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.

In the following questions, a statement of assertion (A) is followed by a statement of reason (R ) A : LiCl is more covalent than BeCl_(2) R : Li^(*) ion is smaller than Be^(2+) .

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If both ASSERTION & REASON are TRUE and the reason is the CORRECT explanation of the assertion, then MARK (1)
If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2)
If Assertion is true statement but Reason is false, then mark (3)
If both Assertion and Reason are false statements , then mark (4).

Answer :D
2.

In the following questions, a statement of assertion (A) is followed by a statement of reason (R ) A : Lattice energy ofCaO is higher than LiCl. R : Lattice energy of ionic compound is directly prportional to the product of charges of ion.

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If both Assertion & Reason are TRUE and the reason is the CORRECT explanation of the assertion, then MARK (1)
If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2)
If Assertion is true statement but Reason is false, then mark (3)
If both Assertion and Reason are false statements , then mark (4).

Answer :A
3.

In the following questions, a statement of assertion (A) is followed by a statement of reason (R ) A : Equal number of sigma and pi bonds are present in ethyne. pi Bond is stronger than sigma bond.

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If both ASSERTION & Reason are TRUE and the reason is the correct explanation of the assertion, then mark (1)
If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2)
If Assertion is true STATEMENT but Reason is false, then mark (3)
If both Assertion and Reason are false STATEMENTS , then mark (4).

Answer :D
4.

In the Arrhenius for a certain reaction, the value of A and E_(a) (activation energy) are 4 xx 10^(13) sec^(-1) and 98.6 "kJ mol"^(-1), respectively. If the reaction is of first order, the temperature at which its half-life period is 10 minutes is

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`280 K`
`290 K`
`311.35 K`
`418.26 K`

SOLUTION :Calculation of `k`:
We know that `k = (2.303)/(t) "LOG"((a)/(a-x))`
`(t_(1//2)=10xx60 SEC)=1.1558xx10^(-3)`
According to ARREHENIUS equation,
`log k = log A - (Ea)/(2.303 RT)`
Substituting the various values in the above equation
`log 1.155 xx 10^(-3) = log 4 xx 10^(13) - (98.6)/(2.303 xx 8.314 xx 10^(-3) xx T)`
On usual calculation, `T = 311.35 K`
5.

In the arrhenius equation, what does the factor e^(-E_(a)//RT) correspond to ?

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Solution :The quantity `e^(-E_(a)//RT)` CORRESPONDS to the fraction of MOLECULES that have KINETIC energy greater than `E_(a)`.
6.

In the Arrhenius equation equation,the Boltzmann factor e^(Ea//RT) represents the…………..of the molecules possessing energ in excess of activation energy

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number
FRACTION
weight
PERCENTAGE

ANSWER :B
7.

In the arrangement given below 20 mole of N_(2) and 5 mole of He are present above water. The total pressure above the water is 60 atm. If K_(H) of N_(2) is 10^(5) atm. How many moles of N_(2) are dissolved in water? give your answer after multiplying by 10^(3) The vapour pressure of water is 10 atm.

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ANSWER :8
8.

In the Arrhenius equation , K=Ae^(-Ea//RT), Arrheniuns constant

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`E_(a)=0`
`T=OO`
`T=0`
`E_(a)=oo`

ANSWER :A::C
9.

In the Arrhenius equation k=A e^(-E_a//RT), A may be termed as rate constant at …………. .

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SOLUTION :INFINITE TEMPERATURE.
10.

In the anion RCOO^(-), the two carbon -oxygen bonds are found to be of equal length. What is the reason for it?

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The anion is obtained by removal of a proton from the ACID molecule.
Electronic orbitals of carbon ATOM are hybridised.
The C=O bond is weaker than the C-O bond.
The anion `RCOO^(-)` has TWO resonating STRUCTURES.

Solution :The anion `HCOO^(-)` has two equivalent resonating structures :
11.

In the analysis of basic radicals, the group reagent H_(2)S gas is generally used in the groups

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I and II groups
II and III groups
III and V groups
II and IV groups

Answer :D
12.

In the analysis of basic radicals, NH_(4)OH and NH_(4)Cl give precipitate with

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I-group radicals
II-group radicals
III-group radicals
IV-group radicals

Solution :III-Group radical GIVES ppt. with `NH_(4)OH` and `NH_(4)CL`
13.

In theanalysis of a 0 .50 -gsample of feldspar, mixture, of thechlorides ofNa and Kis obtained, whichweight 0.1180 g . Subsequenttreatmentof themixedchlorides. with silver nitrate gives 0.2451 g of AgCl . Whatis the percentage of sodiumoxideand potassiumoxide in feldspar ?

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SOLUTION :Suppose the weight of NaCl is x g
`{:(Na_(2)O + K_(2) O overset("Step I") to NaCl + KCI underset(AgNO_(3))overset("Stet II")to AgCl),("(in feldspar)x g (0.1180 - x)0.2451 g "):}`
Applying POAC for Clatoms in Step IIto calculate x,
molesof Cl in NaCl + moles of Cl in KCI = moles of Cl in AgCl
` 1 xx` moles of NaCl ` + 1 xx ` molesof KCl` = 1 xx ` moles of AgCl
`(x)/( 58 . 5) + (0.1180)/( 84.5) = (0.2451)/( 143.5) , x = 0.0345 g`
(NaCl = 58.5, KCI = 74. 5 and Ag Cl = 143 . 5 )
`:.` wt. of NaCl = 0.0343 g,
wt. of KCI = 0.0837 g.
Again, applying POAC for Naand Katoms to CALCULATETHE weight of `Na_(2)O and K_(2)O`respectively ,
we GET,
`2 xx ` moles of `Na_(2) O` = moles ofNaCl
and` 2 xx` moles of `K_(2)O` = molesof KCI
`:. "wt. of " Na_(2) O = (1)/(2)xx ("wt. of Na Cl")/("mol. wt. of NaCl") xx "mol. wt. of " Na_(2) O `
and wt. of `K_(2) O = (1)/(2) xx ("wt. of KCI")/( "mol . wt. of KCI") xx "mol. wt. of " K_(2)O`
`:. ` wt. of ` Na_(2) O = (1)/(2) xx (0.0343)/( 58.5) xx 62 = 0.018 g `
and wt. of `K_(2) O = (1)/(2) xx (0.0837)/( 74.5) xx 94 = 0.053 g`
`:. % of Na_(2) O = (0.018)/(0.50) xx 100 = 3 . 6 %`
and `% of K_(2) O = (0.053)/( 0.50) xx 100 = 10 . 6% `
14.

In thealuminothermiteprocess,aluminium actsas

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an oxidising AGENT
a FLUX
a REDUCINGAGENT
a solder.

ANSWER :C
15.

In the aluminothermic process, aluminum acts as

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a REDUCING AGENT
an OXIDIZING agent
a common solder
a flux

ANSWER :A
16.

In the aluminothermic process

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`Al_2O_3 " is reduced by " CR`
`Cr_2O_3 " is reduced by " Al`
`Al_2O_3 " is reduced by " C`
NONE of these corrcet

Answer :B
17.

In the aluminium thermite process, aluminium acts as

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A FLUX
An OXIDISING AGENT
A SOLDER
A REDUCING agent

Answer :D
18.

In the aluminothermic process, aluminium acts as:

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A REDUCING agent
A flux
An OXIDIZING agent
A solder

Answer :A
19.

In the alkylation of benzene , stable alphacomplex can be

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ANSWER :A, B, C, D
20.

In the alpha-halogenation of aliphatic acid (HVZ reaction ) the catalyst used is:

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ANSWER :B
21.

In the Alksline earth metals, the element forming predominantly covalent compound is

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Be
Mg
Sr
Ca

SOLUTION :Be to Ba IONIC CHARACTER INCREASING
22.

In the adsorption of oxalic acid by activated charcoal, the activated charcoal is known as:

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ADSORBENT
ADSORBATE
ABSORBER
absoibate

SOLUTION :adsorbent
23.

In the adsorption of oxalic acid by activated charcoal, the activated charcoal is known as :

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Adsorbent
Adsorbate
Absorber
None

Answer :A
24.

In the adsorption of a gas on solid, Freundlich isotherm is obeyed. The slope of the plot is zero. The extent of adsorption is

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<P>DIRECTLY proportional to the pressure of the gas
inversely proportional to the pressure of the gas
directly proportional to the square root of the pressure of the gas
independent of the pressure of the gas

Solution :`(x)/(m) = KP ^(1//n)`
ln `(x)/(m ) K + (1)/(n) ln p`
Slope `= 1/n =0`
Thus, `x/m = kp ^(0)`
25.

In the addition of HBr to propene in the absence of peroxides the first step involves the addition of :

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`H^+`<BR>Br
`H^o`
Br

Answer :D
26.

In the acidic medium MnO_(4)^(-) acts as an oixidising agent. MnO_(4) + 8H^(+) + 5e^(-) to Mn^(2+) + 4H_(2)O If the [H^(+)] is reduced to half of its original value, the electrode potential of the half cell MnO_(4), Mn^(2+) |Pt (log 2=0.3010)

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INCREASES by 14.23 MV
DECREASES by 14.23 mV
increases by 28.46 mV
decreases by 28.46 mV

ANSWER :D
27.

In the addition of HBr to propene in the absence of peroxides, the first step involves the addition of:

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`H^(+)`<BR>`Br^(-)`
`H^(**)`
`Br^(**)`

Solution :Th ADDITION is electrophilic in which the electrophile is `H^(+)`.
28.

In the addition of HBr to propene in the absence of peroxides, the first step involves the addition of

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`H^(+)`
`Br^(-)`
`H^(**)`
`Br^(**)`

Solution :In absence of PEROXIDES, +vely CHARGED electrophile, i.e., `H^(+)` adds FIRST.
29.

In the acid base titration [H_3PO_4(0.1 M)+NaOH(0.1 M)] e.m.f of the solution is measured by coupling this electrodes with suitable reference electrode.When alkali is added pH of solution is in acoodance with equation E_(Cell)=E_(Cell)^(@)+0.059 pH For H_3PO_(4) Ka_(1)=10^(-3) , Ka_(2)=10^(-8), Ka_(3)=10^(-13) What is the cell e.m.f at the lind end point of the titration if E_(cell)^(@) at this stage is 1.3805 V.

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SOLUTION :`PH=(pKa_2+pKa_3)/2=10.5`
`E_(CELL)=E_(cell)^@+0.059 pH=1.3805+0.059xx10.5=2V`
30.

In the acid - base reaction,HCl + CH_(3)COOH hArr Cl^(-)+CH_(3)COOH_(2)^(+)the conjugate acid of acetic acid is

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`CH_(3)COOH_(2)^(+)`
HCl
`H_(3)O^(+)`
`CL^(-)`

Answer :A
31.

In the acid hydrolysis of an ester what is the time taken for complete hydrolysis ?

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8 hours
45 minutes
Both (a) and (b)
None

Answer :A
32.

In the acetylation of glucose, which group is involved in the reaction

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CHO group
gtC=O group
alocholic OHgroup
all of these

Answer :C
33.

In the above transaformation 'X' could be

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`NaBH_(4)|C_(2)H_(5)OH|H_(3)O^(+)`
`N_(2)H_(4)|OH^(-)|"Glycol"|Delta`
`LiAlH_(4)|Et_(2)O|H_(3)O^(+)`
`DIBAL-H|THE|H_(3)O^(+)`

ANSWER :C
34.

In the above set of redox reactions, where MnO_(4)^(-1)is the only oxidising agent, order of the required molar quantities of the reducing agents is :

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`n_(1) LT n_(2) lt n_(3)`
`n_(2) lt n_(1) lt n_(3)`
`n_(1) lt n_(3) lt n_(2)`
`n_(2) lt n_(3) lt n_(1)`

Solution :For fixed quantity of `MnO_(4)^(-1)` the MOLAR quantity of the reducing AGENTS will be inversely proportional to their n-factor `(n_(f))`
`{:(n_(f),FeSO_(4),Na_(2)C_(2)O_(4),FeC_(2)O_(4)),(,1,2,3):}`
`therefore` ORDER : `n_(2) lt n_(1) lt n_(3)`
35.

In the above sequence Y is:

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SOLUTION :
36.

In the above sequence X is

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nitrochloromethane
chloropicrin
ethane nitrile
none of the above.

Solution :`{:(""CL),("|"),(Cl-C-NO_(2)),("|"),(""Cl):}` (CHoloropicrin).
37.

In the above reaction, X stands for

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`SnCl_(2)`
`CL`
`NH_(2)`
`NH_(3)^(+)Cl^(-)`.

Solution :
38.

In the above sequence, II is

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`BETA`-ALANINE
`ALPHA`-alanine
ethylenediamine
`GAMMA`-AMINOBUTYRIC acid.

Answer :A
39.

In the above sequence, II is :

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`BETA`-ALANINE
`ALPHA`-alanine
ethylenediamine
`GAMMA`-AMINOBUTYRIC acid.

Solution :
40.

In the above reaction the major product is shown, which is formed through the intermediate (carbocation given below:Which bond will migrate to form above product ?

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<P>p
q
r
S

Answer :B
41.

In the above reaction sequence X & Y are.

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Identical
Homologs
Structural isomers
Stereisomers

Answer :B
42.

In the above reaction sequence, A , B and C respectively are

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BENZENE, NITROBENZENE, aniline
benzene ,m-dinitrobenzene , m-nitroaniline
touluene, m-nitrotoluene, m-toluidine
benzene, nitrobenzene, HYDRAZOBENZENE.

ANSWER :D
43.

In the above reaction (in Q.99) Na_(2)S_(2)O_(4) acts as a

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2 ELECTRON REDUCING agent
1 electron reducing agent
3 electron reducing agent
4 electron reducing agent

Answer :A
44.

In the above question, the order w.r.t. B is:

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3
2
1
ZERO

Solution :d) In the second experiment,
Rate `propto [A]^(x)[B]^(y)`
`8 propto [2]^(3)[2]^(y)`
y=0.
ORDER w.r.t.B is zero.
45.

In the above question, solution is opticaly inactive when :

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`r_(t)=a`
`r_(t)=0`
`r_(t)=X`
`r_(t)=(a+x)`

ANSWER :B
46.

In the above question, the velocity acquired by the electron will be:

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`SQRT(V/m`
`sqrt((EV)/m`
`sqrt((2EV)/m`
None

Answer :C
47.

In the above question if cone, of the reactant is less than 1 M, then

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`r_1=r_2=r_3`
`r_1ltr_2ltr_3`
`r_1gtr_2gtr_3`
NONE of these

Answer :C
48.

In the above question

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All CL atoms show PRIMARY valencies
Two Cl atoms show primary valencies & one Cl atom shows primary as WELL as secondary valency
All the Cl atoms show secondary valencies.
all the above

Answer :C
49.

In the above problem if concentration of reactant is less than 1 M then:

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`r_(1)=r_(2)=r_(3)`
`r_(1)GT r_(2) gtr_(3)`
`r_(1)LT r_(2)lt r_(3)`
All of these

Solution :If `[A] lt 1,`
`r_(1) gt r_(2) gt r_(3)`
50.

In the above problem if concentration of reactant is 1 M then:

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`r_2=r_2=r_3`
`r_1gtr_2gtr_3`
`r_1ltr_2ltr_3`
All

Answer :A