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.

Empirical formula and molecular formula of monomer of natural rubber are respectively

Answer»

`C_(5)H_(8),C_(5)H_(8)`
` C_(5)H_(8)( C_(5)H_(8))N`
` C_(4)H_(8),C_(4)H_(6)`
` C_(5)H_(12), C_(5)H_(8)`

ANSWER :A
Discussion
2.

Emperical formula of compound is SF_(4). At 27^(@)C 0.1gm of gaseous compound occupies a volume of 22.2 ml and exert a pressure 1 atm. What is molecular formula of gas. [Take R=0.08atm lit k^(--1) "mol"^(-1)]

Answer»

`SF_(4)`
`S_(2)F_(8)`
`S_(3)F_(12)`
`S_(4)F_(6)`

SOLUTION :PV=nRT M.F= `(SF_(4))_(X)`
`(1.02)xx22xx10^(-3)`
`(0.1)/((108)x)xx0.0821xx293`
`x=(0.1xx0.0821xx293)/((108)(1.02)(22.1)xx10^(-3))=1`
M.F. `=SF_(4)`
Discussion
3.

E_(MnO_4^(-)// Mn^(+2), H^(+)) can be 0.48 V greaterthan from E^@ value if the [MnO_4^(-)] is equal to [Mn^(+2)] at a P^(H) of ___________

Answer»

Solution :`MnO_4 + 8 H^(oplus) + 5e^(-) to Mn^(+2) + 4H_2O , E= E^(0) - (0.0591)/(5) log""([Mn^(+2)])/([MnO_4^(-)][H^(oplus)]^(8))`
` (E-E^(0)) = (-0.0591)/(2) log [H^(oplus)]^(-8) , 0.48 = -(0.0591)/(5) xx (-8) log[H^(oplus)]`
` P^(H) - log[H^(oplus)] = [(0.048 xx 5)/(0.0591xx 8)] = 5.076`
Discussion
4.

E_(Mn)^(@)+7|""_(Mn)+2 = 1.5 V E_(Mn)^(@)+4|""_(Mn)+2 = 1.2 V, then E_(Mn)^(@)+7|""_(Mn)+4 is

Answer»

A) `0.3 V`
B) `0.1 V`
C) `1.7 V`
D) `2.1 V`

Solution :(c): `Mn^(7+)+5e^(-) to Mn^(2+) , E^(@)` = 1.5 V
`DeltaG_(1)^(@) = -NFE^(@) = -5 xx F xx 1.5 rArr -7.5` F...(i)
`Mn^(4+) + 2e^(-) to Mn^(2+), E^(@) = 1.2` V
`DeltaG_(2)^(@) = -nFE^(@) = -2xxFxx1.2 = -2.4` F...(ii)
On subtracting eqn (ii) from (i), we get,
`Mn^(7+) + 3E^(-) to Mn^(4+), E^(@) =` ?
`DeltaG_(3)^(@) = DeltaG_(1)^(@) - DeltaG_(2)^(@)`
`-3FE^(@) = -7.5 F-(-2.4 F) rArr -3FE^(@) = -5.1 F`
`:. E^(@)` = 1.7 V
Discussion
5.

Emission of beta -particle by an atom of an element results in the formation of its

Answer»

Isotope
Isomer
Isomorph
Isobar

Solution :`._(1)H^(3) rarr ._(2)He^(3) + ._(-1)E^(0)`
`._(1)H^(3) and ._(2)He^(3)` are isobars (same mass no.)
Discussion
6.

Emission of beta-particle by an atom of an element results in the formations of:

Answer»

isobar
isomer
isotope
isotone

Answer :a
Discussion
7.

Emission of beta -particle is equivatent to:

Answer»

INCREASE of ONE PROTON only
decrease of one NEUTRON only
both (a) and (b)
none of these

Answer :C
Discussion
8.

Emission is caused by the transformation of one neutron into a proton. This results in the formation of a new element having

Answer»

Same nuclear charge
Very lower nuclear charge
Nuclear charge higher by one unit
Nuclear charge lower by one unit

Solution :`._(0)n^(1) RARR ._(+1)p^(1) + ._(-1)E^(0)` (`beta` -particle COMES out)
Discussion
9.

Emission of a beta- particle by an atom of the element results in the formation of:

Answer»

ISOTOPE
ISOBAR
ISOMER
isomorph.

Answer :B
Discussion
10.

E.m.f. of the cell Zn|Zn^(2+)(1)||Cu^(2+)(1M)|Cu is 1.1 volt. If the standard reduction potential of Zn^(2+)|Znis -0.78 volt, what is the oxidation potential of Cu|Cu^(2+) ?

Answer»

`+1.86V`
`0.32V`
`-0.32V`
`-1.86V`

ANSWER :C
Discussion
11.

emf value of spontaneous redox reaction will be. . . .

Answer»

Zero
Negative
POSITIVE
none of above

Solution :Positive.
Discussion
12.

Emil Fischer was awarded Nobel Prize for his work on :

Answer»

SUGARS and PURINE synthesis
Ammonia discovery
Optical activity
Alkaloid synthesis

Answer :A
Discussion
13.

Emf of the cell Pt. H_(2)("1 atm")|H^(+) (aq)||AgCl|Ag is 0.27 V and 0.26 V at 25^(@)C and 35^(@)C. Heat of reaction occuring inside the cell at 25^@C is

Answer»

`-54.8kJ`
`26.05kJ`
`-26.05kJ`
`+54.8kJ`

ANSWER :A
Discussion
14.

E.M.F of the cell reaction , 2Ag^(+) + Cu to 2Ag + Cu^(2+)is 0.46 V .If E_(Cu^(2+) // Cu)^(0) is | 0.34 V , E_(Ag^(+)//Ag)^(0) is

Answer»

0.80 V
0.12 V
0.40 V
1.60 V

Answer :A
Discussion
15.

Emf of given Daniell cell at 298 K is E_(1). Zn|ZnSO_(4)(0.01M)||CuSO_(4(1.0M))|Cu. The emf changed to E_(2) when concentration of ZnSO_(4) solution is 1.0 M and CuSO_(4) solution is 0.01 M, then what is the relation between E_(1) and E_(2) ?

Answer»

`E_(1) GT E_(2)`
`E_(1) LT E_(2)`
`E_(1)=E_(2)`
`E_(2)=0 ne E_(1)`

Solution :`Zn+CU^(2+) to Zn^(2+)+Cu`
`E=E^(@)-(0.059)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
`E_(1)=E^(@)-(0.059)/(2)"log"(0.01)/(1.0) THEREFORE E_(1)=E^(@)+0.059`
`E_(2)=E^(@)-(0.059)/(2)"log"(1.0)/(0.01)therefore E_(2)=E^(@)-0.059`
`therefore E_(1) gt E_(2)`.
Discussion
16.

Emf of hydrogen electrode in tern of pH is (at 1 atm pressure):

Answer»

`E_(H_(2)) = (RT)/(F) XX pH`
`E_(H_(2)) = (RT)/(F)(1)/(pH)`
`E_(H_(2)) = (2.303RT)/(F) pH`
`E_(H_(2)) = -0.0591 pH`

ANSWER :D
Discussion
17.

EMF of hydrogen electrode in terms of pH is (at 1 atm pressure)

Answer»

`E_(H_(2)=(RT)/(F)xxpH`
`E_(H_(2))=(RT)/(F)=(1)/(pH)`
`E_(H_(2))=(2.303 RT)/(F)pH`
`E_(H_(2))=-0.591 pH`

Solution :`2H^(+)+2e^(-)rarrH_(2)`
Accordingto nernst equation
`E=E^(@)+(2.303 RT)/(NF) log (1)/(H^(+)]^(2)`
`=-(2.303 RT)/(2xxF)-log (H^(+))^(2)=(2.303RT)/(F)pH`
Discussion
18.

EMF of Daniell cell was found using different concentrations of Zn^(2+) ion and Cu^(2) ion. A graph was then plotted between E_(cell) and log([Zn^(2+)])/([Cu^(2)]). The plot was found to be linear with intercept on. E_(cell) axis equal to 1.10 V. Calculate E_(cell) for Zn|Zn^(2+)(0.1M)||Cu^(2+)(0.01M)|Cu

Answer»

Solution :For DANIELL cell, `Zn+Cu^(2+)TOZN^(2+)+Cu`
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"([Zn^(2+)])/([Cu^(2+)])`
it is the equation of straight line (y=c+mx).
INTERCEPT`=E_(cell)^(@)=1.10V` (GIVEN)
`thereforeE_(cell)^(@)=1.10-(0.0591)/(2)"log"(0.1)/(0.01)=1.10-0.0295=1.0705`V
Discussion
19.

E.M.F. of following cell is 0.265 V at 25^@C and 0.2595 V at 35^@C. Calculate heat of the reaction taking place at 25 ^@C Pt(H_2) | HCl (g) | | AgCl | Ag (s)

Answer»


ANSWER :14.227 KJ
Discussion
20.

emf of cell Ni, Ni^(2+)(1.0M) || Au^(3+)(1.0M), Au is …………If E^@ for Ni^(2+)|Ni is 0.25V, E^@ for Au^(3+)| Au is 1.50 V.

Answer»

`+1.25V`
`1.75V`
`+1.75V`
`+4.0V`

ANSWER :C
Discussion
21.

EMF of a cell whose half cells are given below is Mg^(2+)+2e^(-)toMg(s),E=-2.37V Cu^(2+)+2e^(-)toCu(s),E=+0.33V

Answer»


SOLUTION :`E_(CELL)^(o)=E_("CATHODE")^(o)-E_("ANODE"),""E_(cell)^(o)=0.34-(-2.37)`
`E_(cell)^(o)=2.71V`
Discussion
22.

EMF of a cell in terms of reduction potential of its left and right electrodes is:-

Answer»

`E=E_("left")-E_("right")`
`E=E_("left")+E_(right)`
`E=E_(right)-E_("left")`
`E-(E_(right)+E_("left"))`

SOLUTION :(Reduction POTENTIAL of CATHODE)-(reduction potential of ANODE).
Discussion
23.

EMF of a cell in terms of reduction potential of its left and right electrodes is

Answer»

`E = E_("LEFT") - E_("RIGHT")`
`E=E_("left") + E_("right")`
`E=E_("right") - E_("left")`
`E=-(E_("right") + E_("left"))`

ANSWER :C
Discussion
24.

Emery consists of

Answer»

IMPURE corundum
Impure carborundum
Impure graphite
Purest FORM of iron

Answer :A
Discussion
25.

Elucidate the structure of glucose.

Answer»

Solution :(i) Its moleculer formula is found to be `C _ 6 H_ (12) O _ 6 `.
(ii) On heating with HI it forms N-hexane suggesting that all the six carbon atoms are linked in a straight chain.
` overset(CHO) overset(|) underset(CH_2OH) underset(|) ((CHOH)_4) overset(HIDelta)tounderset("n- hexane")(CH _3 - CH _2 - CH _2- CH _2- CH_ 2- CH _3 ) `
(iii) Glucose reacts with hydroxylamine to form an oxime. This show it contain carbonyl group.
 `overset(CHO)overset(|) underset(CH_2OH) underset(|)((CHOH)_4) overset(NH_2OH) to overset(CH= N -OH)overset(|) underset("Oxime")underset(CH_2OH) underset(|)((CHOH)_4)`
(v)Inpressureofbrominewater,glucoseundergoesoxidationforminggluconicacid.Thisshowscarbonyl groupisanaldehyde group.
`overset(CHO) overset(|) underset( CH_2 OH) underset(|) ((CHOH)_4) overset(Br_2 "water")to overset(COOH) overset(|)underset(CH_2OH) underset(|) ((CHOH) _4)`
(v) ACETYLATION of glucose with acetic anhydride gives glucose penta acetate which confirms the presence of five OH groups.
(vi) On oxidation with strong oxidising AGENT. like Cone. `HNO_3` it gives saccharic acid. This indicate the presence ofprimaryalcoholic group.
Based on the above facts Fischer gave open and ring STRUCTURE for glucose.
Discussion
26.

Elucidate the structure of fructose.

Answer»

Solution :Fructose is the sweetest of all known sugars. It is readily SOLUBLE in water. Fresh solution of fructose has a specific rotation -`133^(@)` which changes to - `92^(@)` at equilibrium due to mutarotation. Similar to glucose, the structure of fructose is deduced from the following facts.
(i) Elemental analysis and molecular weight determination of fructose show that it has the molecular formula `C_(6)H_(12)O_(6)`.
(ii) Fructose on reduction with HI and red phosphorus gives a mixture of n - hexane (major product) and 2 - iodohexane (minor product). This reaction INDICATES that the six carbon atoms in fructose are in a straight chain.
`"Fructose"overset(HI//P)underset("Reduction")rarrCH_(3)(--CH_2--)_(4)CH_(3)+CH_(3)-underset(I)underset(|)"CH"(--CH_(2)--)_(3)CH_(3)`
(iii) Fructose reacts with NH20H and HCN. It shows the presence of carbonyl groups in the molecule of fructose.
(iv) Fructose reacts with acetic anhydride in the presence of PYRIDINE to form penta acetate. This reaction indicates the presence of fivehydroxyl groups in a fructose molecule.
(v) Fructose is not oxidized by brom~ne water. This rules out the possibility of presence of an aldehyde (-CHO) group.
(vi) Partial reduction of fructose with sodium amalgam and water produces mixtures of sorbitol and mannitol which are epimers at second carbon. New asymmetric carbon isformed at C-2. This confirms the presence of keto group.

(vii) On oxidation with nitric acid, it gives GLYCOLIC acid and tartaric acids which contain smaller number of carbon atoms than in fructose.

This shows that a keto group is presence of 10 alcoholic groups at C-1 and C-6.
Discussion
27.

Ellution is the process for :

Answer»

Crystsllisation of COMPOUND
Separation of compound
Extraction of compound
Distillation of compound

ANSWER :B
Discussion
28.

Ellingham diagram represents a graph of

Answer»

`DeltaGvsP`
`DELTAS VS P`
`DeltaG^(@) vs T`
`DeltaG vs T`

Solution :Ellingham diagram - PLOT of `DELTA G^(@) ("Kj")/("mol")` vs temperature `.^(@)C)` where `Delta G^(@)` is for the reaction :
Metal + `O_(2) RIGHTARROW` oxide (s)
Discussion
29.

Ellingham diagram represents :

Answer»

change of `/_\G` with temperature
change of `/_\H` with temperatrue
change of `/_\G` with pressure
change of `(/_\G-T/_\S)` with temperature

Solution :Ellingham diagram REPRESENTS change in VALUE of `DELTAG` with temperature (in K).
Discussion
30.

Ellingham Diagram is useful

Answer»

To know thetempreature where phase TRANSFORMATION ocuurs
To know the temperature where METAL oxide DECOMPOSE on its own
To select suitable REDUCING agent for REDUCTION of metal oxide
All of the above

Answer :4
Discussion
31.

Ellingham diagram is given below for the formation of some oxides. Then select the correct combination

Answer»

Below
`T_(2)Al_(2)O_(3)+ 3Mg to 3MgO+2Al, DELTAG= -ve`
Below
`T_(3)MgO+ CO to CO_(2)+ Mg, DeltaG= -ve`
Below
`T_(3)MgO+ CO to CO_(2)+ Mg, DeltaG= -ve`
all of the above

Solution :If `DeltaG =-ve`the process is SPONTANEOUS
Discussion
32.

Ellingham diagram is given below for the formation of some oxides. The select the correct combination

Answer»

Below `T_(2).Al_(2)O_(3)+3Mg® MgO+2Al,DG=-ve`
Below `T_(3),MgO+CO® CO_(2)+Mg,DG=-ve`
Above `T_(4),Al_(2)O_(3)+3CO® 2Al+3CO_(2) DG=-ve`
Below`T_(1)` MgO is in liquid state

Answer :1
Discussion
33.

Ellingham diagram can be drawn in case of -

Answer»

`sulphide
oxide
halide
all

Answer :D
Discussion
34.

Ellingham diagram does not represents

Answer»

Change of `DELTAG` with TEMPERATURE
Change of `DELTAH` with temperature
Change of `DeltaG` with pressure
Change of `(DeltaG-TDeltaS)` with temperature

Solution :Ellingham diagram gives change of `DeltaG` with temperature
Discussion
35.

Elimination reaction (especially beta-elimination) are as common as the nucleophilic substitution reaction in case of alkyl halides. Specify the reagents used in both cases.

Answer»

Solution :Both elimination reactions (particularly `beta`-elimination reactions) and nucleophilic substitution reactions occur simultaneously. However, by a proper choice of reagents and REACTION conditions, we can make one reaction to predominate over the other. Usually strong and bulkier bases and high temperature FAVOUR elimination reactions whereas weaker and smaller bases and lower temperature favour substitution reactions. For example, ethyl bromide on heating with alcoholic KOH (which contains the stronger base `C_(2)H_(5)O^(-)` ion) at about 473-523K undergoes elimination to give ethene while with aqueous KOH at about 373K it gives ethanol.
`UNDERSET("Ethanol")(CH_(3)CH_(2)OH) underset("373K (Substitution)")overset(KOH(aq))to underset("Ethyl bromide")(CH_(3)CH_(2)Br) underset("473-523K (Elimination)")overset(KOH(alc))to underset("Ethene")(CH_(2)=CH_(2))`
Discussion
36.

Elimination of bromine and hydrogen from 2-bromobutane results in the formation of

Answer»

PREDOMINANTLY TRANS 2-butyne
equimolar MIXTURE of 1-butene and 2-butene
predominantly CIS 2-butyne
predminantly 1-butyne

Answer :A
Discussion
37.

Elimination of bromine from 2-bromobutane results in the formation of

Answer»

equimolar mixture of 1 and 2-butene
PREDOMINANTLY 2-butene
predominantly 1-butene
predominantly 2-butyne

Solution :`overset(beta)(H_(3))underset(Br)underset(|)overset(ALPHA)(C) - C overset(beta)(H) - CH_(2) - CH_(3) underset(-KBr, -H_(2)O)overset(alc.KOH)rarr underset(("Major 80%"))underset("But-2-ene")(H_(3)C - CH )=CH - CH_(3) + underset(("MINOR 20%"))underset("But-1-ene")(CH_(2) = CH - CH_(2)) - CH_(3)`
Discussion
38.

Elevation in p.b of a molar glucose solution (d=1.2g mL^(-1)) is :

Answer»

`0.98K_(B)`
`K_(b)`
`1.20K_(b)`
`1.02K_(b)`

ANSWER :A
Discussion
39.

Elevation of boiling point is directly proportional to

Answer»

MOLALITY of the solution
DEPRESSION of FREEZING POINT in the same solution
both of these
none of these

Solution :It is a FACT.
Discussion
40.

Elevation in b.p of an aqueous urea solution is 0.52^(0). (K_(b)=0.52^(0)"mol"^(-1)kg). Hence, mole -fraction of urea in this solution is :

Answer»

0.982
0.0567
0.943
0.018

Answer :D
Discussion
41.

Elevation in b.p of a 0.5 molar glucose solution (d=1.2gmL^(-1)) is :

Answer»

`0.45K_(B)`
`K_(b)`
`1.20K_(b)`
`1.02K_(b)`

ANSWER :A
Discussion
42.

Elevation in boililng point was 0.52^(@)C when 6 gm of a compound X was dissolved in 100 gm of water. Molecular weight of X is (K_(b) for water is 0.52 per 1000 gm of water)

Answer»

120
60
180
600

Solution :`DELTA T_(b)=(100xx K_(b)xx W)/(m xx W) THEREFORE 0.52=(100xx5.2xx6)/(m xx 100)`
`m=(100xx5.2xx6)/(0.52xx100)=60`.
Discussion
43.

Elevation in boiling point was 0.52^@C when 6 g. of a compound X was dissolved in 100 g of water. Molecular weight of X is-

Answer»

120
60
180
342

Answer :B
Discussion
44.

Eletrolysis of aqueous solution of sodium salt of acids giving hydrocarbon is ________ reaction.

Answer»

Knoevenagal
HVZ
Kolbe's
none of the above

Answer :C
Discussion
45.

Elements of group 11 and 12 are:

Answer»

NORMAL elements
Transition elements
Alkaline EARTH metals
Alkali metals

Answer :B
Discussion
46.

Elements X, Y and Z have atomic numbers 19,37 and 55 respectivley. Which of the following statements is/are true about them ?

Answer»

Their ionization POTENTIAL would INCREASE with INCREASING atomic number
Y would have an ionization potential between those of X and Z
Z would have the highest ionization potential
Y would have the highest ionization potential.

Solution :Elements X, Y and Z with atomic numbers 19, 37, 55 lie in group 1 (alkali metals). WITHIN a group, IE decrease from top to bottom. Therefore, IE of Y could be between those of X and Z.
Discussion
47.

Elements X and Y form two different compounds. In the first, 0.324 g of X is combined with 0.471 g of Y. In the second, 0.117 g of X is combined with 0.509 g of Y. Show that these data illustrate the Law of Multiple Proportions.

Answer»

SOLUTION :CALCULATE the MASSES of Y which combine with 1 g of X in each case which are 1.454g and 435 g, i.e., in the RATIO `1:3`
Discussion
48.

Elements with their electronic configuration are given below: Answer the following questions:

Answer»

`I LT III lt IV lt V`
`V lt IV lt III lt I`
`I lt IV lt V lt III`
`V lt IV lt I lt III`

ANSWER :D
Discussion
49.

Elements with their electronic configuration are given below: Answer the following questions:

Answer»

I and IV
I and V
III and IV
III and V

Answer :D
Discussion
50.

Elements with their electronic configuration are given below: Answer the following questions:

Answer»

I
II
III
IV

Answer :B
Discussion