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.

Haemoglobin, a complex containing iron is a constituent of blood. The oxidation state of iron in the complex is

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`+1`
`+2`
`+3`
`+4`

ANSWER :B
2.

Haematite is reduced in a blastfurnance by

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CO
`CO_2`
`SO_2`
`C`

SOLUTION :In blast furnance , following REACTION occurs
`3Fe_(2)O_(3) +CO overset(300-400^(@)C)to 2Fe_(3)O_(4)+CO_(2)`
`Fe_(3)O_(4) +4CO overset(500-600^(@)C)to 3FeO+4CO+(1)/(4)O_(2)`
`FeO+CO overset(700^(@)C)to Fe+CO_(2) `
3.

Haematite is Fe_(2)O_(3)/Fe_(3)O_(4)

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SOLUTION :`Fe_(2)O_(3)`
4.

Haematite is an ore of :

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Boron
Iron
Manganese
Copper

ANSWER :B
5.

Haematite and magnetic are ores of iron.

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ANSWER :1
6.

Haematite is an ore of ...............

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ANSWER :IRON
7.

Haematite, an ore ofiron, firstconcentratedand thencalcined. Thecalcinedore isthen heated in ablastfurnace inpresence ofcoke thelimestone and atthesame timea blastof hot air(pre heated to1000 K ) isintroduced intothe furnace throughtuyeres.Two products(X) and(Y) are obtained. Now answer thefollowingquestions. (a) Whatmethodis used forconcentrationof haematite ore? (b) Whathappensduringcalcination ? (c )Identiythe twoproductsand mention their uses.

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Solution :(a) Sincehaematite`(Fe_2O _ 3 ) `isanoxideore, it isconcentrated bygravityseparationprocess in which the crushedore iswashed in astreamof waterwhen lighersand and clayparticlesare washed awaywhilethe heavierore particlessettledown.
(b) Duringcalcination, theimpuritiesof sulphur, phosphorus andarsenicescape as theirvolatile oxides.
` S_8 +8 O_ 2 to8 SO _ 2uparrow, P _ 4+5 O _ 2 toP _ 4O _(10) uparrow ,2 As+5 O _ 2toAs_ 2O_5uparrow `
(c)The variousreactionstakingplacein theblastfurnace are :
`CaCO_ 3` decomposes to formCaOand`CO_ 2`. The CaO thus formed actsas aflux andcombines withsilica(present as an IMPURITY ) to formfusiblecalcium silicateslag (N).
` underset("Limestone")(CaCO_3) (s)overset(1123K)toCaO(s) +CO_ 2 (g)`
`CaO(s)+SiO_ 2tounderset("Calciumsilicate (slag)") underset (N) (CaSiO_ 3 ) `
Calciumsilicateslag is usedas a buildingmaterialsuchascement.
`C (s)+CO_ 2(g) to2CO(g) `
Both COandcokereduce`Fe_2 O _ 3`toFemetal(X).
`Fe_2O_ 3 (s) +CO(g) overset (773 -873 K )to2Fe O (s) +CO _ 2(g) `
` FeO (s)+C(s)overset( gt 1073K )tounderset ("Iron metal (M)")(Fe(s//l)) +CO(g) `
`Fe_ 2O_ 3(s)+3 C (s) overset ( 1073K ) tounderset("Iron metal (M)")( 2 Fe (s//l)) +3 CO(g) `
The ironthusobtainedfromthefurnacecontainsabout 4% carbonand manyimpurities(e.g., S, P, Si, MN) insmaller amount. This is calledpig iron and iscast intoavariety of shapes.
8.

Haberprocessis usedfortheproduction ofwhich ofthe following ?

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` NH_ 3 `
` HNO_3`
`H _ 2 SO_4`
` O _3`

SOLUTION :Haber processis usedfor theproductionofammoniagas.
`N _ 2(g )+3 H _2 (g ) hArr2 NH_ 3( g ) `
9.

Haber's process is used to prepare

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`O_2`
`N_2`
`NH_3`
`PH_3`

ANSWER :C
10.

HA is a weak acid. The pH of 0.1 M HA solution is 2. What is the degree of dissociation (alpha) of HA

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0.5
0.2
0.1
0.301

Solution :`pH = 2, :. [H^(+)] = 10^(-2) = 0.01`.
For weak acid, `[H^(+)] = alpha. C`
or, `0.01 = alpha xx 0.1 rArr alpha = 0.1`.
11.

H_(5)IO_(6) is a

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STRONG REDUCING agent
Strong base
Strong OXIDIZING agent
Weak base

Answer :C
12.

H_3PO_4 is tribasic acid and oneof its salts is NaH_2PO_4..What volume of 1 M NaOH should be added to 12 g NaH_2PO_4 (mol.wt.120) to exactly convert it into Na_3po_4.

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100ml
300 ml
200 ml
80 ml

Answer :C
13.

H_(3)PO_(4) is a tribasic acid and one of its salts of NaH_(2)PO_(4). What volume of 1 M NaOH should be added to 12 g NaH_(2)PO_(4) (molecular mass = 120 g/mole) to exactly convert it into Na_(3)PO_(4)

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100 ML
200 mL
80 mL
300 mL

Answer :B
14.

H_3PO_4 consist of

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<P>two P-OH bonds and 2 P=O bonds
ONE P-OH bond and 3 P =O bonds
2 P-OH bonds and one P-O bond
3 P-OH bonds and one P=O bond

Answer :D
15.

H_(3)PO_(3) is diprotic (or dibasic). Why ? Or What is the basicity of H_(3)PO_(3) and why ?

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Solution :Its structure is

Since it CONTAINS only TWO ionizable H-atoms which are present as OH GROUPS, it BEHAVES as a dibasic ACID.
16.

H_(3)PO_(3) is diprotic (or dibasic). Why?

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SOLUTION :Its structure is

Since it contain only two ionizable H - ATOMS which are present as OH GROUPS, it behaves as a dibasic acid.
17.

H_3PO_3+HNO_3+(NH_4)_2MoO_4to Oxidation number of Mo in product is x. Cr_2O_7^(2-)+H^(+)+NaCl to Oxidation state of Cr in product is y. SCN^(-)+MnO_2toNumber of pi bonds in sulphur containing product is z. Find x+y+z

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SOLUTION :x=+6`(NH_4)_3 PMo_12O_40`
y=+6 `CrO_2Cl_2`
z=4
`ZtoN^(@)C-S-S-C^(@)N`
18.

H_3PO_3 is:

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A TRIBASIC acid
A DIBASIC acid
Neutral
Basic

Answer :B
19.

H_(3)PO_(3) disproportionates to

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`H_(3)PO_(4) and PH_(3)`
`HPO_(3) and PH_(3)`
`H_(3)PO_(2) and PH_(3)`
`P_(2)O_(5) and PH_(3)`

Answer :A
20.

H_(3)PO_(2) overset(140^(@)) rarrA overset(220^(@)C) toB overset(320^(@)C)rarrC. Compound ( C) is :

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`H_(2)PO_(3)`
`H_(3)PO_(3)`
`(HPO_(3))_(N)`
`H_(4)P_(2)O_(7)`

Answer :C
21.

H_(3)PO_(2) is the formula for one of the phosphorus acids. Its name and basicity are respectively

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PHOSPHOROUS acid and TWO
hypophosphorous acid and two
hypophosphorous acid and ONE
hypophosphoric acid and two

Answer :3
22.

H_3PO_2 is monoprotic acid. Explain.

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ANSWER :1
23.

H_3PO_2 has the name and basicity respectively:

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PHOSPHORUS ACID and two
Hypophosphoric acid and two
Hypophosphoric acid and one
Hypophosphoric acid and two

Answer :C
24.

H_(3)CNH_(2)overset((CH_(2))_(2)O)to(A). What is (A)

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`H_(3)C-NH-CH_(2)-CH_(2)-OH`

`H_(3)C-NH-CH_(2)-O-OH`

Solution :
25.

[H_(3)C(CH_(2))_(15)-overset(CH_(3))overset("|")underset(CH_(3))underset("|")"N "-CH_(3)]^(+)Br^(-) the given structure is for

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Cetyltrimethyl ammonium BROMIDE
Linear ALKYL bromide
Ester of STERIC acid and polyethylen glycol
Cetyl tribromide AMONIA 

Solution :Cetyltrimethyl ammonium bromide
26.

H_3C-undersetunderset(CH_3)|CH-overset+CH=CH_2+HBr to A A (Predominantly ) is

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`CH_3-undersetunderset(BR)|CH-undersetunderset(CH_3)|CH-CH_3`
`CH_3-undersetunderset(CH_3)|CH-undersetunderset(Br)|CH-CH_3`
`CH_3-undersetunderset(CH_3)(|)CH-CH_2-CH_2Br`
`CH_3-undersetunderset(CH_3)(|)oversetoverset(Br)(|)C-CH_2CH_3`

SOLUTION :
27.

H_(3)C-underset(CH_(3))underset(|)(C)H-CH=CH_(2)+HBr to A A (predominantly) is

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`CH_(3)-UNDERSET(Br)underset(|)(C)H-underset(CH_(3))underset(|)(C)H-CH_(3)`
`CH_(3)-underset(CH_(3))underset(|)(C)H-underset(Br)underset(|)(C)H-CH_(3)`
`CH_(3)-underset(CH_(3))underset(|)(C)H-CH_(2)-CH_(2)Br`
`CH_(3)-underset(CH_(3))underset(|)overset(Br)overset(|)(C)-CH_(2)CH_(3)`

Solution :
THUS, option (d) is correct.
28.

H_(3)C underset(CH_(3))underset(|) overset(CH_(3)) overset(|) (--) overset(C_(2)H_(5))overset(|)CH-underset(CH_(3))underset(|)(CH)-OH underset(/_\) overset(H^(o+))toP Product mixture, the product's mixture contains

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`H_(3)C-underset(CH_(3))underset(|)(CH)-underset(OH)underset(|)(CH)-overset(CH_(3)) overset(|)(CH)-C_(2)H_(50`
`H_(3)C-underset(CH_(3))underset(|)overset(CH_(3)) overset(|)(C)-underset(C_(2)H_(5))underset(|)(C)=underset(CH_(3))underset(|)(C)-H`
`H_(3)C-underset(CH_(3))underset(|)overset(CH_(3)) overset(|)(C)-underset(C_(2)H_(5))underset(|)(C)=underset(CH_(3))underset(|)(C)-CH_(3)`
`H_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-underset(C_(2)H_(5))underset(|)(CH)-overset(H) overset(|)(C)=CH_(2)`

SOLUTION :
29.

H_(3)C-underset(CH_(3))underset("|")(CH)-CH=CH_(2)+HBrrarrA A (predominantly) is :

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`CH_(3)-underset(CH_(3))underset(|)overset(Br)overset(|)(C)-CH_(2)CH_(3)`
`CH_(3)-underset(Br)underset(|)(CH)-underset(CH_(3))underset(|)(CH)-CH_(3)`
`CH_(3)-underset(CH_(3))underset(|)(CH)-underset(Br)underset(|)(CH)-CH_(3)`
`CH_(3)-underset(CH_(3))underset(|)(CH)-CH_(2)-CH_(2)Br`

Solution :`CH_(3)underset(CH_(3))underset("|")(CH)-CH=CH_(2)+HBroverset(-Br^(-))rarrCH_(3)-underset((2^(@)-"CARBOCATION"))underset(CH_(3))underset(|)(CH)-overset(+)(C)H-CH_(3)underset("shif")overset("1,2-hydride")rarrCH_(3)underset(3^(@)"Carbocation (more stable)")underset(CH_(3))underset(|)overset(+)(C)-CH_(2)-CH_(3)overset(+Br^(-))rarrCH_(3)-underset(CH_(3))underset(|)overset(Br)overset(|)(C)-CH_(2)-CH_(3)`
30.

H_(3)C-underset(CH_(3))underset("|")("C ")H-CH=CH_(2)overset(HBr)rarr

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`CH_(3)-UNDERSET(CH_(3))underset("|")("C ")H-underset("BR ")underset("| ")("C ")H-CH_(3)`
`CH_(3)-overset("Br ")overset("| ")underset(CH_(3))underset("|")("C ")-CH_(2)-CH_(3)`
`CH_(3)-underset(CH_(3))underset("|")("C ")H-CH_(2)CH_(2)Br`
`CH_(3)-overset("Br ")overset("| ")underset(CH_(3))underset("|")("C ")-CH_(2)CH_(2)Br`

Answer :B
31.

H_(3)C-underset(CH_(3)) underset(|)overset(CH_(3)) overset(|)(C)-overset(O)overset(||)(C)-CH_(3) underset((ii)"hydrolysis"//H_(3)O^(+))overset((i)F_(3)C-overset(O)overset(||)(C)-O-overset(18)(O)-H)to Products. Which of the following option is/are correct?

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`(CH_(3))_(3)C-OVERSET(18)(OH)`
`CH_(3)CO overset(18)(O)H`
`(CH_(3))_(3)COH`
`CH_(3)COOH`

Solution :
32.

H_(3)C-overset(O)overset("||")C-OH+C_(2)H_(5)overset(18)(OH)overset(conc. H_(2)SO_(4))rarr(X) H_(3)C-overset(O)overset("||")C-OH+(CH_(3))_(3)C-overset(18)(OH)overset(conc. H_(2)SO_(4))rarr(Y) In the above reaction (X) and (Y) are respectively:

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`CH_(3)-overset(O)overset("||")C-overset(18)(O)-C_(2)H_(5) and CH_(3)-overset(O)overset("||")C-O-overset(CH_(3))overset(|)underset(CH_(3))underset(|)C-CH_(3)`
`H_(3)C-overset(O)overset("||")C-overset(18)(O)-C_(2)H_(5) and H_(3)C-overset(O)overset("||")C-overset(18)(O)-overset(CH_(3))overset(|)underset(CH_(3))underset(|)C-CH_(3)`
`CH_(3)-overset(O)overset("||")C-O-C_(2)H_(5) and H_(3)C-overset(O)overset("||")C-O-overset(CH_(3))overset(|)underset(CH_(3))underset(|)C-CH_(3)`
`CH_(3)-overset(O)overset("||")C-O-C_(2)H_(5)and CH_(3)-overset(O)overset("||")C-overset(18)(O)-overset(CH_(3))overset(|)underset(CH_(3))underset(|)C-CH_(3)`

ANSWER :A
33.

H_(3)C-overset(O)overset("||")C-O-underset(Ph)underset(|)C=CH_(2)+H_(2)O overset(NaOH)underset(H_(3)^(oplus)O)rarrX+Y X and Y are:

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`X " is "CH_(3)-OVERSET(O)overset("||")C-O-CH_(2)-Ph`
`Y " is "HCHO`
`Y " is "H_(3)C-overset(O)overset("||")C-OH`
`X " is "Ph-underset(O)underset("||")C-CH_(3)`

ANSWER :C::D
34.

H_(3)C - overset(O)overset(||)(C) - COOH underset((ii) H_(2)//Pd)overset((i) NH_(3)) rarr (A). Identify (A)

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`CH_(3)CONH_(2)`
`H_(3)C - overset(O)overset(||)(C) - CONH_(2)`
`CH_(3)CH_(2) CONH_(2)`
`CH_(3)CH (NH_(2)) COOH`

SOLUTION :`CH_(3)-CH(NH_(2))COOH`.
35.

H_(3)C-CH_(2)-overset(CH_(3))overset(|)CH-underset(O)underset(||)C-NH_(2)overset(Br_(2)+KOH)underset(Delta)toA overset(1.CH_(3)"I(excess)")underset(2.AgOH,Delta)toB.

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`CH_(3)-CH_(2)-CH=CH_(2)`
`CH_(3)-CH_(2)-OVERSET(CH_(3))overset(|)CH-UNDERSET(O)underset(||)C-N{:(CH_(3)),("/"),("\"),(CH_(3)):}`
`H_(3)C-CH=CH-CH_(3)`

SOLUTION :`H_(3)C-CH_(2)-overset(CH_(3))overset(|)CH-overset(O)overset(||)C-NH_(2)overset(Br_(2)//KOH,DELTA)toH_(3)C-CH_(2)-overset(CH_(3))overset(|)CH-NH_(2)overset(1.H_(3)CI("excesses"))overset(2.AgOH,Delta)toH_(3)C-CH_(2)-CH=NH_(2)`
36.

H_(3)C-CH_(3)overset("Pysolysis")rarrAoverset(1000-200atm)underset("RCOOOCOR")rarrB. Aoverset("ziegler-natta")underset("catalyst")rarrC Here B and C are respectively

Answer»

LDPE, HDPE
HDPE, LDPE
LDPE, LDPE
HDPE, HDPE

Answer :A
37.

H_3BO_3 IS _____ acid ?

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MONOBASIC
diabasic
Tribasic
NONE

Answer :A
38.

H_3BO_3 accepts hydroxide ion from water as shown below H_3BO_3(aq)+H_2O(l) leftrightarrow B(OH)_4^(-) + H^+ Predict the nature of H_3BO_3 using lewis concept.

Answer»

Solution :Boric acid is also calledas hydrogen BORATE or orthoboric acid . It is a weak MONO basic LEWIS acid of BORON and it is written as `B(OH)_3`. It accepts hydroxyl `(OH^-)` ion from water.It does not dissociate to give hydronium `(H_3O^+)` ion rather forms metaborate ion and this ion in turn give `H_3O^+` ion.
`B(OH)_3+H_2O leftrightarrow [B(OH)_4]^-+ H_3O^+`
HENCE it is considered as weak acid.
39.

H_(3)BO_(3) is :

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MONOBASIC and WEAK Lewis ACID.
Monobasic and weak Bronsted acid.
Monobasic and strong Lewis acid.
Tribasic and weak Bronsted acid.

Solution :B in `H_(3)BO_(3)` is electron deficient and therefore, it is a Lewis acid. It ACCEPTS a `OH^(-)` ion from water liberating a proton. Therefore, it behaves as a monobasic acid.
40.

H_(3)BO_(3) acceptshydroxideionfromwatershownbelow, H_(3) BO_(3) (aq) + H_(2)O(l) -> B(OH)_(4)^(-)+ H^(+) Predictthe natureof H_(3)BO_(3) usingLewisconcept.

Answer»

Solution :Boricacidis alsocalledas hydrogenborateor orthoboricacid . ITIS aweakmonobasicLewisacidof borronandwrittenas `B(OH)_(4)` . Itacceptshydroxy`(OH^(-))` ION fromwater. Itdoesnotdissociateto givehydronium`H_(3) O^(+)` ion ratherformsionand thisions inturmgive`H_(3)O^(+)` ions
`B(OH)_(3)+ H_(2)O= (B(OH)_(4))^(-)= H_(3)O^(+)`
Henceit isconsideredas weakacid.
41.

H_3 C CH_2 CO_2 H underset(Delta)overset(NH_3)to X underset(Delta)overset(-OH)toY. What are X and Y ?

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`{:("" X,""Y),(H_3C CH_2 CONH_2,H_3C CH_2CH_2 CO_2 H):}`
`{:("" X,""Y),(H_3C CH_2 NH_2 , H_3C CH_2 CO_2H):}`
`{:("" X,""Y),(H_3C CH_2CONH_2 , H_3C CH_2 CO_2 H):}`
`{:("" X,""Y),(H_3C CH_2 CN , H_3 C CH_2 CONH_2):}`

ANSWER :C
42.

H_3 PO_3 has ……non ionisable P-H bonds

Answer»

NONE
1
2
3

Solution :Non-ionisable
43.

H_(3) C - CH = CH - CH_(2) - underset(NH_(2))underset(|)(CH) - CH_(2) - COOH The IUPAC name of the above compound is

Answer»

5-Amino-hex-2-en carboxylic ACID
5-Amino-2-heptenoic acid
3-Amino-5-heptenoic acid
`beta-`Amino-8-heptenoic acid

Solution :`UNDERSET("3-Amino -5-heptenoic acid")(H_(3)OVERSET(7)(C)-overset(6)(C)H=overset(5)(C)H-overset(4)(C)H_(2)-overset(3)(C)underset(NH_(2))underset(|)H-overset(2)(C)H_(2)-overset(1)(C)OOH)`
44.

CH_3- CH_2-overset(O ) overset(|| ) C - underset(CH_3) underset(|) overset(CH_3) overset(|) C- CH_3overset(PhCO_2 H)toA , Awill be

Answer»




SOLUTION :
45.

H_3 BO_3 accepts hydroxide ion from water as shown below H_3 BO_3(aq)+H_2 O (I) B(OH)_(4)^(-)+H^+ Predict the nature of H_3 BO_3 using Lewis concept.

Answer»

SOLUTION :
46.

H_(2)SO_(5) is

Answer»


ANSWER :B
47.

H_(2)SO_(4), reacts with PCl_(5) to give

Answer»

thionyl chloride
sulphuryl chloride
sulphur tetrachloride
phosphoric acid

Solution :`H_(2)SO_(4)+2PCl_(5) to underset(("Sulphuryl chloride"))(SO_(2)Cl_(2)+2POCl_(3)+2HCL)`
48.

Sulphuric acid reacts with PCl_5 to give

Answer»

Thionyl chloride
SULPHUR MONOCHLORIDE
Sulphuryl chloride
Sulphur ultrachloride

SOLUTION :`HO-SO_2 - OH + PCl_5 to Cl-SO_2 - OH + POCl_3 + HCL`
`HO-SO_2-OH+2PCl_5 to Cl-SO_2 -Cl + 2POCl_3 + 2HCL`
49.

H_2SO_4 reacts with sugar and act as:

Answer»

A DEHYDRATING agent
An OXIDIZING agent
A sulphonating agent
None

Answer :A
50.

H_2SO_4 is labelled as 9.8% by weight. Specific gravity of H_2SO_4 is 1.8. The volume of the acid to be taken to prepare 1000 ml of 0.18M solution is

Answer»

10 ml
100ml
740 ml
360 ml

Answer :B