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.

On cracking petrol we get

Answer»

`CH_4`
`C_3H_6`
Both A and B
`CH_3+CH_4+C_2H_6+ " ALCOHOLS "`

ANSWER :C
Discussion
2.

On cooling hot zinc oxide, the yellow colour become white. Why ?

Answer»

Solution :The composition of `Zn^(2+) and O^(2-)` ions in STOICHIOMETRIC `ZnO` is `n : n`.
On heating ZnO, oxygen is lost reversibly.
A non - stoichiometric COMPOUND with excess `Zn^(2+)`ions is formed possessing a composition of `Zn^(2+) and O^(2-)` ions `n:n-x`.
Due to excess METAL in the interstities, it is YELLOW.
On cooling the non-stoichiometric zinc OXIDE, it gains oxygen. A stoichiometric zinc oxide is formed and hence it is white.
`Zn^(2+) + (1)/(2) O_(2) 2e^(-) to ZnO`
Discussion
3.

On conversion into the Grignard reagent followed by treatment with absolute ethanol, how many isomeric alkyl chlorides would yield 2-methylbutane :

Answer»

2
3
4
5

Answer :C
Discussion
4.

On complete oxidation , 1 mole of an organic compound gave 4 moles of water. The compound should be

Answer»

METHANOL
propyne
ethylene
but-1-ene

Solution :N/A
Discussion
5.

On complete hydrogenation, natural rubber produces

Answer»

ethylene-propylene COPOLYMER
vulcanised RUBBER
polyproylene
polybutylene .

SOLUTION :
Discussion
6.

On complete hydrogenation , natural rubber produces.

Answer»

ethylene - PROPYLENE copolymer
vulcanished RUBBER
POLYPROPYLENE
polybutylene

Solution :
Discussion
7.

On commercial scale, phenol is obtained from chlorobenzene by raschig's process. Which one of the following is Raschig's process?

Answer»




ANSWER :D
Discussion
8.

On combustion of hydrogen in hydrogen fuel cell. . .

Answer»

More pure water generated.
Potential difference PRODUCED between two electrodes.
More heat is produced.
Not given

Solution :Potential difference produced between two electrodes.
Note : `H_(2)O` is FORMED by combustion of `H_(2)` with the help of `O_(2)` in fuel ELL. But the objective of this cell is to convert combusted ENERGY into electrical energy. So potential difference should be generated. So it is right.
Discussion
9.

On combustion of 0.2475 g of an organic compound 0.4950 g of CO_(2) and 0.2025 g of H_(2)O were obtained. Find out the percentage of hydrogen.

Answer»


Solution :`%"HYDROGEN"=(2 XX "WEIGHT of" H_(2)O xx 100)/(18 xx "weight of organic compound")`
Discussion
10.

On combustion, carbon forms two oxides CO and CO_(2), Heat of formation of CO_(2) gas is 94.3 kcal and that of CO is 26.0 kcal. Heat of combustion of carbon is

Answer»

26.0 kcal
`- 94.3` kcal
68.3 kcal
`- 120.3` kcal

Solution :`C+O_(2)rarrCO_(2),DeltaH=-94.3`
This is ALSO heat of FORMATION of `CO_(2)`
`C+(1)/(2)O_(2)rarrCO, DeltaH=-26.0`.
Discussion
11.

On combustion carbon forms two oxides CO and CO_(2). Heat of formation of CO_(2) is -393.5kJ and that of CO is -110.5kJ. Heat of combustion of CO is

Answer»

`-393.5kJ`
`-504.0kJ`
`-283.0kJ`
283.0kJ

Solution :(i) `C + (1)/(2)O_(2)(g) rarr CO, Delta H = - 110.5 kJ`
(ii) `C + O_(2)(g) rarr CO_(2), Delta H = - 393.5 kJ`
SUBTRACT EQ. (i) from eq. (ii),
`CO + (1)/(2)O_(2) rarr CO_(2), Delta H = - 283.0 kJ`
Discussion
12.

On charging lead storage cell. . .

Answer»

Pb of electrode utilize.
Solution becomes diluted
`H_(2)SO_(4)` of solution is utilized.
`PbO_(2)` is DEPOSITED on one electrode.

Solution :On charging of lead storage CELL, `PbO_(2)` is obtained at anode on oxidation REACTION.
`PbSO_(4(S))+2H_(2)O_((l))underset("Charging")overset("DC voltage") toPbO_(2(S))+SO_(4(aq))^(2-)+4H_((aq))^(+)+2e^(-)`
Where, (A), (B) and (C) is INDICATION of discharging reaction of storage cell.
Discussion
13.

On bromination, propionic acid yields two isomeric 2-bromopropionic acids. This pair is an important example of

Answer»

CHAIN isomers
Optical isomers
Cis-trans isomers
Position isomers

Answer :B
Discussion
14.

On catalytic hydorgenation how many isomeric alkene will give 2-Methyl butane.

Answer»


SOLUTION :`CH_(3)-underset(CH_(2))underset(||)(C )-CH_(2)-CH_(3), CH_(3)-underset(CH_(3))underset(|)(C )-CH-CH_(3), underset(CH_(3))underset(|)(CH )-CH=CH_(2)`
Discussion
15.

On calculating the strength of current in amperes if a charge of 840 C (coulomb) passes through an electrolyte in 7minutes, it will be

Answer»

1
2
3
4

Answer :B
Discussion
16.

On bombardin g._(8)O^(16) with deutrons, the nuclei of the product formed will be

Answer»

`._(9)F^(18)`
`._(9)F^(17)`
`._(8)O^(17)`
`._(7)N^(14)`

Solution :`._(8)O^(16) + ._(1)H^(2) rarr ._(9)F^(18)`
Discussion
17.

On boiling with concentrated hydrobromic acid, phenyl ethyl ether will yield

Answer»

PHENOL and ETHYL bromide
Phenol and ethane
Bromobenzene and ethanol
Bromobenzene and ethane

Solution :
Discussion
18.

On boiling with concentrated HBr phenyl ethyl ether will give.

Answer»

Phenol and ethyl BROMIDE
Bormobenzene and ethanol
Phenol and ETHANE
Bromobenzene and ethane.

Solution :`C_6H_5-O-CH+`conc. `Hbroverset(DELTA)toC_6H_5-OH+CH_2Br`
Discussion
19.

On boiling with concentrated hydrobromic acid pheyl ethyl ether yields:

Answer»

PHENOL and bromide
Bromobenzene and ethanol
Phenol and ethane
Bromobenzene and ethane

Answer :A
Discussion
20.

On boiling with concentrated HBr, phenyl ether will give:

Answer»

Phenol and ETHYL bromide
Bromobenzene and ethanol
Phenol and ETHANE
Bromobezene and ethane.

SOLUTION :Br ATOM in HBr prefers to combine with smaller gorup `(C_(2)H_(5))` to avoid steric hidracne.
Discussion
21.

On boiling tin with alkali, the product is :

Answer»

`SnO_3^(2-)`
`SnO_2`
`SN(OH)_4`
`Sn_2O_3^(2-)`

ANSWER :D
Discussion
22.

On boiling the egg, what structural changes are taking place in the egg white ?

Answer»

The colour of the EGG changes from colourless to white.
`2^(@)` and `3^(@)` structures are destroyed but `1^(@)` structure remains intact.
`1^(@),2^(@)and 3^(@)` structures of egg are destroyed.
A reversible change TAKES PLACE which can be reversed by DECREASING the temperature .

Solution :During denaturation of proteins, `2^(@) and 3^(@)` structures are destroyed but primary structure remainsintact.
Discussion
23.

On boiling an aqueous solution of KClO_3 with iodine the product formed is:

Answer»

`KIO_3`
`KClO_4`
`KIO_4`
KCl

Answer :A
Discussion
24.

On being strongly heated, a blue salt leaves a black residue. Which of the following cations can be present in the salt?

Answer»

`FE^(2+)`
`Fe^(3+)`
`Cu^(2+)`
`ZN^(2+)`

ANSWER :C
Discussion
25.

On being strongly heated, which of the following substances will leave a black residue

Answer»

`CuSO_(4), 5H_(2)O`
`ZnCO_(3)`
`PbCO_(3)`
`MnSO_(4)`

SOLUTION :CuS formed
Discussion
26.

On being slowly passed through water, PH_(3) forms bubbles but NH_(3) dissolves. Why is it so?

Answer»

Solution :N-H bond is more polar than P-H bond. HENCE, `NH_(3)` forms hydrogen BONDS with `H_(2)O` MOLECLES and hence dissolves in it whereas `PH_(3)` does not dissolve and hence forms bubbles.
Discussion
27.

On being slowly passed through water, PH_3 forms bubbles but NH_3 dissolves. Why is it so ?

Answer»

Solution :Ammonia DISSOLVES in water due to hydrogen BONDING. Phosphine being LESS soluble, escapes out in the form of BUBBLES.
Discussion
28.

On being heated with conc. HNO_(3) and ammonium molybdate solution, the salt solutions gives a yellow precipitate. The salt(s) may be

Answer»

`Na_(2)HPO_(4)`
`As_(2)O_(3)`
`FeSO_(4)`
`BaCl_(2)`

Solution :`Na_(2)HPO_(4)` FORMS YELLOW COLOURED AMMONIUM phosphomolybdate precipitate `[(NH_(4))_(3)PO_(4).12MoO_(3)].As_(2)O_(3)` forms yellow precipitate of ammonium arsemomolybdate `[(NH_(4))_(3) AsO_(4). 12MoO_(3)]`
Discussion
29.

On being heated with alcoholic KOH, neopentyl bromide gives mainly :

Answer»

but-2-ene
2-methyl but-1-ene
2-Methyl but-2-ene
2, 2-dimethyl but-1-ene

ANSWER :C
Discussion
30.

On being heated in oxygen, 3.120 g of a metal M converts to 4.560 g of oxide (atomic weight of M= 52.0). Mark the correct statement(s).

Answer»

Equivalent wt. of metal M= 17.33
Number of EQUIVALENTS of oxygen reacted with metal = 0.09
Metal M FORMS halide `MCl_(2)`
The simplest formula of the metal oxide which it forms is `M_(2)O_(3)`

SOLUTION :Weight of oxygen reacted
`=4,560 = 3,120 = 1,440`
No. of equivalents = `(1,440)/8 = 0.18`
Equivalent weight of metal `=3.12/1.44 xx 8 = 17.33`
Valency `=52/17.33 =3`
Discussion
31.

On applying pressure to the equilibrium, ice hArr water which phenomenon will happen:

Answer»

More ICE will be formed
More WATER will be formed
Equiliberium will not be disturbed
Water will equilibrium

Answer :B
Discussion
32.

On analysis, a substance was found to have the following percentage composition: K=31.84, Cl=28.98 and O=39.18 Calculate its molecular formula if its molecular mass is 122.5.

Answer»

SOLUTION :`KClO_(3)`
Discussion
33.

On analysis of an impure sample of sodium chloride. The percentage of chlorine was found to be 45.5 What is the percentage of pure sodium chloride in the given sample?

Answer»

SOLUTION :The MOLECULAR mass of pure sodium chloride (NACL)
=At. Mass of Na+ At. Mass of chlorine.
=(23+35.5)=58.5
% of chlorine in pure NaCl
`=(35.5)/(58.5)xx100=60.6`
Thus,
% of PURITY of NaCl in the sample `=(45.5)/(60.6)xx100=75`.
Discussion
34.

On analysis, a sample of uranium was found to contain 0.277g of ""_(82)Pb^(20) and 1.667g of ""_(92)U^(238). The half-life period of ""_(92)U^(238) is 4.51xx10^(9) years. If all the lead were assumed to have come from decay of ""_(92)U^(238), what is the age of the earth?

Answer»


ANSWER :`1.143xx10^(9)` YEARS
Discussion
35.

On analysing an impure sample of sodium chloride, the percentage of chlorine was found to be 45.5. What is the percentage of pure sodium chloride in the sample?

Answer»

SOLUTION :Molar MASS of pure NaCl = `23+35.5=58.5"g mol"^(-1)`
`% " of Cl in pure NaCl "=(35.5)/(58.5)xx100=60.6`
Thus, 60.6 parts of Cl are PRESENT in 100 parts of pure NACl
45.5 parts of Cl will be present in pure NaCl `=(100)/(60.6)xx45.5="75 parts"`
The remaining `25%` is impurity present in impure sample.
HENCE, `%` of pure NaCl in impure sample = `75%`.
Discussion
36.

on aldol condensation followed by heating gives

Answer»




SOLUTION :
Discussion
37.

On addition of small amount of KMnO_(4) to concentrated H_(2)SO_(4) , a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following :

Answer»

`Mn_(2)O_(7)`
`MnO_(2)`
`MnSO_(4)`
`Mn_(2)O_(3)`

Solution :`2KMnO_(4) + 2H_(2)SO_(4) ( "CONC.") rarr Mn_(2)O_(7) + 2KHSO_(4) + H_(2)O`
Discussion
38.

On addition of small amount of KMnO_(4) toconcentrated H_(2)SO_(4), a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following

Answer»

`Mn_(2)O_(7)`
`MnO_(7)`
`MnSO_(4)`
`Mn_(2)O_(3)`

Solution :`2KMnO_(4) + underset("(conc.)")(2H_(2)SO_(4)) rarr underset("(Explosive)")underset("Green OILY compound")(Mn_(2)O_(7)) + 2KHSO_4 + H_2O`
Discussion
39.

On addition of small amount of KMnO_(4) to concentrated H_(2)SO_(4), a green oily compound is obtained which is highly explosive in nature. Identify the compound from the following.

Answer»

`Mn_(2)O_(7)`
`MnO_(2)`
`MnSO_(4)`
`Mn_(2)O_(3)`

Answer :a
Discussion
40.

On addition of one ml solution 10% NaCl to 10 ml gold sol in the presence of 0.0250g of starch , the coagulation is just prevented starch has the gold number :

Answer»

`2.5`
25
`0.25`
250

Solution :Gold no. `=` amount of protective colloid in MG to be added to `10ml` of gold sol to PREVENT its COAGULATION on add of `1 ml` of `10% NaCl` solution of Gold no. `=0.0250xx1000=25`
Discussion
41.

On addition of one mL solution of 10% NaCl to 10 mL gold sol in the presence fo 0.0250 g of starch, the coagulation is just prevented. Starch has the following gold number

Answer»

`0.025`
`0.25`
`2.5`
`25.`

SOLUTION :AMOUNT of starch in mg that prevent COAGULATION by 1 ml of `10%` NaCl solution `=0.025xx1000=25.` Hence, gold number =25.
Discussion
42.

On addition of one mL of 10% NaCl solution to 10 mL gold sol in the presence of 0.25 g of starch, the coagulation is just prevented. Starch has the gold number

Answer»

0.025
0.25
2.5
none of these.

Solution :By DEFINITION, gold NUMBER of starch is the amount of starch in MILLIGRAM added to 10 mL standard gold solution which prevents the coagulation of gold on adding 1 mL of 10% NaCl solution. So the amount of starch is 0.25 g = 250 mg. Hence, gold number is 250.
Discussion
43.

On addition of increasing amount of AgNO_(3)(g)to 0.1 M each of NaCl and NaBr in a solution, what % of Br^(-)ion gets precipitated when Cl^(-)ion starts precipitating? Given: K_(sp)(AgCl) = 1.0 xx 10^(-10), K_(sp) (AgBr) = 1 xx 10^(-13).

Answer»


SOLUTION :`[AG^(+)]` required for commencement of percipitation of AgCl `=(K_(sp)(AgCl))/([Cl^(-)])= (1.0 xx 10^(-10))/0.1`
`=1.0 xx 10^(-9)` M
`[Br^(-)]` remaining at this stage
% of `Br^(-)` remaining unpredicted
% of `Br^(-)` precipitated =100-0.1 = 99.9
Discussion
44.

On addition of excess of sodium hydroxide solution to stamnous chloride solution, we obtains:

Answer»

`Sn(OH)_(2)`
`SnO_(2),H_(2)O`
`Na_(2)SnO_(2)`
None of these

Solution :`SnCl_(2)+2NaOH to Sn(OH)_(2)+2NaCl`
`Sn(OH)_(2)+2NaOH to Na_(2)SnO_(2) +2H_(2)O`
Discussion
45.

On addition of cone. H_2SO_4 to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because ……...

Answer»

`H_2SO_4` REDUCES HI to `I_2`
HI is of violet colour
HI gets oxidised to `I_(2)`
HI changes to `HIO_(3)`

Solution :`H_(2)SO_(4)` oxidising HI to `I_(2)`
`2Nal + H_(2)SO_(4) to Na_(2)SO_(4) + underset(2H_(2)O + SO_(2) + I_(2))underset(darr H_(2)SO_(4))(2HI)`
Discussion
46.

On addition of conc. H_2SO_4 to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because

Answer»

`H_2SO_4` REDUCES HI to `I_2`
HI is of VIOLET colour
HI gets oxidised to `I_2`
HI changes to `HIO_3`

Solution :HI being a stronger reducing AGENT than `H_2SO`., reduces `H_2SO_4 " to " SO_2`is itself oxidised to I. On the other hand, HCL is RELATIVELY weaker reducing agent and docs not reduce `H_2SO_4 " to " SO_2`and itself is not oxidised to `Cl_2.`
Discussion
47.

On addition of conc. H_(2)SO_(4) to a chloride salt, colourless fumes are evolved but in case of iodide salt, fumes come out. This is because

Answer»

`H_(2)SO_(4)` reduces `HI` to `I_(2)`
`HI` is of VIOLET colour
`HI` gets oxidised to `I_(2)`
`HI` CHANGES to `HIO_(3)`

Solution :`HI` being a stronger reducing agent than `H_(2)SO_(4)` reduces `H_(2)SO_(4)` to `SO_(2)` and itself gets oxidised to `I_(2)`.
Discussion
48.

On addition of conc. H_(2)SO_(4) to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because

Answer»

`H_(2)SO_(4)` reduces HI to `I_(2)`
HI is of violet colour
HI gets oxidised to `I_(2)`
HI changes to `HIO_(3)`

SOLUTION :HI being a stronger reducing AGENT than `H_(2)SO_(4)`, reduces `H_(2)SO_(4)` to `SO_(2)` and is itself oxidised to `I_(2)`.
Discussion
49.

On addition of an inert gas at constant volume to the reaction N_(2)+3H_(2)hArr2NH_(3) at equlibrium

Answer»

The reaction remains unaffected
FORWARD reacton is favored
The reaction halts
Backward reaction is FAVOURED

Solution :Addition of an inert gas of constant VOLUME CONDITION to an equlibrium has no EFFECT.
Discussion
50.

On addition of aqueous NaOH to a salt solution, white gelatinous precipitate is formed, which dissolves in excess alkali. The salt solution contains

Answer»

CHROMIUM ions
Aluminium ions
Barium ions
Iron ions

Solution :`AL^(+3)+(OH^(-))_(3)to UNDERSET(("WHITE ppt."))(Al(OH)_(3)`
Which is soluble in excess alkali.
Discussion