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 acid catalysed hydrolysis of an organic compound 'A' at 303 K has a time for half change of 100 minute when carried out in a buffer solution at pH=5 and 10 minute when carried out at pH=4. Both times of half change are independent of the initial concentration of A. If the rate constant 'K' is given by (-d[A])/(dt) = K[A]^(X)[H]^(Y). The value of 'y' is

Answer»


Solution :The rate equation is `(-d[A])/(dt) = K[A]^(x)[H^(+)]^(Y)`
During any EXPERIMENT, ph is constant, hence
`(-d[A])/(dt) = K[A]^(y)` where `K^(.) =K[H^(.)]^(y)`
Given that - life is independent of the INITIAL concentration of A, hence x=1. Consequently K. is a first ORDER rate constant is given by
`K.= (0.693)/t_(1//2)`, therefore,
`((t_(1//2))_(2))/(t_(1//2))_(2) = K_(2)^(.)/K_(1)^(.) = (K[H^(+)]_(2)^(1))/(K[H^(+)]_(1)^(y)) = [H^(+)]_(2)^(y)/[H^(+)]_(1)^(y)`
`100/10 =(10^(-4))/(10^(-6)) RARR y=1`
Discussion
2.

The acid catalysed hydration of alkenes proceeds by formation of ……

Answer»

CARBENE INTERMEDIATE
CARBOCATION intermediate
Carbanion intermediate
FREE RADICAL intermediate

Solution :Carbocation intermediate
Discussion
3.

The acetylation of amines is carried out in the presence of a base

Answer»

STRONGER than the amine
as STRONG as the amine
weaker than the amine.
very WEAK base

Answer :A
Discussion
4.

The acetylation and benzoylation of phenol are called ………………. .

Answer»

Dow's process
Schotten-Baumann reaction
Reimer-Tiemann reaction
WILLIAMSON ETHER SYNTHESIS

SOLUTION :Reimer-Tiemann reaction
Discussion
5.

The acdicaqueoussolutionof ferrous ionforms a browncomplex in thepresence of NO_(3)^(Θ)by thefollowingtwosteps [Fe(H_(2)O)_(6)]^(2+) + NO_(3)^(Theta) + H^(o+)rarr ….+ [Fe(H_(2)O)_(6)]^(3+) + H_(2)O[Fe(H_(2)O)_(6)]^(3+) + ..... rarr ....+ H_(2)O Complex andbalance the equations.

Answer»

Solution :`3[Fe(H_(2)O)_(6)]^(2+) +NO_(3)^(Theta) +4H^(o+) RARR NO +3 [Fe(H_(2)O)_(6)]^(3+) +2H_(2)O`
`[Fe(H_(2)O)_(6)]^(2+) +NO rarr [Fe(H_(2)O)_(5)NO]^(2+) +H_(2)O`
Discussion
6.

The acetaldehydecan be prpared from acetyl chloride in the presence of Pd and BaSO_(4) this reaction is called

Answer»

rosenmund reaction
WOLF kishner REDUCTION
clemmensen reduction
PERKIN reaction

ANSWER :A
Discussion
7.

The acetylation and benzoylation of phenol are called ……………………….. .

Answer»

SOLUTION :Schotten-Baumann REACTION
Discussion
8.

The accumulation of molecualr species at the surface rather than in the bulk of a solid or liquid is termed adsorption. Write the mathematical expression of Freundlich adsorption isothem.

Answer»


Answer :`(x)/m=kp^((1)/N) or log(x)/m = (1)/nlogp+logk`.
`xrarr` Amount of gas absorbed by 'm' GRAM of the adsor bent at a pressure P. K, n are constants.
Discussion
9.

The accompanying represents a reversible camot cycle for an ideal gas: What is the entropy increase during isothemal expansion at 1000 K?

Answer»

`0.15 JK^(-1)`
`150 JK^(-1)`
`150 JK^(-1)`
`750 JK^(-1)`

SOLUTION :`DeltaS =q_("rev")/T =q_(H)/T_(n) =(150 xx 10^(3))/1000`
`=150 JK^(-1)`
Discussion
10.

The accumulation of molecualr species at the surface rather than in the bulk of a solid or liquid is termed adsorption. What is asorption isotherm?

Answer»


Answer :A PLOT between the AMOUNT of gas absorbed PER gram of absorbent (x/m) and the pressur of the gas at CONSTANT temperature is called Adsorption ISOTHERM.
Discussion
11.

The absorption maxima of several octahedral complex ions are as follows: He crystal field splitting is maximum for :

Answer»

`[CO(H_(2)O)_(6)]^(3+)`
`[Co(CN)_(6)]^(3-)`
`[Co(Nh_(3))_(6)]^(3+)`
All the complex ions have the same spliting

Answer :B
Discussion
12.

The absolute value of which of the following cannot be determined?

Answer»

EMF of the cell
Quantity of electricity
Single ELECTRODE potential
Free ENERGY change

Answer :C
Discussion
13.

The absolute temperature of a gas is increases 3 times. The root - mean square velocity of the molecules will be

Answer»

3 TIMES
9 times
`SQRT(3)` times
`1//3` times

Answer :C
Discussion
14.

The absolute enthalpy of neutralisation of the reaction MgO(s)+2HCl(aq)rarrMgCl_(2)(aq)+H_(2)O(l) will be

Answer»

Less than -57.33 kJ `mol^(-1)`
`-57.33 kJ mol^(-1)`
GREATER than `-57.33 kJ mol^(-1)`
`57.33 kJ mol^(-1)`

SOLUTION :Heat of neutralization will be less than `-57.33 kJ//"MOLE"` because some amount of this ENERGY will be required for the DISSOCIATION of weak base (MgO)
Discussion
15.

The absolute enthalpy of neutralisation of the reaction: MgO(s) + 2HCl(aq)toMgCl_2(aq) + H_2O(l)will be

Answer»

`-57.33 kJ "mol"^(-1)`
greater than `-57.33 kJ"mol"^(-1)`
less than `-57.33 kJ "mol"^(-1)`
`57.33 kJ "mol"^(-1)`

Solution : MgO is the oxide of weak BASE and we know that heat of neutralisation of 1 eq. of STRONG ACID with strong base is - 57.33 kJ/mol. With weak base some heat is ABSORBED in DISSOCIATION of weak base. Therefore, the heat of neutralisation of weak base with strong acid will be less than -57.33 kJ/mol.
Discussion
16.

The absolute confirgurations of the chiral centres 1,2 and 5 in the following molecule are

Answer»

1R, 2R, 5R
1S, 2S, 5S
1R, 2S, 5R
1S, 2R, 5S

Answer :C
Discussion
17.

The absolute enthalpy of neutralisation of the reaction MgO(s) + 2HCl(aq) to MgCl_(2)(aq) + H_(2)O(l) will be

Answer»

less than -57.33 KJ `"mol"^(-1)`
`-57.33 KJ "mol"^(-1)`
greater than `-57.33 KJ "mol"^(-1)`
57.33 KJ `"mol"^(-1)`

Solution :HEAT of neutralisation of strong acid and strong base is-57.33 KJ. MGO is weak base while HCL is strong acid. So the heat of neutralisation of MgO and HCl is lower than `-57.33 KJ ` because MgO requires some heat for ionisation, therefore the net RELEASED amount of heat is decreased.
Discussion
18.

The absolute configuration of the compound is …….

Answer»

E
R
S
Z

Solution :
Discussion
19.

The absolute configuration ofis …..

Answer»

R, S
S, R
S, S
R, R

SOLUTION :
Change S - R on the R.H.S CHIRAL carbon since H the atom of LOWEST PRIORITY is above the PLANE of the paper.
Discussion
20.

The absolute configuration of is :

Answer»

(2S , 3S)
(2R ,3 R)
(2 R , 3 S)
(2 S , 3 R)

Solution :
Discussion
21.

The absolute configuration of

Answer»

(2R, 3R)
(2R, 3S)
(2S, 3R)
(2S, 3S)

SOLUTION :
Discussion
22.

The absolute configuration of

Answer»

(2S, 3R)
(2S, 3S)
(2R, 3R)
(2R, 3S)

ANSWER :A
Discussion
23.

The absolute configuration of:

Answer»

(2R,3S)
(2S,3R)
(2S,3S)
(2R,3R)

Solution :.
At C-2
Without changing the position of `NH_(2)` group, rotate the remaining atoms/groups in ANTICLOCKWISE DIRECTION.

At C-3
REPEAT the same proceduct in this CASE also
.
Discussion
24.

The absolute configuration of -

Answer»

(2R, 3R)
(2R, 3S)
(2S, 3R)
2S , 3S)

ANSWER :C
Discussion
25.

The above reaction is known as

Answer»

CURTIUS reaction
Schmidt Reaction
Hoffmann's reaction
DOW's PROCESS.

Solution :Dow's process.
Discussion
26.

The above transformation can be done by using

Answer»

BAEYER's Reagent
Tollen's reagent
Pyridinium dichloromate in `CH_(2)Cl_(2)`
JONE's reagent

Answer :C
Discussion
27.

The above structure (I) is an example of ......alcohol.

Answer»

ALLYLIC TERTIARY
Secondary
Allylic secondary
Allylic primary

Answer :A
Discussion
28.

The above structures are related to each other as

Answer»

IDENTICAL substance
diastereomers
enantiomers
epimers

Answer :B
Discussion
29.

The above reaction is known as

Answer»

CURTIUS reaction
Schmidt reaction
Hoffmann's reaction
DOW's PROCESS

Solution :Dow's process.
Discussion
30.

The above reaction is called

Answer»

HOFFMANN MUSTARD OIL REACTION
Vilsmeyer reaction
Darzen reaction
Arndt Eistert synthesis.

Solution :Vilsmeyer reaction.
Discussion
31.

The above reaction is called

Answer»

WOHL Zeigler bromination
Streacker's synthesis
Tischenko REACTION
Diel's Alder reaction

Solution :This reaction is called Wohl Zeigler bromination (ALLYLIC bromination).
Discussion
32.

The above process in which alpha and beta form remain in equilibrium with acyclic form and a change in optical rotation is observed which is called as -

Answer»

MUTAROTATION
Epimerisation
Condensation
Inversion

Solution :The above PHENOMENON is CALLED as mutarotation
Discussion
33.

The above phenomenon can be explained as the basis of

Answer»

ionic POTENTIAL of MEAL
ELECTRON affinity
ionization potential
None of these

ANSWER :A
Discussion
34.

The above is a plot of binding energy per nucleon E_(b), against the nuclear mass M, A, B, C, D, E, F correspond to different nuclei. Consider four reactions (i) A + B rarr C + varepsilon "(ii)"C rarr A + B+ varepsilon (iii) D + E rarr F + varepsilon"and(iv) "F rarrD+ E + varepsilon Wherevarepsilon is the energy released ? In which reactions is  positive.

Answer»

(i) and (IV)
(i) and (III)
(II) and (iv)
(ii) and (iii)

ANSWER :A
Discussion
35.

The above pair represents

Answer»

ENANTIOMERS
diastereomers
identical compounds
positional isomers

Answer :C
Discussion
36.

The above conversion can be done by

Answer»

`CO+HC l //AlCl_(3)`
`HCN+HCl//AlCl_(3)`
`CHCl_(3)//NaOH`
All the above

Answer :D
Discussion
37.

High boiling point of water is due to :

Answer»

SMALL size
presence oflone pair of electronon OXYGEN
strong INTERMOLECULAR HYDROGEN bonding
presence of Vander Waal's forces

Solution :`H_2O` is associate molecule due to intermolecular H-bonding which increases B.P.
Discussion
38.

The above answer is correct because the chosen halide has

Answer»

Minimum IONIC character
Maximum ionic character
Highest oxidising POWER
LOWEST polarity

Answer :B
Discussion
39.

The ability of an ion to bring about coagulation of a given colloid depends upon

Answer»

the size of the ION
the SIGN of the CHARGE ALONE
the magnitude of the charge
both magnitude and charge

Solution :both magnitude and charge
Discussion
40.

The ability of an ion to bring about coagulation of a given colloid depends upon ...........

Answer»

its size
MAGNITUDE of its CHARGE
the SIGN of its charge
both the magnitude and sign of the charge

Solution :both the magnitude and sign of the charge
Discussion
41.

The ability of a substance to assume two or more crystalline structures is called

Answer»

Isomerism
polymerphism
ISOMORPHISM
AMORPHISM

ANSWER :B
Discussion
42.

The abbreviation PDI refers to:

Answer»

Name of the polymer
POLY dispersity index of polymer
Application
Poly DIAGONAL index

Answer :B
Discussion
43.

The ABCABCABC type of arrangement is reffered as_________ structure.

Answer»

OCTAHEDRAL CLOSE packed
hexagonal close packed
tetragonal close packed
cubic close packed

Answer :D
Discussion
44.

The A^(+) ion and B^(-) have radii 88 pm and 200 pm respectively. In the close packed crystal of compound AB, predict the coordination number of A^(+).

Answer»

SOLUTION :In the compound AB, A+ ions occupy the close packed lattice formed by `B^(-)`ions. The radius RATIO is:
`(r^(+))/(r^(-))= (88)/(200)= 0.44`
Since radius ratio LIES between 0-414 to 0.732.
`:.` Co-ordination of `A^(+) = 6`
Discussion
45.

The ""_(92)^(238) U disintegrates to give 4 alpha and 6 beta- particles . The atomic number of daughter element obtained is 10 x. x is ___

Answer»


Solution :`""_(92) U^(238) to ""_(Z) X^(A) + 4 (""_(2) He^(4)) + 6 (""_(-1) beta^(0)), 238 = 16+ A + 0 implies A = 238 - 16 = 222`
`92 = 8-6 + Z , 92 = 2 + Z implies Z = 90`
Discussion
46.

The ._(88)Ra^(226) is

Answer»

<P>N-mesons
u-mesons
Radioactive
Non-radioactive

Solution :`._(88)Ra^(226)` is radioactive because `(n)/(p)` ratio for it is 1.56 which is greater than 1.5
Discussion
47.

The ""_(92)^(235) U disintegrates to give 4 alpha and 6 alpha particles . The atomic number of the end product is

Answer»

92
96
84
90

Solution :`""_(92) U^(235) to ""_(Z)X^(A) + 4 (""_(2) He^(4)) + 6 (""_(-1) BETA^(0)) , 235 = A + 16 + 0 implies A = 235 - 16 = 219`
`92 = Z + 8 - 6 implies Z = 92 - 2 = 90`
Discussion
48.

The 8 : 8 type of packing is present in:

Answer»

CSCL
KCL
NACL
`MgF_(2)`

Answer :A
Discussion
49.

The 8 : 8 type of packing is present in :

Answer»

`MgF_(2)`
`CsCI`
KCI
NaCI

Solution :The 8 : 8 TYPE of packing is present in cesium chloride (CSCL). In this structure each `Cs^(+)` ION is surrounded byions and each ions is also surrounded by 8 `Cs^(+)`ions.
Discussion
50.

The ._(6)C^(14) and ._(6)C^(12) ratio in a piece of woods is 1//16 part of atmosphere. Calculate the age of wood. t_(1//2) "of" C^(14) is 5577 years?

Answer»


ANSWER :22308 YEAR
Discussion