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.

In the electrolysis of aqueous sodium chloride solution, which of the half cell reaction will occur at anode ?

Answer»

`Cl_((aq))^(-)to(1)/(2)Cl_(2)+e^(-),"" E_(cell)^(@)=1.36" volts"`
`2H_(2)O_((l))to O_(2)+4H^(+)+4e^(-) , E_(cell)^(@)=1.23" volts"`
`Na_((aq))^(+)+e^(-) toNa_((s)) , "" E_(cell)^(@)=-2.71" volts"`
`H_((aq))^(+)+e^(-) to (1)/(2)H_(2) , "" E_(cell)^(@)=0.00" volts"`

Solution :(a) It is the correct answer. For details, CONSULT section 3.15 (Electrolysis of aqueous NaCl solution).
Discussion
2.

In the following, the element with the highest ionisation energy is

Answer»

`[Ne]3s^(2)3P^(1)`
`[Ne]3s^(2)3p^(3)`
`[Ne]3s^(2)3p^(2)`
`[Ne]3s^(2)3p^(4)`

Solution :Half-filled orbitals and FULLY filled orbitals are more stable and HENCE they have higher ionisation energies. So `[Ne] 3s^(2)3p^(3)`has the highest ionisation ENERGY.
Discussion
3.

In the electrolysis of aq. CuSO_(4) solution with Pt electrodes, using 1 F of electricity, which of the following processes shall occur?

Answer»

1 eq. of CU is deposited at the cathode
1 eq. of `O_(2)` is liberated at the anode
1 eq. of `OH^(-)` is discharged at the anode
1 eq. of `H_(2)SO_(4)` is produced

Answer :A::B::C::D
Discussion
4.

In the following structure,the double bonds are marked as I, II, III and IV Geometrical isomerism is not possible at site (s):

Answer»

I
III
I and III
I and IV

Answer :A
Discussion
5.

In the electrolysis of an aqueous solution of a salt, the pH in the space near one of the electrodes increased. A solution of which of the following salts was electrolysed ?

Answer»

`CU(NO_(3))_(2)`
`CuCl_(2)`
KCl
none of these

Answer :C
Discussion
6.

In the electrolysis of aqueous NaCl solution, Cl_(2) is produced at the anode and not O_(2). This is due to____shown by water for oxidation to O_(2).

Answer»

SOLUTION :OVERVOLTAGE
Discussion
7.

In the electrolysis of an aqueous solution of NaF , the product obtained at the cathode and anode respectively are

Answer»

`Na, F_(2)`
`Na, O_(2)`
`H_(2) , O_(2)`
`H_(2) , F_(2)`

Solution :At CATHODE , `H_(2)O` is more easily REDUCED than `Na^(+)` to GIVE `H_(2)`
Discussion
8.

In the electrolysis of an aqueous Cr_(2)(SO_(4))_(3) solution using a current of 2 amp, the mass of cathode is increased by 8 g. How long was electrolysis conducted ?

Answer»


ANSWER :6.19 H
Discussion
9.

In the following statement, which combination of true (T) and false (F) options in correct? (A). Ionic mobility is highest of I^(-) in water as compared to other halides. (B). IF_(5) is square pyramidal and IF_(7) is pentangonal bipyramidal in shape (C). Reactivity order is F_(2)ltCl_(2)ltBr_(2)ltI_(2) (D). Oxidising power order is F_(2)ltCl_(2)ltBr_(2)ltI_(2)

Answer»

TFTF
TTFT
TTTT
TTFF

Solution :Fluorine is most REACTIVE AMONGST halides and has HIGHEST oxidising POWER.
Discussion
10.

In the electrolysis of alumina, cryolite is added to alumina to

Answer»

higher the m.p. of alumina
increase the electrical CONDUCTIVITY
MINIMIZE the ANODE effect
remove impurities from alumina

ANSWER :B
Discussion
11.

In the electrolysis of alumina cryolite is added to:

Answer»

lower the melting POINT of ALUMINA
increase the electrical conductivity
both
REMOVE impurities from alumina

Solution :Cryolite is added to lower the mpt of `Al_(2)_O_(3)` and to increase the electrical conductivity of `Al_(2)O_(3)`.
Discussion
12.

In the following, strongest acid is

Answer»

`CH_3CH_2COOH`
`CH_3COOH`
`C_6H_5COOH`
`C_6H_5CH_2COOH`

ANSWER :C
Discussion
13.

In the following strcuture anomeric carbon is

Answer»

1
2
3
4

Solution :C-1 is an anomeric CARBON and DUE to this it exists in TWO FORMS , `alpha` and `BETA`
Discussion
14.

In the following sets reactants which two sets best exhibits the amphoteric character of Al_(2)O_(3)*xH_(2)O? Set 1: Al_(2)O_(3)*xH_(2)O and OH^(-) Set 2: Al_(2)O_(3)*xH_(2)O and H_(2)O (aq) Set 3:Al_(2)O_(3)*xH_(2)O and H^(+)(aq) Set 4:Al_(2)O_(3)*xH_(2)O and NH_(3)(aq)

Answer»

1 and 2
1 and 3
2 and 4
3 and 4

Solution :Aluminium oxide is AMPHOTERIC oxide becouse it shows the properties of the both basic and ACIDIC oxides. It react with bothe acids and bases to FORM salt and water.
`Al_(2)O_(3)*xH_(2)O+2NaOH to NaAlO_(2)+H_(2)O`
`Al_(2)O_(3)*xH_(2)O+HCl to AlCl_(3)+H_(2)O`
Discussion
15.

In the electrolysis of alumina, cryolite is added to :

Answer»

Lower the melting POINT of alumina
Increase the electrical conductivity
Both (a) and (b)
REMOVE IMPURITIES from alumina

Answer :C
Discussion
16.

In the following species, the one which is likely to be the intermediate during benzoin condensation of benzaldehyde, is

Answer»

`Ph-Coverset((+))(-=)O`


`Ph-overset((-))(C )=O`

SOLUTION :
Discussion
17.

In the electrolysis of alumina , cryolite and CaF_(2) areadded to

Answer»

increase the emf of CELL
decrease the emf of cell
decrease the melting point
Both (b) and (c)

SOLUTION :CRYOLITE and `CaF_2`(fluorspar) are added to alumina in its electrolysis to decrease its melting point and to increase the electrical conductivity .
Discussion
18.

In the following species, the one which is likely to be intermediate during benzoic condensation of benzaldehyde, is

Answer»

`Ph-C-=overset(+)(O)`


`Ph-overset(-)(C)=O`

SOLUTION :
Discussion
19.

In the electrolysis of alkaline water , a total of mass 1 mole of gases is volved . The amount of water decomposed is

Answer»

1 mole
2 moles
2/3 mole
1/3 mole

SOLUTION :`H_(2) O to H_(2) + (1)/(2) O_(2)` .
SUPPOSE `H_(2)O` decomposed = x moles . Then `H_(2) = x , O_(2) = (1)/(2) x , i.e., x + (1)/(2) x = 1`
or `(3)/(2) x = 1 or x = (2)/(3)`
Discussion
20.

In the following series of reactions the major productsP and S are respectively.

Answer»




SOLUTION :
Discussion
21.

In the electrolysis of alkaline water, a total of 1 mole of gases in evolved. The amount of water decomposed in

Answer»

1 mole
2 MOLES
`(1)/(3)` mole
`(2)/(3)` mole.

Solution :`H_(2)OrarrH_(2)+1//2O_(2)`
1 mole of water gives on electrolysis `(3)/(2)` moles of gases
`THEREFORE ` To PRODUCE 1 mole of gases water electrolysed `=(2)/(3)` mol
Discussion
22.

In the following sets of compounds, the one which contains only medicinal compounds is:

Answer»

Dodecyl benzene, lauryl alcohol
Aspirin, alitame, phenolphthalein
 Boric acid, CHLORAMPHENICOL ,asprin.
pentaerythrityl STEARATE ,boric acid , morphine.

Answer :C
Discussion
23.

In the electrolysis of acidulated water, it is desired to obtain hydrogen at the rate of 1 cc per second at NTP condition. What should be the current passed?

Answer»

Solution :`2H^(+)+2E^(-)toH_(2)`
Thus, 1 mole of `H_(2)` i.e., 22400 cc at NTP requires 2F=`2xx96500`coulombs
`therefore`1cc at NTP requires `=(2xx96500)/(22400)xx1=8.616C`
As `Q=Ixxt""thereforeI=(Q)/(t)=(8.616)/(1s)=8.616`ampere
Discussion
24.

In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second uner S.T.P. condition. The current to be passed is

Answer»

A) 9.65 A
B) 19.3 A
C) 0.965 A
D) 1.93 A

Solution :No of MOLES of `H_(2)=1.12/22400`
`:.` No. of EQUIVALENTS of hydrogen
`=(1.12xx2)/(22400)=10^(-4)`
No. of Faradays required `=10^(-4)`
`:.` Current to be passed in 1 sec. `=96500xx10^(-4)`
`=9.65A`
Discussion
25.

In the following series of chemical reactions identify Z C_(3)H_(7)OH underset(160-180^(@)C)overset(Conc. H_(2)SO_(4))to X overset(Br_(2))to Y underset(Alc. KOH)overset("Excess of ") to Z

Answer»

`CH_(3)-UNDERSET(NH_(3))underset(|)CH-underset(NH_(2))underset(|)CH_(2)`
`CH_(3)-underset(OH)underset(|)CH-underset(OH)underset(|)CH_(2)`
`CH_(3)-underset(OH)underset(|)C=CH_(2)`
`CH_(3)C-=CH`

Solution :`CH_(3)CH_(2)CH_(2)OH underset(160-180^(@)C)overset("conc. " H_(2)SO_(4))to CH_(3)CH=CH_(2)overset(Br_(2))to CH_(3)-underset(Br)underset(|)CH-underset(Br)underset(|)Chunderset("In EXCESS")overset("alc." KOH)to underset("PROPYNE")(CH_(3)-C=-CH)`
Discussion
26.

In the electrolysis of acidulated water, it is desired to obtain 1.12 cc of hydrogen per second under S.T.P. condition. The current to be passed is

Answer»

9.65A
19.3A
0.965A
1.93A

Solution :No of moles of `H_(2)=(1.12)/(22400)`
`THEREFORE` No. of equivalents of hydrogen`=(1.12xx2)/(22400)=10^(-4)`
No. of FARADAYS REQUIRED`=10^(-4)`
`therefore`Current to be passed in 1 sec. `=96500xx10^-4`
`=9.65`A
Discussion
27.

In the following sequential transformation, considering only the major products formed in each step, what is the correct statement with respect to product 'Y' is?

Answer»

It GIVES a positive TOLLEN's test and is a FUNCTIONAL isomer of 'X'
It gives a positive Tollen's test and is a GEOMETRICAL isomer of 'X'
it gives a positive IODOFORM test and is a functional isomer of 'X'
It gives a positive iodoform test and is a geometrical isomer of 'X'

Solution :N//A
Discussion
28.

In the electrolysis of acidified water, it is desired to obtained 1.12 cc of hydrogen per second under STP condition. The current that should be passed is:

Answer»

`9.65` A
`19.3` A
`0.965` A
`1.93` A

Answer :D
Discussion
29.

In the following sequences of reactions. 'A' overset("Reaction")to'B' overset(HNO_(2))toCH_(3)CH_(2)OH The compound 'A' is:

Answer»

PROPANE NITRILE
ETHANE nitrile
nitromethane
methyl isocyanate

Answer :B
Discussion
30.

In the electrolysis of a metal oxide, name a substance that helps to lower the melting point and to increase the conductivity.

Answer»

SOLUTION :CRYOLITE
Discussion
31.

In the following sequences of reaction CH_(3)Broverset(KCN)toAoverset(H_(3)O^(+))toBunderset(ether)overset(LiAlH_(4))toC The end product (C) is.

Answer»

acetone
methane
acetaldehyde
ethyl alcohol.

Solution :.
Discussion
32.

Identify (Z) in the series. C_(3)H_(7)OH underset(443K)overset("conc."H_(2)SO_(4))(rarr)(X)overset(Br_(2))(rarr)(Y) underset("alc. KOH")overset("Excess of ")(rarr)(Z)

Answer»

`CH_3-UNDERSET(NH_2) underset(|)CH-underset(NH_2)underset(|)CH_2`
`CH_3-underset(OH)underset(|)CH-underset(OH)underset(|)CH_2`
`CH_3-underset(OH)underset(|)C=CH_2`
`CH_3C-=CH`

ANSWER :D
Discussion
33.

In the electrolyses of CuSO_4 the reaction Cu^(2+)+2^(@-)rarr Cu , Takes place at :

Answer»

Anode
Cathode
In solution
None

Answer :B
Discussion
34.

In the following sequence the product D is, CH-= CH overset(HBr)rarr A overset(HBr)rarr B overset(KOHalc)rarr Coverset(NaNH_2)rarr D

Answer»

Ethanol
Ethane
Ethyne
Ethanal

Answer :C
Discussion
35.

In the electrodeposition of Ag , the silver ions are

Answer»

REDUCED to ANODE
reduced to anode
oxidised to anode
oxidised to cathode

SOLUTION :`Ag^(+) + E to Ag`
Discussion
36.

In the electro-deposition of Ag, the silver ions are :

Answer»

REDUCED at anode
Reduced at cathode
Oxidised at anode
Oxidised at cathode

Answer :B
Discussion
37.

In the electrochemical cell Zn|ZnSO_(4)(0.01M)||CuSO_(4)(1.0M)|Cu, the emf of this Daniel cell is E_(1). When the concentration of ZnSO_(4) is changed to 1.0M and that of CuSO_(4) changed to 0.01M, the emf changes to E_(2). From the followings, which one is the relationship between E_(1) and E_(2) (given,(RT)/(F)=0.059)

Answer»

`E_(1) lt E_(2)`
`E_(1)gtE_(2)`
`E_(2)=0neE_(1)`
`E_(1)=E_(2)`

SOLUTION :For cell
`Zn|ZnSO_(4)(0.01M)||CuSO_(4)(1M)|Cu`
Cell reaction `toZn+Cu^(+2)toZn^(+2)+Cu`
`E_(1)=E^(o)-(0.059)/(2)"log"(Zn^(+2))/(Cu^(+2))`
`E_(1)=E^(o)-(0.059)/(2)"log"(0.01)/(1)`
`=E^(o)-(0.059)/(2)"log"(1)/(100)`. . . .(i)
Fore cell
`Zn|ZnSO_(4)(1M)||CuSO_($)(0.01M)|Cu`
`E_(2)=E^(o)-(0.059)/(2)"log"(1)/(0.01)`
`=E^(o)-(0.059)/(2)log100` . . . . .ltBrgt `E_(1)gt E_(2)`
Discussion
38.

In the following sequence the product ( C) is :CH_3CHOoverset(H_2)underset(Pd)rarr(A)overset(Na)rarr(B)overset(CH_3l)rarr(C)

Answer»

ALCOHOL
Ether
Alkene
None

Answer :B
Discussion
39.

In the electrochemical cell: Z|ZnSO_(4)(0.01M)||CuSO_(4)(1.0M)|Cu, the e.m.f. this daniel cell is E_(1). When the concentration of ZnSO_(4) is changed to 1.0 M and that of CuSO_(4) is changed to 0.01 M, the e.m.f. changes to E_(2). from the following, which one is the relationship between E_(1) and E_(2)? (Given, (RT)/(F)=0.059)

Answer»

`E_(1)=E_(2)`
`E_(1) gt E_(2)`
`E_(1) gt E_(2)`
`E_(2)=0neE_(2)`

Solution :The cell reaction is
`Zn+CuSO_(4)toZnSO_(4)+CU,n=2`
`E=E_(cell)^(@)-(2.303RT)/(2F)"log"([ZnSO_(4)])/([CuSO_(4)])`
In 1 st CASE, `E_(1)=E_(cell)^(@)-(2.303RT)/(2F)"log"(0.01)/(1)`
In 2nd case, `E_(2)=E_(cell)^(@)-(2.303RT)/(2F)"log"(1)/(0.01)`
Evidently, `E_(1)gtE_(2)`.
Discussion
40.

In the following sequence of reactions, Zoversett(PCl_(5))to X overset("Alc. KOH")to Y underset((ii)" "H_(2)O", boil")overset((i)" conc. "H_(2)SO_(4))to Z is

Answer»

`CH_(3)CH_(2)CH_(2)OH`
`CH_(3)CHOHCH_(3)`
`CH_(3)CH_(2)CH_(2)CH_(2)OH`
`(CH_(3))_(3)C CH_(2)OH`

Solution :Since alkenes give Markovnikov's ADDITION products, therefore, `1^(@)` ALCOHOLS (a), (c) and (d) cannot be obtained by hydration. Thus, Z must be 2-propanol, i.e.,
`underset((Z))(CH_(3)CHOHCH_(3))overset(PCl_(5))to underset((X))(CH_(3)CHClCH_(3))underset(-HCL)overset("Alc. KOH")to underset((Y))(CH_(3)CH=CH_(2)) underset((ii)" "H_(3)O^(+))overset((i)" conc. "H_(2)SO_(4))to underset((Z))(CH_(3)CHOHCH_(3))`
Discussion
41.

In the electrochemical cell : Zn|ZnSO_4 (0.01M)|CuSO_4 (1.0 M)|Cu, the emf of this Daniel cell is E_1. When the concentration of ZnSO_4, is changed to 1.0 M and that CuSO_4 changed to 0.01M, the emf changes to E_2. From the following , which one is the relationship between E_1 and E_2 ?

Answer»

`E_1 LT E_2`
`E_1 GT E_2`
`E_2 = 0 UARR E_1`
`E_1 = E_2`

Solution :`E_("cell") = E_("cell")^(@) - (0.0591)/(2) "LOG" ([zn^(2+)])/([Cu^(2+)])`
`E_1 = E_("cell")^(@) - (0.0591)/(2) "log" (10^(-2))/(1) "" Zn (S) to Zn^(2+)(aq) + 2e^(-)`
`E_1 = E_("cell")^(@) + 0.0591 ………..(1) "" Cu^(2+)(aq) + 2e^(-) to Cu(s)`
`E_2 = E_("cell")^(@) - (0.0591)/(2) log 1/(10^(-2)) "" Zn(s) + Cu^(2+)(aq) to Zn^(2+)(aq) +Cu(s)`
`E_2 = E_("cell")^(@) - 0.0591 .............(2)`
`:. E_1 > E_2`.
Discussion
42.

In the electrochemical cell- Zn|ZnSO_(4)(0.01M)||CuSO_(4)(1.0M)|Cu, the emf of this Daniel cell is E_(1). When the concentration ofZnSO_(4) is changed to 1.0M and that of CuSO_(4) changed to 0.01M, the emf changes to E_(2). From the following , which one is the relationship between E_(1) and E_(2) ? (Given , (RT)/(F) = 0.059)

Answer»

`E_(1) =E_(2)`
`E_(1)LT E_(2)`
`E_(1) gt E_(2)`
`E_(2) = equalnot E_(1)`

Answer :C
Discussion
43.

In the following sequence the compound (D) is , (##MBD_HKR_CHE_XII_P02_C12_E07_425_Q01##)

Answer»

ACETIC anhydride
Acetone
Ethanol
Acetaldehyde

Answer :B
Discussion
44.

In the following sequence of reactions, what is D?

Answer»

PRIMARY amine
an amide
phenyl isocyanate
a CHAIN LENGTHENED hydrocarbon

Answer :C
Discussion
45.

The emf of a Daniell cell at 298 K is E_(1) Zn|ZnSO_(4)(0.01 M)||CuSO_(4)(1.0M)|Cu When the concentration of ZNSO_(4) is 1.0 M and that of CuSO_(4) is 0.01 M, the emf changed to E_(2). What is the relationship between E_(1) and E_(2) ?

Answer»

`E_1ltE_2`
`E_1gtE_2`
`E_2=0neE_1`
`E_1=E_2`

ANSWER :B
Discussion
46.

In the following sequence of reactions, what is D ?

Answer»

Primary amine
An amide
Phenyl isocyanide
Chain lengthened hydrocarbon.

Solution :`underset("Toluene")(C_(6)H_(5)CH_(3)) underset("OXIDATION")OVERSET(O)(rarr) underset(underset((A))("Benzoic (acid)"))(C_(6)H_(5)COOH) underset(-SO_(2), -HCl)overset(SOCl_(2))(rarr) underset((B))underset("Benoyl chloride")(C_(6)H_(5)COCL) underset(-NaCl)overset(NaN_(3))(rarr) underset(underset((C))("BENZOYL azide"))(C_(6)H_(5)CON_(3)) underset(underset("rearrangemet)")("(CURTIUS"))overset("Heat")(rarr) underset((D))underset("Phenyl isocynanate")(C_(6)H_(5)-N=C=O)`
Discussion
47.

In the electrochemical cell: Znabs(ZnSO_(4)(0.01M))abs(CuSO_(4)(1.0M))Cu, the emf of this Daniel cell is E_(1). When the concentration of ZNSO_(4) is changed to 1.0 M and that CuSO_(4) changed to 0.01M, the emf changes to E_(2). From the followings, which one is the relationship between E_(1) " and "E_(2)?

Answer»

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

Answer :B
Discussion
48.

In the following sequence of reactions Tolueneoverset(KMnO_4)to A overset(SOCl_2)to B underset(BaSO_4)overset(H_2//Pd)to C The product C is

Answer»

`C_6H_5COOH`
`C_6H_5CH_3`
`C_6H_5CH_2OH`
`C_6H_5CHO`

SOLUTION :
Discussion
49.

In the eextraction of copper from its sulphide ore, th metal is finally obtained by the reduction of cuprous oxide with

Answer»

Copperf (1) SULPHIDE `(CU_(2) S)`
Carbon MONOXIDE (CO)
IRON sulphide (FeS)
Sulphur dioxide `(SO_(2))`

Answer :A
Discussion
50.

In the electro refining process, the impure metal is made of

Answer»

Anode
Cathode
Both
None

Answer :A
Discussion