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.

[Fe (NO_(2))_(3) Cl_(3)] and [Fe (O-NO)_(3) Cl_(3)] shows

Answer»

Linkage ISOMERISM
Geometrical isomerism
Optical isomerism
Hydrate isomerism

Solution :`[FE(NO_(2))_(3) Cl_(3)]` and `[Fe(O-NO)_(3)Cl_(3)]` contain an ambidentate ligand. HENCE, the compound EXHIBITS linkage isomerism.
Discussion
2.

Fe drops of conc. HNO_3 are added to group II filtrate before proceeding for group III in order to :

Answer»

MAKE acidic medium
convert `Fe^(2+) " to " Fe^(3+) `
convert `Fe^(3+) " to " Fe^(2+) `
boil off `H_2S` GAS from the filtrate .

Solution :Conc.`HNO_3` is added after group II to convert `Fe^(2+) " to " Fe^(3+)` so that it may be PRECIPITATED as `Fe(OH)_3`
Discussion
3.

Fe + conc. underset((gt80%))(HNO_(3))rarr X . Then X will be

Answer»

`Fe_(2)O_(3)`
`FEO`
`Fe_(3)O_(4)`
None of these

Answer :c
Discussion
4.

Fe CI_(3) gives positive test with

Answer»


Acetylacetone
Acetoacetic ester
All of the above

Answer :D
Discussion
5.

Fe, Al and Cr are grouped together in qualitative analysis, because

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these have three electrons in VALENCY shell.
their hydroxides are insoluble
their valency is three
their SULPHIDES are soluble in water

Solution :`M(OH)_(3)` are insoluble
Discussion
6.

Fe^(+3)oxidises NH_(2) to ........

Answer»

`NO_2`
`N_2O`
`N_2`
NO

Solution :`4FeCl_(3)+RNH_2OH RARR 4FeCl_(2)+N_(2)O+H_(2)O+4HCl` ( `Fe^(3+)` OXIDES `NH_2OH ` into `N_2O`)
Discussion
7.

Fe , Al and Cr are grouped together in qualitative analysis because :

Answer»

these have three electrons in the valence SHELL
their valency is three
their HYDROXIDES are insoluble in NH3
their sulphides are soluble in water .

Answer :C
Discussion
8.

Fe^(+2) reduces NH_(2)OH to ......

Answer»

`NH_(3)`
`N_3H`
`N_(2)_(4)`
`N_2`

Solution :`2FE(OH)_2+NH_2 OH +H_(2)Orarr 2Fe(OH)_3+ul(NH_3)` This shows unidising NATURE of hydroxylamine.
Discussion
9.

FCC structure possesses the most close packing of atoms in it.

Answer»

If both Statement- I and Statement- II are TRUE, and Statement - II is the correctexplanation of Statement– I.
If both Statement - I and Statement - II are true but Statement - II is not thecorrect explanation of Statement – I.
If Statement - I is true but Statement - II is FALSE.
If Statement - I is false but Statement - II is true.

ANSWER :A
Discussion
10.

Favourable conditions for physical adsorption are

Answer»

<P>LOW T, HIGH P
High T, high P
Low T, low P
Low T, low P

Answer :A
Discussion
11.

Favourable conditions for electrovalency are

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Low charge on ions, LARGE cation, small anion
High charge on ions, small cation, large AMINO
High charge on ions, large cation, small anion
Low charge on ions, small cation large amino

ANSWER :A
Discussion
12.

Fats, on alkaline hydrolysis , gives

Answer»

Oils
Soaps
Detergents
Glycol + acid

Solution :Fats are esters of higher fatty acids with GLYCEROL , hence on alkaline HYDROLYSIS they give back glycerol and SODIUM or POTASSIUM salt of acid (this is called soap).
`{:(CH_(2)OCOR""CH_(2)OH),(|""|),(CHOCOR+3NaOH to CHOH + underset("Soap")(3RCOONa)),(|""|),(underset("Fat")(CH_(2)OCOR)""CH_(2)OH):}`
Discussion
13.

Fats on alkaline hydrolysis give ………………………. .

Answer»

Oil+SOAP
Soap +GLYCOL
Soap +ESTER
Soap +GLYCEROL

SOLUTION :Soap +Glycerol
Discussion
14.

Fats contain higher percentage of

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UNSATURATED FATTY acids
saturated fatty acids
free fatty acids
glycerol

Answer :B
Discussion
15.

Fats are ester of

Answer»

sugar
glycerol
tributyrine
polypeptide

SOLUTION :Oil and FATS are glyceride ESTER of higher CARBOXYLIC acids. e.g. palmitin.`{:(CH_2OCOC_15H_31),(|),(CHOCOC_15H_31),(|),(underset"Palmitin (fat)"(CH_2OCOC_15H_31)):}`
Discussion
16.

Fats contain higher percentage of :

Answer»

UNSATURATED FATTY acids
Saturated fatty acids
Free fatty acids
Glycerol

Answer :B
Discussion
17.

Fats and oils are mixture of

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GLYCERIDES and SATURATED FATTY acids
Glycerides and UNSATURATED fatty acids
Glycerides of saturated and unsaturated fatty acids
Only saturated and unsaturated fatty acids

Answer :C
Discussion
18.

Fats and oils are formed from respectively.

Answer»

glycerol and long CHAIN UNSATURATED acids only
glycerol and long chain: saturated acids only
glycerol and long chain saturated acids and unsaturated acids
ethylene glycol and long chain unsaturated and saturated acids

Answer :C
Discussion
19.

Fats and oils are

Answer»

ALDEHYDES and ketones
esters
acids
alcohols

Answer :B
Discussion
20.

Fats and oils.

Answer»

Solution :The NATURALLY OCCURRING triesters of glycerol in which all the three `-OH` groups are esterified with long chain fatty acids are called FATS and oils.
They are ALSO called triacylglycerol (TAG) or triglycerides.
Discussion
21.

Fats and oils are :

Answer»

Polyhydric alcohols
Solid and liquid HYDROCARBONS
LIPIDS
Weak CARBOXYLIC acids

ANSWER :C
Discussion
22.

"Fat"overset("Hydrolysis")rarr"Carboxylic acid "+" Alcohol." Which of the following will make the reaction to occur very rapidly ?

Answer»

LIPASE
Emulsin
INVERTASE
MALTASE

SOLUTION :Lipase
Discussion
23.

Fats and oil are formed from

Answer»

Giycerol and long CHAIN unsaturated acids only
Glycerol and long chain SATURATED acids only
Glycerol and long chain saturated and unsaturated acids
Ethylene glycol and long chain saturated and unsaturated acids

Answer :C
Discussion
24.

Fat consists of

Answer»

monohydroxy carboxylic acid
monohydroxy aliphatic carboxylic acid
monohydroxy, aliphatic, saturated carboxylic acid
dihydroxy aliphatic carboxylic acid

Solution :FAT CONSIST of monohydroxy, aliphatic, saturated carboxylic acid (where a chain has more than 12 C atoms.)
Discussion
25.

Fast and oils serve as :

Answer»

RESERVE FOOD for the body
Immediate SOURCE of energy
Nitrogeneneous food
Control matrial of METABOLISM

ANSWER :A
Discussion
26.

Fats and oils belong to the class of:

Answer»

ALCOHOLS
Acids
Esters
Hydrocarbons

Answer :C
Discussion
27.

Farnesence is a compound found in the waxy coating of apples. On hydrogenation it gives 2,6,10-Trimethyl dodecane. On ozonolysis it gives one mole acetone, one mole of formoaldehyde, one mole of 2-Methylpentanedial and one mole of 4-Oxopentanal. The structure proposed for Farnesence may be

Answer»




SOLUTION : , `OVERSET(O_(3)//ZN)RARR 2-"methylpentanedial+4-oxopentanal+formaldehyde"`.
Discussion
28.

Faraday's laws of electrolysis are related to the

Answer»

Atomic NUMBER of the reaction
Atomic number of the anion
Equivalent weight of the electrode
Speed of the cation.

Solution :`(W_(1))/(W_(2))=(E_(1))/(E_(2))=(Z_(1)It)/(Z_(2)It) therefore(Z_(1))/(Z_(2))=(E_(1))/(E_(2))`.
Here `E_(1) and E_(2)` are equivalent weight of the IONS.
Discussion
29.

Faraday's law of electrolysis is not applicable when

Answer»

temperature is INCREASED
INERT electrodes are used
a mixture of electrolytes is used
in case of NON electrolyte

SOLUTION :Faraday's laws are independent of all other external factors and not applicable for non-electrolyte .
Discussion
30.

Faraday's laws hold good at

Answer»

all PRESSURES
only at 298 K
indifferent solvents
only at 1 atm pressure

Solution :Faraday's laws are independent of EXTERNAL factors .
Discussion
31.

Faraday.s laws hold good at:

Answer»

All pressures
Only at 298 K
In DIFFERENT solvents
All of these

ANSWER :A
Discussion
32.

Faraday.s law of electrolysis fails when:

Answer»

TEMPERATURE is increased
Inert ELECTRODES are used
A MIXTURE of ELECTROLYTES is used
In NONE of these cases

Answer :D
Discussion
33.

Faraday's laws of electrolysis are related to

Answer»

ATOMIC NUMBER of the cation
atomic number of the anion
equivalent WEIGHT of the electrolyte
speed of the cation

Answer :C
Discussion
34.

Faraday's first law of electrolysis can be expressed as

Answer»

`W alpha Q`
`W alpha 1//Q`
`W alpha Q^(2)`
`W alpha Q^(3)`

SOLUTION :When Q is charge FLOWN through electrolyte and W is the AMOUNT deposited .
Discussion
35.

Faraday's law of electrolysis are related to

Answer»

ATOMIC NUMBER of the cation
Atomic number of the anion
Equivalent WEIGHT of the electroyte
Speed of the cation

Answer :C
Discussion
36.

Faraday's first law of electrolysis can be expressed as :

Answer»

`W PROP Q`
`W prop 1//Q`
`W prop Q^2`
`W prop Q^3`

ANSWER :A
Discussion
37.

Faraday constant is defined as

Answer»

Charge CARRIED by 1 electron
Charge carried by one mole of ELECTRONS
Charge required to DEPOSIT one mole of substance
Charge carried by TWO moles of electrons

Solution :1F is the charge carried by 1 mole of electrons.
Discussion
38.

Faradays constant is defined as

Answer»

CHARGE CARRIED by 1 electron
charge carried by one mole of electrons
charge required to DEPOSIT one mole of substance
charge carried by`6.22 times 10^(10)` electrons.

Answer :B
Discussion
39.

Faraday has the dimension of

Answer»

Coulombs
Coulomb equivalent
Coulomb PER equivalent
Coulomb per DEGREE kelvin

Answer :C
Discussion
40.

Faraday is equal to :

Answer»

`96.5` COULOMB` "equivalent"^(-1)`
`96.5xx10^3` coulomb `"equivalent"^(-1)`
`96.5xx10^(10)` coulomb `"equivalent"^(-1)`
`96.5xx10^(23)` coulomb `MOL^(-1)`

Answer :B
Discussion
41.

Faraday.

Answer»

Solution :It is DEFINED as the quantity of the ELECTRIC charge on one MOLE of electrons. It has value, 1 F = 96500 C/mol.
Discussion
42.

False statement is

Answer»

Chloroform is heavier than WATER
`C Cl_(4)` is non-inflammable
Vinyl chloride is more reactive than allyl chloride
`Br^(-)` is a good NUCLEOPHILE as compared to `I^(-)`

Solution :Vinyl chloride is less reactive than allyl chloride DUE to resonance effect.
ORDER of nucleophilicity amongst the halide ion are as
`I^(-) GT Br^(-) gt Cl^(-)`.
Discussion
43.

Factor that can affect the rate of a chemical reaction between a solid and a solution include all of the following except the :

Answer»

CONCENTRATION of the REACTANTS in solution
volume of the container.
size of the SOLID particles.
temperature.

Answer :C
Discussion
44.

Factose is a :

Answer»

reducing sugar
disaccharide
component of MILK PRODUCTS
COMPOUNDS capable of SHARING mutarotation

Solution :N//A
Discussion
45.

Facial and meridionalisomerism will be exhibitedby:

Answer»

`[CO(NH_(3))_(3)Cl_(3)]`
`[Co(NH_(3))_(4)Cl_(2)]CL`
`[Co(en)_(3)]Cl_(3)`
`[Co(NH_(3))_(5)Cl]Cl_(2)`

Solution :Complexes of the type `MA_(3)B_(3)` exhibit MERIDIONAL and facial ISOMERISM.
Discussion
46.

Face centred cubic crystal lattice of copper has density of 8*966" g. cm"^(-3). Calculate the volume of the unit cell. Given molar mass of copper is 63*5"g. mol"^(-1) and Avogadro number N_(A) is 6*022xx10^(23)"mol^(-1).

Answer»

Solution :Density of the unit CELL
`(d)=("Maxx of the unit cell"(M))/("Volume of the unit cell"(V))`. . . (i)
Let number of atoms in the unit cell be Z and mass of each atom m then `M=Zxxm`
Mass of each atom
`(m)=("Atomic mass")/("Avogadro's no.")`
Thus from equation (i)
`d=(ZXX"Atomic mass")/("Avogadro's no." xx V)` . . . .(ii)
For FACE centered cubic lattice (FCC), the TOTAL no. of atoms in the unit cell
`(Z)=(1)/(8)xx8+(1)/(2)xx6`
`=1+3=4`
So, `8*966"g cm"^(-3)=(4xx63*5"g mol"^(-1))/(6*022xx10^(23)"mol^(-1)xxV)`
or, Volume of the unit cell
`(V)=(254)/(53*99xx10^(23))cm^(3)`
`=4*7xx10^(-23)cm^(3)`
Hence, the volume of the unit cell is `4*7xx10^(-23)cm^(3)`.
Discussion
47.

Face - centred cubiccrystallattics of copperhasdensityof 8.966g cm^(-3). Calculatethe volumeof the unitcell . Givenmolarmass ofcopperis 63. 5 g mol^(-1)and AvogadronumberN_(A)is 6.022xx 10^(23) mol^(-1)

Answer»

SOLUTION :volumeof unitcell= 4.702 `xx 10^(23) CM^(-3)`
Discussion
48.

Fac-mer isomerism is shown by……… .

Answer»

`[Co(EN)_(3)]^(3+)`
`[Co(NH_(3))_(4)(Cl)_(2)]^(+)`
`[Co(NH_(3))_(3)(Cl)_(3)]`
`[Co(NH_(3))_(5)Cl]SO_(4)`

ANSWER :C
Discussion
49.

Fac-Mer isomerism is associated with which one of the following complexes ? (M = central metal)

Answer»

`[M(A A)_(2)]`
`[MA_(3)B_(3)]`
`[M(A A)_(3)]`
`[MABCD]`

ANSWER :B
Discussion
50.

Fac-mer isomerism is shown by

Answer»

`[Co(en)_(3)]^(3+)`
`[Co(NH_(3))_(4)(CL)_(2)]^(+)`
`[Co(NH_(3))_(3)(Cl)_(3)]`
`[Co(NH_(3))_(5)Cl]SO_(4)`

Solution :`[Co(NH_(3))_(3)(Cl)_(3)]`
Discussion