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.

Schottky defect is observed in crystal when ______

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some CATIONS move from their LATTICE sites to interstitial sites
equal NUMBER of cations and anions are missing from the lattice.
some lattice sites are occupied by electrons
some impurity is present in the lattice.

ANSWER :b
2.

Schottky defect is observed in crystal when ………

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some CATIONS move from their lattice sites to interstitial sites.
equal NUMBER of cations and anions are missing from the lattice.
some lattice sites are occupied by ELECTRONS.
Some IMPURITY is present in the lattice.

Answer :B
3.

Schottky defect generally appears in

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CSCL
KCL
NACL
All

Solution :Schottky defect is common in highly ionic compounds with high coordination number and SIZE of two ions (cations and anions) are almost same. e.g. NaCl, CsCl, KCl and KBr.
4.

Schotten Baumann reaction is

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`"Phenol + BENZOYL CHLORIDE" overset(NaOH)rarr "Phenyl banzoate"`
`"Sodium phenate + methyl iodide" overset(-HCl)rarr "Anisole"`
`"Phenol + Chloroform" overset(NaOH)rarr "SALICYLADEHYDE"`
`"Benzene DIAZONIUM chloride + Phenol"rarr beta-"Hydroxyazobenzene"`

SOLUTION :The reaction of phenols with benzoyl chloride in the presence of aq. NaOH is known as Schotten Baumann reaction.
5.

Schiff's base and Schiff's reagent are

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STRUCTURAL isomers
ENTIRELY DIFFERENT COMPOUNDS
Anomers
Disteromers

Solution :Entirely different compounds
6.

Schiff's and Piria method is used for the estimation of:

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nitrogen
sulphur
halogens
oxygen

Answer :C
7.

Scattering of light takes place in

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electrolysis
collodial solutions
electroplating
SOLUTION of electrolytes.

Solution :Scattering of light TAKES PLACE in case of colloidal solution (Tyndall effect ).
8.

Saturatedsolutionof crudeNaCl overset("x ") toPureNaCl.In thereaction ,X is___

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HCIgas
`NHO_(3)` gas
`NH_(3)` gas
`H_(2) SO_(4) `gas

Answer :A
9.

Satement 1 A solution of FeSO_(54) can be stored in a copper vessel Statement 2 2E^(@) "of" Cu "lt" E^(@) of Fe

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Statement 1 is true statement 2 is true , statement 2 is a corrrect EXPLANATION for statement 1
Statement 1 is true statement 2 is true statement 2 is not a correct explanation for statement 1
Statement 1 is true statement 2 is false
Statement 1 is false statement 2 is true

SOLUTION :Correct statement 2 `E^(@)` of Fe is lower than `E^(@)` of Cu
10.

Sate Gay Lussac's Law. Calculate the pressure exerted by 4 mole of agas occuping a volume of 1.5m^(3) at 100 K. Given R=8.314 J/k/mol.

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Solution :GAY Lussac.s law (PRESSURE-temperature relationship)
Gay Lussac.s law states that at constant volume, pressure of a fixed amount of a GAS varies DIRECTLY with the temperature.
Given `n=4` moles
`V=1.5m^(3),T=100klR=8.314`J/kmol, `P=?`
`P=nRTimplies(nRT)/V=(4"moles"xx8.314"Jk mol"xx100k)/(1.5m^(3))=8269Pa`
11.

Saponification of ethyl benzoate with caustic soda as alkalie, gives

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Benzyl alochol, ETHANOIC ACID
Sodium benzoate, ethanol
Benzoic acid, Sodium ethoxide
Phenol, ethanoic acid

Solution :`underset("Acetaldehyde")(C_(6)H_(5)OVERSET(O)overset(||)(C) - OC_(2)H_(5))overset(NaOH)rarr underset("Ethanol")(C_(2)H_(5)OH) + underset("Sodium benzoate")(C_(6)H_(5)COONA)`
12.

Saponification means hydrolysis of an ester with

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DIL. NAOH
`dil.H_(2)SO_(4)`
enzymes
none of the above

Solution :dil. NaOH.
13.

Saponification of an oil results in the formation of

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Glycol
GLYCEROL
Sodium SALT of an ACID
Glycerol and SOAP

Solution :Glycerol and soap
14.

Sample of water has hardness 77.5ppm Ca^(2+). If this is passed through an ion exchange column where Ca^(2+) is replaced by H^(+), what is the pH of water after it has been so treated

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1.4
4.4
2.4
5.4

Answer :C
15.

Sample of different gases are given at different conditions in column-I and column-II consisting of translational kinetic energy of these gases at given conditions. {:(,"Column-I",,"Column-II"),("(P)",2mole SO_(2)(g)at 700K,,"(1)Maximum K.E. per gram"),("(Q)",1 mole SO_(3)(g)at 400K,,"(2)Maximum total K.E."),("(R)",4 mole CH_(4)(g)at 300K,,"(3)Maximum K.E. per gram"),("(S)",2.5 mole He(g)at 450 K,,"(4)Minimum K.E. per molecule"):}

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`{:("PQRS"),("4312"):}`
`{:("PQRS"),("2314"):}`
`{:("PQRS"),("1234"):}`
`{:("PQRS"),("2341"):}`

ANSWER :(d)
16.

Same volume of two monoatomic gases A and B is mixed at same temperature. ........ .. Is (C_p)/( C_v) = ….......

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`0.83`
`1.67`
`3.3`
`3.30`

SOLUTION :`C_(p) = (5)/(2) R, C_(v) = (3)/(2) R`
`therefore (C_p)/( C_v) = (5)/(3) = 1.67`
17.

Same reasons are given reagarding the limited use of H_(2) as fuel (I) Its calorific value is low (II) It availability in free state is less (III) It is difficult to store (IV) Its transportation is easy The correct statement are

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I,II and III
II,III and IV
All are CORRECT
II and III

Answer :D
18.

Same mass of diamond and graphite ( bothbeing carbon) are burnt in oxygen. Will be heat produced be same or different ? Why ?

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Solution :Heat EVOLVED will be difficult . This is because they have DIFFERENT crystal STRUCTURE.
19.

Same functional group containing different organic compounds possesses………chemical reaction

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not equal
sam
same and not same
not decided

Answer :B
20.

Same amount of metal combines with 0.1g of oxygen and lg of a halogen. Hence the equivalent mass of halogen is :

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9
35.5
80
127

Solution :0.1 GM `rarr` 1 gof halogen
`8 G O_(2) rarr ? = 80`
The amount of the substance which combines with the 8gm of oxygen is CALLED its eqwt.
21.

Samarium-62 is the member of.....

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s-block element
p-block element
d-block element
f-block element

Answer :d
22.

samall quantities of solutionof componds TX ,TYand TZ are put in to separate test tubes cntaining x y and z solutoin Tx does not react with any of these TY reacts oth with x and z tz rectswithx th drcasing order of ase of oxidation of the anons X^(-) Y^(-0) and Z^(-1) id

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`Y^(-),Z^(-),X^(-)`
`Z^(-),X^(-),Y^(-)`
`Y^(-),X^(-),Z^(-)`
`X^(-),Z^(-),Y^(-)`

Solution :Since TX does not react with any of the solution it means that `X^(-)` is least EASILY oxidizedsince TY reacts with both X and Z i.r
`Y^(-) +XrarrX^(-)+Y` and `Y^(-)+ZrarrZ^(-)+Y` THEREFORE `Y^(-)` is mosteasily oxidizedsince tZ reacts with ony X i.r `Z^(-)`is more easily oxidized than `X^(-)` combining all the result the decreasing order of OXIDATION of anions follows the order
`Y^(-)GTZ^(-)gtX^(-)`
23.

Salts given in column-I reacts with the excess of reagents given in column-II and form white/coloured precipitates. Select the correct options for the salts given in column-I with the reagent(s) given in the column-II.

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SOLUTION :`ZN^(2+)` is soluble in excess `NaOH` and `NH_(3)` solution.
`CU^(2+)` is soluble in `NH_(3)` solution (excess).
24.

Salt (W) gives golden yellow colouration in Bunsen flame. When (W) is oxidised by acidified K_(2)Cr_(2)O_(7), a colourless gas (X) is evolved which turns lime water milky and the milkiness disappears on passing excess of gas (X). However gas (X) does not decolourise pink solution of acidified KMnO_(4). when aqueous solution of salt (W) is added slowly to hydrated MnO_(2), red coloured solution is obtained due to formation of soluble complex (Y) along with evolution of gas (X). The red coloured complex (Y) is:

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`[Mn(C_(2)O_(4))_(3)]^(2-)`
`[Mn(C_(2)O_(4))_(3)]^(3-)`
`[MnCI_(4)]^(-)`0
`[Mn(NH_(3))_(6)]^(2+)`

Solution :`underset((W))(Na_(2)C_(2)O_(4))+K_(2)Cr_(2)OP_(7) rarr underset((X))(CO_(2)),`
`7C_(2)O_(4)^(2-)+2MnO(OH)_(2)darr +8H^(+) rarr 2[Mn(C_(2)O_(4))_(3)]^(3-)+2CO_(2)uarr +6H_(2)O`
25.

Salt (W) gives golden yellow colouration in Bunsen flame. When (W) is oxidised by acidified K_(2)Cr_(2)O_(7), a colourless gas (X) is evolved which turns lime water milky and the milkiness disappears on passing excess of gas (X). However gas (X) does not decolourise pink solution of acidified KMnO_(4). when aqueous solution of salt (W) is added slowly to hydrated MnO_(2), red coloured solution is obtained due to formation of soluble complex (Y) along with evolution of gas (X). The salt (W) is:

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`HCOONa`
`CH_(3)COONa`
`Na_(2)C_(2)O_(4)`
`Na_(2)SO_(3)`

Solution :`underset((W))(Na_(2)C_(2)O_(4))+K_(2)Cr_(2)OP_(7) rarr underset((x))(CO_(2)),`
`7C_(2)O_(4)^(2-)+2MnO(OH)_(2)darr +8H^(+) rarr 2[Mn(C_(2)O_(4))_(3)]^(3-)+2CO_(2)uarr +6H_(2)O`
26.

Salt (W) gives golden yellow colouration in Bunsen flame. When (W) is oxidised by acidified K_(2)Cr_(2)O_(7), a colourless gas (X) is evolved which turns lime water milky and the milkiness disappears on passing excess of gas (X). However gas (X) does not decolourise pink solution of acidified KMnO_(4). when aqueous solution of salt (W) is added slowly to hydrated MnO_(2), red coloured solution is obtained due to formation of soluble complex (Y) along with evolution of gas (X). Gas (X) is:

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`CO_(2)`
`SO_(2)`
`SO_(3)`
`NO_(2)`

Solution :`underset((w))(Na_(2)C_(2)O_(4))+K_(2)Cr_(2)OP_(7) rarr underset((x))(CO_(2)),`
`7C_(2)O_(4)^(2-)+2MnO(OH)_(2)DARR +8H^(+) rarr 2[Mn(C_(2)O_(4))_(3)]^(3-)+2CO_(2)uarr +6H_(2)O`
27.

Salt solution of 'A' + dil. HCI rarr only 'B' gas{:(("suffocating odour of burning sulphur")),(""darrBaCI_(2)),('C'["white ppt"]rarr"Soluble in dilHCI and gives 'B' gas"),(""darr"Reagent M"),("White ppt.of"[D]("insoluble in dilHCI")):} Find the total number of reagents 'M': (conc. HNO_(3)),O_(2),H_(2)O_(2),N_(2),NH_(3),(Br_(2)+H_(2)O)

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<BR>

Solution :`HNO_(3),O_(2),H_(2)O_(2),(Br_(2)+H_(2)O)`
28.

Salt is a mixed form of acidic and basic radical. The confirmatory test of basic and acidic radicals by using chemical reagent is called qualitative analysis. Precipitation reaction is used for the confirmatory test of radicals. Which of the following is an example of IIA group basic radical sulphide?

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`MNS`
`FES`
`CDS`
`Na_(2)S`

Answer :C
29.

Salt is a mixed form of acidic and basic radical. The confirmatory test of basic and acidic radicals by using chemical reagent is called qualitative analysis. Precipitation reaction is used for the confirmatory test of radicals. Which of the following iodide has scarlet red colour?

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`PbI_(2)`
`HgI_(2)`
`AgI`
All of these

Answer :B
30.

Salt bridge maintains ……………………in the solution of two half cells

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ANSWER :electroneutraliy
31.

Salt (A) on heating gives a colourless neutral gas which supports combustion. From the aforesaid, flow diagram, answer the following questions. Which of the following statement is correct?

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Salt (A) gives yellow PRECIPITATE with chloroplantinic acid as well as with SODIUM cobaltinitrite.
The brown RING is formed DUE to the formation of nitroso ferrous sulphate `[Fe(NO)]^(2+) SO_(4)^(-)`.
Salt C reacts with silver nitrate solution to form white precipitate
(a) and (b) both

Solution :(A) Contains `NO_(3)^(-) (C) rarr NaNO_(3)`
`(B) rarr NH_(3) (C) [Fe(H_(2)O)_(5)NO] SO_(4)` or `FeSO_(4).NO`
`(A) rarr NH_(4)NO_(3)`
32.

Salt (A) on heating gives a colourless neutral gas which supports combustion. From the aforesaid, flow diagram, answer the following questions. The compound (A) contains the following acidic radical:

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`NO_(2)^(-)`<BR>`NO_(3)^(-)`
`Br^(-)`
`SO_(3)^(2-)`

Solution :(A) Contains `NO_(3)^(-) (C) RARR NaNO_(3)`
`(B) rarr NH_(3) (C) [Fe(H_(2)O)_(5)NO] SO_(4)` or `FeSO_(4).NO`
`(A) rarr NH_(4)NO_(3)`
33.

Salt (A) on heating gives a colourless neutral gas which supports combustion. From the aforesaid, flow diagram, answer the following questions. The basic radical of salt (A) and gas (B) both gives brown precipitate with Nessler's reagent. the composition of the brown precipitate is:

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`(NH_(4))_(2)[HgI_(4)]`
`Hg(NH_(2))NO_(3)`
`HgO.Hg(NH_(2))I`
`(NH_(4))_(3)[Co(NO_(2))_(6)]`

SOLUTION :(A) Contains `NO_(3)^(-) (C) rarr NaNO_(3)`
`(B) rarr NH_(3) (C) [FE(H_(2)O)_(5)NO] SO_(4)` or `FeSO_(4).NO`
`(A) rarr NH_(4)NO_(3)`
34.

Salol is prepared from

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SALICYLIC ACID and phenol
Salicylic acid and METHYL alcohol
Both
None

Solution :
35.

Salol is

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ACETYL SALICYLIC acid
Phenol benzoate
Acetyl SALICYLATE
PHENYL salicylate

Solution :SALOL is phenyl salicylate.
36.

Salol can be used as

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Anseptic
Antipyretic
Both
None

Solution :As INTERNAL ANTISEPTIC
37.

Saline hydrides are known to react violently with water producing fire, Can CO_(2), a well known extinguisher, be used in this case? Explain.

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SOLUTION :Whenever a saline hydride `(NaH or CaH_(2))` reacts with water, the reaction is so HIGHLY exothermic that the hydrogen evolved catched fire. For example,
`NaH(s)+H_(2)O(aq) to NaOH(aq)+H_(2)(g)+"heat"`
`CaH_(2)(s)+2H_(2)O(aq) to Ca(OH)_(2)(aq)+2H_(2)(g)+"heat"`
`CO_(2)` which is normally used as fire EXTINGUISHER cannot be used in this case beacuse it will react with the hydroxide formed in the reaction to form a carbonate. This will INCREASE the rate of the forward reaction in which heat is evolved.
`2NaOH(aq)+CO_(2)(g) to Na_(2)CO_(3)(aq)+H_(2)(aq).`
38.

Saline hydrides are known to react with water violently producing fire. Can CO_2 a well known fire extinguisher be used inthis case? Explain.

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Solution :No.The FIRE produced by the action of water on SALINE hydrides cannot be put out by `CO_2`because the HOT METAL hydride reduces `CO_2` readily. `NaH+CO_2rarrHCOONa`(spadium formate)
39.

Saline hydrides are known to react with water violently poducing fire. Can CO_(2) , a well known fire entingiusher, be used in this case? Explain .

Answer»

Solution :Saline hydrides (such as NaH, `CaH_(2)`, etc.) , react with water violently to form the corresponding metal hydroides with the evolution of dihyrogen .
(`NaH(s) + H_(2)O(l) to NaOH(aq) + H_(2)(g)`
`CaH_(2)(s) + 2H_(2)O(l) to Ca(OH)_(2) (aq) +2H_(2)(g)`
These REACTIONS are so much exothermic that the evolved `H_(2)` catches fire. The fire so produced cannot be extinguished by`CO_(2)` because it gets reduced by the HOT metal hydride to form sodiym FORMATE.
`NaH+ CO_(2) to HCOONa`
`However, sand is useful since it is a highly STABLE solid
40.

Saline hydrides are known to react with water violently producing fire can CO_2, a well known fire extinguisher be used in this case ? Explain.

Answer»

Solution :Saline hydrides react violently with WATER PRODUCING DIHYDROGEN gas
`NaH_((s)) + H_2O_((aq)) to NaOH_((aq)) + H_(2(g))`
But A well known fire extinguisher `CO_2` is used then FOLLOWING red will be obtained
`NaH+CO_2 to HCOONa`
41.

Salient features of Molecules Orbital Theory(MOT)

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Solution :1. The electrons in an atom are found in atomic orbtials, the electrons, in a MOLECULE are found in molecuar orbitals.
2. The molecule orbitals are formed by the COMBINATION of atomic orbitals of comparable energies ande proper symmetry.
3. The BMOhas LOWER energyand hence GREATER STABILITY than the corresponding b ABMO.
4 The molecular orbitals are filled by electrons in accordance with Aufau principle, Pauli.s exclustion principle and the Hung.s rule.
42.

Salicylic acid is prepared from phenol by

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REIMER Teimann reaction
Kolbe's reaction
Kolbe-electrolysis reaction
None of these

Solution :REFER to A-Level INFORMATION of PHENOLS.
43.

Salicylic acid is a stronger acid than p-hydroxybenzoic acid due to

Answer»

STERIC hindracne
Hygrogen bonding
Mesomeric effect
Solvation energy

Answer :B
44.

Salicyladehyde can be prepared from

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PHENOL and CHLOROFORM
phenol, chloroform and SODIUM hydroxide
phenol, CARBON TETRACHLORIDE and NaOH
None

Solution :
45.

Salicyaldehyde involves________hydrogen bonding .

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SOLUTION :INTRAMOLECULAR
46.

Sacchrin an artificial sweetener, is manufactured from :

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CELLULOSE
starch
cyclohexane
toluene

Answer :D
47.

Saccharin, an artificial sweetener, is manufacturered from :

Answer»

cellulose
toluene
cyclohexane
starch

Answer :C
48.

S_((8)) + 8O_(2) rarr 8SO_(2), DeltaH = -QKJ, then DeltaE

Answer»

`-8`
`-Q`
`(-Q)/(8)`
`(+Q)/8`

ANSWER :C
49.

S_(4)N_(x) in vapour state is subjected to decompoition totally to get octaatomic sulphur vapour and nitrogen gas . Under similar conditions the volume of products is 2.5 times to the reactants. Deduce the molecular formula of the nitride of sulphur. Calculate the weight percentage of nitrogen.

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Solution :The decompoition is given as `UNDERSET((G))(S_(4)N_(x))to underset((g))((1)/(2))S_(8) +underset((g))((x)/(8))N_(2)`
By analysing the given data , 1+x=5:x=4
MOLECULAR formula =`S_(4)N_(4)`
Weight percentage of NITROGEN `(4xx14)/(4(32+14))xx10=30.43%`
50.

S^(2-) and SO_(3)^(2-) ions can be distinguished by using

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`(CH_(3)COO)_(2) Pb`
`Na_(2) [Fe(CN)_(5) NO]`
Both (a) & (B)
None of these

Solution :Distinction can be made by both `(CH_(3) C O O)_(2)Pb` and `Na_(2) [Fe(CN)_(5)NO]`
`Pb(CH_(3)COO)_(2)` gives a BLACK ppt with `S^(2-)` ions (e.g., `H_(2)S`) while no precipitate is formed with `SO_(3)^(2-)` ions.
`Pb(CH_(3)COO)_(2) + H_(2)S rarr PBS rarr + underset(("Black ppt"))(2CH_(3) COOH)`
`Na_(2) [Fe(CN)_(5)NO]` gives a purple colouration with `S^(2-)` ions `(Na_(2)S)` while no precipitate is formed with `SO_(3)^(2-)` ions.
`Na_(2) [Fe(CN)_(5) NO] + Na_(2)S rarr Na_(4) [Fe(CN)_(5)NOS]`