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.

Purification of aluminium by electrolytic refining is carried out by :

Answer»

Hoope process
Hall process
Baeyer process
Serpeck process

Solution :Hoope's process is a metallurgical process used to obtain the aluminium metal of very high purity by electrolytic refining. In this process, alumunium ions from the FUSED mixture are discharged which at the cathode as pure metal rises to the surface while impurities remain at the BOTTOM. The aluminium metal OBTAINED by this process is `99.98%` pure.
2.

Purification of alumina is essential because :

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Impure ALUMINA is a very POOR conductor of electricity
Impure alumina has a very HIGH MELTING point
Impure alumina cannot react with the oxidizing agent
It is difficult to purify it purify aluminium metal

Answer :D
3.

Purest form of iron is cast iron. True or False

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SOLUTION :WROUGHT IRON
4.

Purest form of silica is :

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Quartz
Flint
Sandstone
Kieselguhr

Answer :A
5.

Purest form of iron is called as

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SOLUTION :WROUGHT IRON
6.

Write the name of purest form of iron.

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CAST iron
Wrought iron
Steel
Pig iron

Answer :B
7.

Purest form of iron is

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CAST iron
wrought iron
steel
pig iron

Answer :B
8.

Purest form of iron is :

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Cast IRON
HARD STEEL
Stainless steel
WROUGHT iron

Answer :D
9.

Pure water is kep in a vessel and it remains exposed to atmospheric CO_(2) which is absorbed, then its pH will be

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GREATER than 7
Less than 7
7
Depends on IONIC product of water

Solution :`CO_(2)` is acidic OXIDE which on dissolution in water develops acidic nature.
10.

Pure water is saturated with pure solid AgCl, a silver electrode is placed in the solution and the potential is measured against normal calomel electrode at 25^(@)C. This experiment is then repeated with a saturated solution of Agl. If the difference in potential in the two cases is 0.177V. What is the ratio of solubility product (solubility) of AgCl and Agl at the temperature of the experiment?

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`10^(3)`
`10^(6)`
`10^(2)`
`10^(2)`

SOLUTION :`AgtoAg^(+)+e^(-)`
`E_(1)+E_("oxid")+E_("calomel")`
`=E'-(0.0591)/(1)logK_(sp_(1))+E_("calomel")`
`E_(2)=E'-(0.0591)/(1)logK_(sp_(2))+E_("calomel")`
`E_(2)-E_(1)=0.177=0.0591log((K_(sp_(1)))/(K_(sp_(2))))`
`(K_(sp_(1)))/(K_(sp_(2)))=10^(3)`
11.

Pured Cl_2 is prepared on heating

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NACL
`PtCl_4`
`CuCl_2`
All

Answer :B
12.

Pure water is saturated with pure solid AgCl, a silver electrode is placed in the solution and the potential is measured agains normal calomet electrode at 25^(@)C. This experiment is then repeated with a saturated solution of AgI. If the difference in potential in the two cases is 0.177V. What is the ratio of solubility product (solubility) of AgCl and AgI at the temperature of the experiment?

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`10^(3)`
`10^(6)`
`10^(2)`
`10^(4)`

Answer :A
13.

Pure water is kept in a vessel and it remains exposed to atmospheric CO_2 which is absorbed, then its pH will be…….

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greater than 7
less than 7
equal to 7
depends on ionic PRODUCTION of WATER

Solution :`CO_2` is acidic OXIDE which on DISSOLUTIONS in water DEVELOPS acidic nature.
14.

Pure water freezes at 273 K and 1 bar. The addition of 34.5g of ethanol to 500g of water changes the freeing point of the solution. Use of f.p. depression constant of water as 2Kg mol^(-1). The figures shown below represents plot of vapour pressure (V.P.) versus temperature (T) (mol wt. of ethanol =46 g mol^(-1)). Among the following the option representing change in the f.p. is

Answer»




ANSWER :A
15.

Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression contant of water as 2 K kg mol^(-1). The figures shown below represent polts of vapour (V.P.) versus temperature (T). [Molecular weight of ethanol is 46g mol^(-1)] Among the following, the option representing change in the freezing point is

Answer»




ANSWER :A
16.

Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water at "2 K kg mol"^(-1). The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T) [Molecular weight of ethanol is "46 g mol"^(-1)]. Among in the freezing point is

Answer»




Solution :As T increases, V.P. increases. HENCE, option (c) and (d) are WRONG.
`DeltaT_(F)=K_(f)xxm`
`Delta_(f)^(@)-T_(f)=2xx(34.5//46)/(0.5)`
`273-T_(f)=3 or T_(f)=270`.
Hence, only option (a) is correct.
17.

Pure water freezes at 273 K and 1 bar. The addition of 34.5 g of ethanol to 500 g of water changes the freezing point of the solution. Use the freezing point depression constant of water as 2 K kg mol^(-1).The figures shown below represent plots of vapour pressure (V.P.) versus temperature (T). [molecular weight of ethanol is 46 g mol^(-1) Among the following, the option representing change in the freezing point is

Answer»





Solution :MOLE of `C_(2)H_(5)OH=34.5/46`
`m=34.5/46xx1000/500`

`DeltaT_(f)=34.5/46xx1000/500xx2=3`
`T_(f)^(@)-T_(f)=273-T_(f)=3`
`T_(f)=273-3=270K`
18.

Pure water does notconduct electricity because it is

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BASIC
almost not IONIZED
DECOMPOSED EASILY
acidic

Answer :C
19.

Pure water dissociates to a small extent as per equilibrium 2H_2O_((i)) hArr H_3O_((eq))^(+) +OH_((eq))^(-) The pH of pure water at 298 K is 7, what will be pH of pure water at 310 K?

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0
`LT 7`
`GT 7`
`7`

Answer :B
20.

Pure water does not conduct electricity because it

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SOLUTION :Because it does not have IONS.
21.

Pure water does not conduct electricity because it is :

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NEUTRAL liquid
Low BOILING B. PT. liquid
Almost non-ionised
None

Answer :C
22.

Pure sulphur was burnt. the gaseous products are SO_(2)=60% (mol), SO_(3)=20% (mol) and O_(2)=20% (mol). If initially 50 moles of sulphur was taken then how many moles of O_(2) should be taken.

Answer»

110
68.75
55
50

Solution :Assume that MOLE of PRODUCT mixture =x mole
`n_(a)=50=xxx0.6+x xx0.2-0.8x`
`x=(50)/(0.8)=(500)/(8)=62.5`
`n_(o_(2))` remain=62.5xx0.2-12.5
`n_(o)=2xxn_(SO_(2))+3xxn_(SO_(3))=2xx x xx0.6+3xx x xx0.2`
`n_(o)=1.2x+0.6x=1.8x=1.8xx62.5`
`n_(o_(2))=(1.8xx65)/(2)=0.9xx65=56.25`
`n_(o_(2))` TOTAL require `=56.25+12.5=68.75`
23.

Pure solvent A has freezing point 16.5^(@)C. On dissolving 0.4 g of B in 200 g of A, the solution freezing at 16.4^(@)C and on dissolving 2.24 g of C in 100 g of A, the solution has freezing point of 16.0^(@)C. If the molar mass of Bis "74 g mol"^(-1), what is the molar mass of C?

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Solution :For solute B, dissolved in solvent A,
`DELTA T_(F)=(1000K_(f)w_(2))/(w_(1)M_(2)),"i.e., 0.1"=(1000xxK_(f)xx0.4)/(200xx74)or K_(f)="3.7 K kg MOL"^(-1)`
For solute C dissolved in the same solvent A, `M_(2)=(1000K_(f)w_(2))/(w_(1)DeltaT_(f))=(1000xx3.7xx2.24)/(100xx0.5)=165.8`
Alternatively,`""((DeltaT_(f))_(B))/((DeltaT_(f))_(C))=(w_(B))/(w_(A)M_(B))xx(w_(A)'xxM_(C))/(w_(C))`
24.

Pure silicon is an insulator. Silicon doped with phosphorus is a semi-conductor. Silicon doped with gallium is also a serni-conductor. What is the difference between two doped silicon semi- conductors ?

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Solution :Pure silicon has a network or lattice in which all the four valence electrons are bonded to four other atoms. Therefore, it is an insulator. However, when silicon is doped with phosphorus (`NS^(2) NP^(3)`) having five valence electrons, the impurity leads to an EXCESS of electron after forming four covalent bonds like silicon. The extra electrons SERVE to conductivity and silicon doped with phosphorus behaves as semi-conductor.
When silicon is doped with gallium (`ns^(2) np^(1)`) having three valence electrons, the bonds formed create electron deficient sites called holes. Under the influence of applied electric field, one electron from neighbouring atoms moves to fill hole but creates another hole at its own place. Therefore, the electrical conductance is due to the movement of positive hole. The basic DIFFERENCE between the two doped semi- conductors is that phosphorus doped silicon semi- conductor is n-type semi-conductor while gallium doped silicon-conductor is p-type. In n-type, the conduction is due to movement of electrons while in p-type, the conduction is due to movement of positive hole.
25.

Pure (R) Mandelic acid has specific rotation of 150. Ifa sample contains 60% of the R and 40% of its enantiomer, the [alpha] of his solution is.

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ANSWER :30
26.

Pure silicon is:

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Insulator
Semi-conductor
Conductor
NONE of these.

Answer :A
27.

Pure primary amines underline("can be prepared ")by the action of ammonia on alkyl halides.

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SOLUTION : PURE primary amines cannot be PREPARED by the ACTION of ammonia on alkyl halides.
28.

Pure ozone is ……………gas.

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SOLUTION :PALE BLUE
29.

Pure ozone is ………………. .

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YELLOW GAS
BLUE gas
PALE blue gas
bright blue gas

Solution :Pale blue gas
30.

Pure phosphoric acid is very viscous, because :

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It is a strong acid
It is TRIBASIC acid
It is hygroscopic
It has `PO_(4)^(3-)` groups which are BOUNDED by MANY hydrogen bonds

Answer :D
31.

Pure phosphine is not combustible while impure phosphine is combustible, this combustibility is due to presence of:

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`P_2H_4`
`N_2`
`PH_5`
`P_2O_5`

ANSWER :A
32.

Pure ozone is ……………

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YELLOW GAS
BLUE gas
PALE blue gas
bright blue gas

Solution :Pale blue gas
33.

Pure nitric acid is colourless, on standing it becomes yellow. Justify your answer.

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Solution :Nitric ACID DECOMPOSES on exposure to sunlight or on being heated, into nitrogen dioxide, water and oxygen.
`4HNO_(3)rarr4NO_(2)+2H_(2)O+O_(2)`
Due to this reaction pure acid or its CONCENTRATED solution becomes yellow on standing.
34.

Pure NaCl is prepared by saturating a cold saturated solution of common salt in water with HCl gas. The principle used is

Answer»

Le-Chatelier principle
Displacement law
COMMON ION EFFECT
Fractional DISTILLATION

SOLUTION :Common ion effect.
35.

Pure O_(2) instead of air is used to oxidise the pig iron because:

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MOLTEN metal took up small amount of nitrogen which makes the steel brittle
Air is not as efficient to OXIDISE all the impurities to their respective oxides
Air contains moisture and will precipitate iron as `Fe_(2)O_(3)`
Iron reacts with air to `FeCO_(3)`

ANSWER :A
36.

Pure nitrogen gas is obtained from

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`NH_(3)+NaNO_(2)`
`NH_(4)Cl+NaNO_(2)`
`N_(2)O+Cu`
`(NH_(4))_(2)Cr_(2)O_(7)`.

Solution :AMMONIUM NITRITE `(NH_(4)Cl+NaNO_(2))` gives `N_(2)` on HEATING.
37.

Pure N_(2) gas is obtained from

Answer»

`NH_(3)+NaNO_(2)`
`NH_(4)Cl+NaNO_(2)`
`N_(2)O+Cu`
`(NH_(4))_(2)Cr_(2)O_(7)`

Solution :`underset("(AQ)")(NH_(4)Cl)+ underset("(aq)")(NaNO_(2)) to NH_(4)NO_(2) +NaCl`
`NH_(4)NO_(2) overset("HEAT")tounderset("(g)")(N_(2))+underset("(l)")(2H_(2)O)`
38.

Pure m-bromotoluene can be obtained by

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Bromination of toluene
Friedel Crafts REACTION of bromobenzene with `CH_(3)CL`
Bromination of nitrobenzene followed by replacement of NITRO GROUP with methyl group
None of these.

Solution :Here m-bromotoluene can only be prepared as follows :
39.

Pure N_2 can be obtained by:.

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ANSWER :C
40.

Extra pure N_2 can be obtained by heating

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`NH_(4)NO_(3)`
`Ba(N_(3))_(2)`
`(NH_(4))_(2)Cr_(2)O_(7)`
`NH_(3)" with "CUO`

Answer :B
41.

Pure HF does not attack

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GLASS
`SiF_(4)`
`SiO_(2)`
polythene

Answer :D
42.

Pure HF does not attack:

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Glass
Fe
`SiO_2`
Polythene

Answer :D
43.

Pure HBr gas may be obtained by heating sodium bromide with syrupy phospheric acid and not with concentrated sulphuric acid because concentrated sulphuric acid is:

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More VOLATILE
LESS stable
A WEAKER acid
an OXIDISING agent

Answer :D
44.

Pure H_2S gas can be obtained by the action of water on :

Answer»

CuS
FeS
Flower of SULPHUR
`Al_2S_3`

ANSWER :D
45.

Pure conc. HNO_(3) makes iron passive as the surface is covered with protective layer of

Answer»

`Fe_(2)O_(3)`
`FEO`
`Fe_(3)O_(4)`
`FE(NO_(3))_(3)`

SOLUTION :The passivity of iron is DUE to the formation of a thin insoluble and INVISIBLE iron film on surface which prevents its further reactions. The film is due to the formation of `Fe_(3)O_(4).`
46.

Pure CO can be obtained from :

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SODIUM oxalate
Nickel tetracarbonyl
Formic acid
Carbon DIOXIDE and hydrogen

Answer :B
47.

Pure H_2 is obtained by the action of:

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Alluminium over KOH
NaH over `H_2O`
ELECTROLYSIS of warm solution of `BA(OH)_2` USING NI electrodes
All of the above

Answer :D
48.

Pure Cl_(2) is prepared by

Answer»

`PtCl_(4)`
NaCl
`CuCl_(2)`
`KCl`

Solution :Pure chloride OBTAINED by heating dry platonic chloride `[PtCl_(4)]` (or ) GOLD chloride `[AuCl_(3)]` in a hard glass tube.
`PtCl_(4) overset( 374^(@))(rarr) PtCl_(2) + Cl_(2) + CuCl_(2) + H_(2) O overset( 582^(@))(rarr) Pt+Cl_(2)`
49.

Pure chloroform may be prepared by :

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chlorination of methane
partial REDUCTION of `C CI_(4)`
the action of BLEACHING powder and alkali on ethanol
DISTILLING CHLORAL hydrate with conc. Aqueous alkali solution

Solution :N//A
50.

pure benzene freezes at 5.3^(@)C. A solution of 0.233 g of phenyl acetic acid (C_(6)H_(5)CH_(2)COOH) freezes at 4.47^(@)C. Form this observation, one can conclude that pbenyl acetic acid

Answer»

exists as such in benzene
undergoes partial IONISATION in benzene.
undergoes complete ionisationin benzene
dimerises in benzene.

Solution :`DeltaT_(f)=5.3-4.47=0.83 K`
`"Molaliy (m)"=((0.223 g)//(0.0044 kg))/((136 g MOL^(-1)))`
`=0.373 mol//kg`
`DeltaT_(f)=iK_(f)m`
`i=(DeltaT_(f))/(K_(f)m)=((0.83K))/((5.12 K mol^(-1))xx(0.373 mol kg^(-1)))`
=0.435.
Since I isclose to 0.5, this means that the acid dimerises in benzene.