpha = SQRT((K_(b))/(C))=sqrt((1.7xx10^(-9))/(01))=sqrt(1.7xx10^(-8))=13.10^(-4)`
`: .` % disssociation (i.e. , % that forms `C_(5)H_(5)overset(+)NH` ion) `=1.3xx10^(-4)xx100=0.013 %`
Discussion
11.

The percentage of pyridine (C_5H_5N) that forms pyridinium ion (C_5H_5N^(+)H) in a 0.1 M aqueous pyridine solution (K_b for C_5H_5N=1.7xx10^(-9) ) is

Answer»

`0.77%`
`1.6%`
`0.006%`
`0.013%`

SOLUTION :Pyridine `(C_5H_5N)` is a WEAK BASE,
`K_b=C alpha^2`
`alpha=sqrt((1.7xx10^(-9))/0.1)=1.30xx10^(-4)`
`%alpha=1.30xx10^(-4)XX100`
`%alpha`=0.013%
Discussion
12.

The percentage of Pi-charaacter in the orbitals forming P-P bonds in P_(4) is

Answer»

25
33
50
75

Solution :In `P_(4)` is `SP^(3)` hybridized. Since p-orbitals form `pi` -BONDS, therefore, `pi` CHARACTER of p-orbitals is 75%.
Discussion
13.

The percentage of p-character of the hybrid orbitals in graphite and diamond are respectively

Answer»

33 and 25
50 and 75
67 and 75
33 and 75

Solution :Hybridisation in graphite = `sp^(2)`
`therefore %` p-character = `(2)/(3) XX100 = 67%`
Hybridisation in diamond = `sp^(3)`
`therefore %` p-character `= (3)/(4)xx100 = 75%` .
Discussion
14.

The percentageof p-character in SF_(6) are

Answer»

`120^(@), 20%`
`90^(@), 33%`
`109^(@), 25%`
` 90^(@), 25%`

Solution :`SF_(6)` involves `sp^(3) d^(2)` hybridization . The shape is
octahedral with each bond angle = `90^(@)`.
% of d-character
`=("2(no. of d-ORBITALS)")/("6(total no.of hybrid orbitals ")XX100 = 33%`.
Discussion
15.

The percentage of oxygen in NaOH is

Answer»

40
6
8
20

Answer :B
Discussion
16.

The percentage of oxygen in a compound is determined by

Answer»

DUMAS method
Kjeldhl's method
CARIUS method
Subtraction the sum of PERCENTAGE of all other ELEMENTS PRESENT from 100.

Answer :D
Discussion
17.

The percentage of nitrogen in urea (NH_(2)CONH_(2)), is

Answer»

38.4
46.6
59.1
61.3

Answer :B
Discussion
18.

The percentage of nitrogen in urea is about......

Answer»

23.23
46.67
75
37.5

Answer :B::D
Discussion
19.

The percentage of nitrogen in urea

Answer»

0.46
0.28
0.85
0.64

Answer :A::D
Discussion
20.

The percentage of lead in lead pencils is

Answer»

0
100
80
50

Solution :LEAD PENCIL = ( GRAPHITE `+` CLAY ) , No lead
Discussion
21.

The percentage of ionic character of LiH is 76.81% and the bond length is 1.596Å. Wha is the value of dipole moment of LiH molecule? [1D=3.335xx10^(-30)C*m]

Answer»

Solution :If the molecule is 100% ionic, then
`mu_("ionic")=qxxd=1.602xx10^(-19)Cxx1.596xx10^(-10)m`
`=2.557xx10^(-29)C*m`
`%` ionic CHARACTER`=(mu_("observed"))/(mu_("ionic"))XX100`
`thereforemu_("observed")=("% ionic character"xxmu_("ionic"))/(100)`
`=(76.81xx2.557x10^(-29))/(100)=1.96xx10^(-29)C*m`
`=(1.96xx10^(-29))/(3.335xx10^(-30))D=5.07D`
Discussion
22.

Thepercentageof ______in thecompoundis usuallyfoundbydifferencebetweenthetotalpercentagecompositon(100)and thesumof thepercentageof all otherelements .

Answer»

nitrogen
carbon
hydrogen

Answer :C
Discussion
23.

The percentage of enol content in the following is in the order I) CH_(3)COCH_(2)COCH_(3)II) CH_(3)COCH_(2)COOCH_(3)III) CH_(3)COCH_(2)CH_(2)CH_(5)

Answer»

`I GT II gt III`
`I gt III gt II`
`III gt II gt I`
`II gt III gt I`

Solution :I is more enole contant when compare II DUE to CROSS conjugation.
Discussion
24.

The percentage of empty space in a body centred cubic arrangement is _____

Answer»

74
68
32
26

Answer :C
Discussion
25.

The percentage of copper in copper (II) salt can be determined by using a thisulphate titration. A 0.305g of copper (II) salt was dissolved in water and added to an excess of potassium iodide solution liberating iodine according to the following equation: 2Cu^(2+)(aq)+4I^(-)(aq)to2Cul(s)+I_(2)(aq) The iodine liberated required 24.5cm^(3) of a 0.1 mole dm^(-3)solution of sodium thiosulphate for titration according to reaction: 2S_(2)O_(3)^(2-)+I_(2)(aq)+S_(4)O_(6)^(2-)(aq) The percentage of copper by mass in the copper (II) salt is (Atomic mass of copper =63.5)

Answer»

64.2
51
48.4
25.5

Answer :B
Discussion
26.

The percentage of chlorine in the chloride of an element is 44.71%. 158.5 g of this chloride on vaporisation occupies a volume of 22.4 litres at NTP. Calculate the atomic weight and valency of the element.

Answer»

Solution :Since 22.4 litres are occupied by 1 MOLE at NTP and WT. of 1 mole is the mol . Wt . In GRAMS ,
`:.` mol . Wt. of CHLORIDE = 158.5
Discussion
27.

The percentage of Carbon, Hydrogeon and Oxygen in an organic substance are 40, 6.666 and 53.34 respectively. Molecular mass of the compound is 180 g mol^(-1). Find the value of integal number.

Answer»

6
2
1
4

Solution :
`:.` EMPIRICAL formula `= CH_(2)O`
and Empirical formula MASS `= C + 2H + O`
`= 12 + 2+ 16 = 30`
Integral no. `= ("molecular mass")/("empirical formula mass") = (180)/(30) =6`
Discussion
28.

the percentage of carbon and hydrogen are estimated simultaneously in an organic cmpound by liebig method

Answer»


ANSWER :t
Discussion
29.

The percentage of C, H and Cl in an organic compounds is 10%, 0.84%, 89.2% respectively. Find out its empirical formula.

Answer»

`C Cl_(4)`
`CHCl_(3)`
`CH_(2)Cl_(2)`
`CH_(3)Cl`

Answer :B::C
Discussion
30.

The percentage of C in methanoic anhydride is ........

Answer»

64.86
32.43
3.243
31.43

Solution :Methanoic anhydride is `H-overset(O) overset(||)C-O-overset(O)overset(||)C-H`
It is `H_(2)C_(2)O_(3)` and its molar mass
`= 2(H) +2(C) +3(O)`
`=2(1)+2(12)+3(16)`
`=2+24+48= 74 g mol^(-1)`
1 MOLE `H_(2)C_(2)O_(3)` has 24 g carbon,
`% C = (24xx100)/(74) = 32.432~~32.43%`
Discussion
31.

The percentage of an element M is 53 in its oxide of molecular formula M_(2)O_(3). Its atomic mass is about

Answer»

45
9
18
27

Solution :100 g of metal oxide contains 53 g of metal,
`:.` EQUIVALENT mass of element`=("Mass of element")/("Mass of oxygen") xx 8 = (53)/(47) xx 8 ~=9`
`:.` ATOMIC mass `=` Equivalentmass of an element `xx` Valency `= 9 xx 3 = 27g`
Discussion
32.

The percentage of all the elements present o in a compound is 95 . What doesit indicate ?

Answer»


Answer :This indicates that the compound CONTAINS OXYGEN. Its percentage is given as 100 - 95 = 5 .
Discussion
33.

The percentage nitrogen in a compound is determined by

Answer»

Nessler's method
Kjeldhahl's method
Carius method
Taking the difference between TOTAL PERCENTAGE and the sum of PERCENTAGES of all other ELEMENTS present.

Answer :B
Discussion
34.

The percentage ionic character of a covalent bond (A-B) is about 50%, the electronegativity difference of the two elements will be :

Answer»

`1.5`
`2.1`
`0.9`
`1.2`

ANSWER :B
Discussion
35.

The percentage composition of the elements of C_(8)H_(9)ON is

Answer»

8:9:1:1
71.1:6.7:11.8:10.4
12:1:16:14
none of these

Answer :B
Discussion
36.

The percentage degree of hydrolysis of a salt of weak acid (HA) and weak base (BOH)in its 0.1 M solutions is found to be 10% If the molarity of the solution is 0.05M, the percentage hydrolysis of the salt should be :

Answer»

`5% `
` 10 %`
` 20%`
NONE of these

Solution :The formula is h =`(h)/(1-h)=sqrt((K_w)/( K_a. K_b))`
In this formula , Conc. TERM is ABSENT i.e., .h.does not DEPEND on concentration .
Discussion
37.

The percentage composition of carbon by mole in methane is:

Answer»

0.75
0.8
0.2
25%.

ANSWER :C
Discussion
38.

The percentage by volume of C_(3)H_(8) in a gaseous mixture of C_(3)H_(8),CH_(4)andCO is 20. When 10 ml of the mixture is burnt in excess of O_(2), the volume of CO_(2) produced is 2xml. Find the value of .x..

Answer»


SOLUTION :20% of 10 ml =2 ml
Let VOLUME of `CH_(4)=x`
`implies CO=(8-x)ml`
`{:(C_(3)H_(8)+5O_(2),rarr,3CO_(2)+4H_(2)O),("2 ml",rarr,"6 ml"),(CH_(4)+2O_(2),rarr,CO_(2)+2H_(2)O),("x ml",rarr,"x ml"),(CO+(1)/(2)O_(2),rarr,CO_(2)),((8-x)ml,rarr,(8-x)ml):}`
`V_(CO_(2))=6+x+(8-x)=14ml`
Discussion
39.

The percentage by mass of C, H, and Cl in a compound are C 52.2%, H 3.7% and Cl 44.1%. How many carbon atoms are in the simplest formular of the compound?

Answer»

3
4
6
7

Answer :D
Discussion
40.

The particle size range from……..in colloidal state

Answer»

1-100nm
200-2000nm
2000-4000nm
0.1-1nm

Solution :PARTICLES in COLLOIDAL STATE should have a SIZE between 1-100 NM
Discussion
41.

The partial pressure of nitrogen in air of 0.76 atm and its Henry's law constant is 7.6 xx 10 ^(4) atm at 300K. What is the mole fraction of nitrogen gas in the solution obtained when air is bubbled through water at 300 K ?

Answer»

`1 xx 10 ^(-4)`
`1 xx 10 ^(-6)`
`2 xx 10 ^(-5)`
`1 xx 10 ^(-5)`

Answer :A::D
Discussion
42.

The partial pressure of dry gas is

Answer»

GREATER than the of WET gas
lesser than that of wet gas
equal to that of wet gas
none of these

Answer :B
Discussion
43.

The partial pressure of carbon dioxide in the reaction CaCO_(3)(s)hArrCaO(s)+CO_(2)(g) is 1.017xx10^(-3) atm at 500^(@)C. Calculate K_(P) at 600^(@)c C for the reaction. DeltaH for the reaction is 181 kJ mol^(-1) and does not change in the given range pf temperature.

Answer»

SOLUTION :`P_(CO_(2))=1.017xx10^(-3)" atm "T=500^(@)C`
`K_(P)=P_(CO_(2))`
`:.L_(P_(1))=1.017xx10^(-3),T=500+273=773K`
`K_(P_(1))=?""T=600+273=873K`
`DeltaH^(@)=181" kJ mol"^(-1)`
`"LOG"((K_(P_(2)))/(K_(P_(1))))=(DeltaH^(@))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
`"log"((K_(P_(2)))/(1.017xx10^(-3)))=(181xx10^(3))/(2.303xx8.314)((873-773)/(873xx773))`
`"log"((K_(P_(2)))/(1.017xx10^(-3)))=(181xx10^(3)xx100)/(2.303xx8.314xx873xx773)`
`(K_(P_(2)))/(1.017xx10^(-3))="anti log of "(1.40)`
`(K_(P_(2)))/(1.017xx10^(-3))=25.12`
`rArr" "K_(P_(2))=25.12xx1.017xx10^(-3)`
`K_(P_(2))=25.54xx10^(-3)`
Discussion
44.

The partial pressure of carbon dioxide in the reaction CaCO_3(s) hArr CaO(s)+ CO_2(g) is 1.017 xx 10^(-3) atm at 500^@C. Calculate K_P" at " 600^@C for the reaction. DeltaH for the reaction is 181 kJ mol^(-1) and does not change in the given range of temperature.

Answer»

Solution :`P_(CO_2) = 1.017 XX 10^(-3)a t m, T = 500^@C`
`K_P = P_(CO_2)`
`:. K_(P_1) = 1.017 xx 10^(-3), T= 500 + 273 = 773 K`
`K_(P_2) = ?, T = 600+ 273 = 873 K`
`DeltaH^@ = 181 kJ mol^(-1)`
`log""((K_(P_2))/K_(P_1)) = (DeltaH^@)/(2.303 R) ((T_2-T_1)/(T_1T_2))`
`log""(K_(P_2)/(1.017 xx 10^(-3)))=(181xx 10^3)/(2.303 xx 8.314) ((873 - 773)/(873 xx 773))`
`log""(K_(P_2)/(1.017 xx 10^(-3))) = (181xx 10^3xx 100)/(2.303 xx 8.314 xx 873 xx 773)`
`K_(P_2)/(1.017xx 10^(-3))`= ANTILOG of (1.40)
`K_(P_2)/(1.017xx 10^(-3)) = 25.12`
`rArr K_(P_2) = 25.12 xx 1.017 xx 10^(-3)`
`K_(P_2) = 25.54 xx 10^(-3)`
Discussion
45.

The pari of compounds that can exist together is

Answer»

`FeCI_(3),SnCI_(2)`
`HgCI_(2),SnCI_(2)`
`FeCI_(2),SnCI_(2)`
`FeCI_(3),KI`

Solution :SINCE CI in `CIO^(-)` has an O.N of +1 which is higherthan its lowest O.N of -1 and lower than its highest O.N of +5 therefore `CIO^(-)` can act as an OXIDISING as well as reducing agent
Discussion
46.

The parameters that describe the gaseous state are _________

Answer»

VOLUME
PRESSURE
TEMPERATURE
all the these

Answer :D
Discussion
47.

The paramagnetic species is

Answer»

CARBONIUM ion
Carbanion
Free RADICAL
None

Answer :B
Discussion
48.

The paramagnetic species is...

Answer»

`KO_(2)`
`S iO_(2)`
`TiO_(2)`
`BaO_(2)`

ANSWER :A
Discussion
49.

The pairs of species of oxygen and their magnetic behaviours are noted below . Which of the following presents the correct description ?

Answer»

`O_(2)^(-), O_(2)^(2)` - BONE diamagnetic
`O^(+) , O_(2)^(2)`- Both paramagnetic
`O_(2)^(+), O_(2)`- Both paramagnetic
`O, O_(2)^(2-)` - Both paramagnetic

Solution :`O_(2)` has two UNPAIRED electrons in `pi_(2p_(x)) and pi_(2p_(y)) `,
i.e.,` pi_(2p_(x))^(**1)pi_(2p_(x))^(**2)`
`O_(2)^(-)` will have one upaired electron , `O_(2)^(2-)` will
have no unpaired electron, `O_(2)^(+)` will have one
unprired electron, `""_(8) O = 1s^(2) 2s^(2) 2p_(x)^(2) 2p_(y)^(1) 2p_(z)^(1)`
and THUS has two unpaired electrons , `O^(+)` will have
one unpaired electron .
Thus , `O_(2)^(+) and O_(2)` contain unpaired electrons and
HENCE will be both paramagnetic .
Discussion
50.

The para magnetic nature of oxygen is best explained by

Answer»

V.B.theory
Hybridisation
M.O.theory
VSEPR theory

Answer :C
Discussion