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.

Di-n-butylphthalate is a .......................

Answer»


ANSWER :PLASTICIZER
Discussion
2.

Dextrorotatory substance or d-Isomer.

Answer»

SOLUTION :An optically active substance (or isomer) which rotates the PLANE of a plane polarized light to the right hand side (RHS) is CALLED dextrorotatory substance (or isomer) and DENOTED by d or `(+)` sign.
Discussion
3.

Dextrorotary alpha- pinenehasaspecific rotation [alpha]_(D)^(20) =+52.3^(@).A sampleofalpha- pinenecontainingboth theenantiomerswasfoundto havea specificrotationa value [alpha]_(D)^(20) = + 30.8^(@). Thepercentagesofthe (+ )and (-)enantiomerspresentin thesamplearerespectively.

Answer»

`70% ` and 30%
80% and 20%
20 % and 80 %
60 % and 40 %

Answer :B
Discussion
4.

Dextro-isomers rotate the plane of polarised light towards right. (R) Dextro-isomers are represented by putting (D) before their name.

Answer»


ANSWER :C
Discussion
5.

Dextron is

Answer»

POLYESTER
POLYAMIDE
polycellulose
POLYPROPYLENE

ANSWER :A
Discussion
6.

Dextro -tartaricacidand meso- tartaric acidare :

Answer»

enatiomers
RACEMIC MIXTURE
DIASTEROMERS
sterosomers

Answer :C
Discussion
7.

Devise a series of reactions to convert benzene into meta-chlorobrombenzene. Select reagents and conditions from the following table, listing them in the order of use {:(,"Compound",,"Compound",,"Compound",),(1,"Sulphuric acid (conc.) heat",5,Mg "in ether",9,Cu_(2)Br_(2) + HBr,),(2,Cl_(2) + FeCl_(3) " and heat",6,PBr_(3),10,(CH_(3)CO)_(2)O + " Pyridine",),(3,NaNO_(2) +H_(3)O^((+)) 0^(@)C,7,H_(3)PO_(2),,,),(4,H_(2) " Pt catalyst",8,HNO_(3) ("conc.") + H_(2)SO_(4) ("conc.") " and heat",,,):}

Answer»

1 then 2 then 6
2 then 8 then 4 then 3 then 9
8 then 4 then 10 then 2 then 3 then 9
8 then 2 then 4 then 3 then 9

Solution :
Discussion
8.

Devise sequences for preparing the following compounds from benzene

Answer»





Solution :(a) The process can be started if we can do the nitration of benzene ring two times and then can reduce one of the group selectively. AMMONIUM sulphide allows selective reduction of one NITRO group in the presence of anothe. The mechanism of this reaction is complex. It is the transformation itself which you need to know.

(b) The target can be achieved if wee can form phenol from benzene. Formation of phenol from benzene is not straightforward. The route shown below involves the nitration, reduction to aniline and diazotisation. "Hydrolysis" of the diazonium ion gives phenol.

The subsequent steps involve temporary introduction of a `-SO_(3)` group as a protecting group of the para-position. This allows selective bromination of the ortho-position.(Note : Which products would you get without the `-SO_(3)` group?)

(c) Alkylation are carried out using Friedel-Crafts method. As carbocations are involved the potential of rearrangements has to be taken into account. In this case, the primary propyl cation obtained from chloropropane could easily rearrange into the more stable SECONDARY carbonication.

This can be avoided using a Friedel-Crafts acylation. The CARBONYL group is easily reduced , using the Clemmensen reduction as shown below:

(d) Iodine and fluorine are halogens which cannot be introduced using conventional Friedel-Crafts halogenation chemistry(why?). Since aniline is very electron rich,it undergoes electrophilic substitution even with relatively poor electrophiles such as iodine.
Discussion
9.

Dextrins (C_6H_10O_5) are used in :

Answer»

MAKING ADHESIVE
Confectionary
Sizing paper
All

Answer :D
Discussion
10.

Dexto rotatory sucrose is named equal as either

Answer»

`ALPHA`-D-glucopyranosyl `BETA`-D-fructofuranoside
`alpha`-D-glucopyranoside `beta`-D-fructofuranosyl
`alpha`-D-fructopyronoside
Both a and b

Answer :D
Discussion
11.

Deviations for ideal behaviour will be less if the gas is subjected to :

Answer»

low temperature and high presure
high temperature & low PRESSURE
low temperature
high temperature

ANSWER :B
Discussion
12.

Deviation from Markovnikov's rule occurs in presence of

Answer»

Zinc
Peroxides
Hg-Zn/HCl
All of these

Solution :
Discussion
13.

Deuterons when bombarded on a nuclide produce ._(18)Ar^(38) and neutrons. The target is

Answer»

`._(17)CL^(35)`
`._(19)K^(27)`
`._(17)Cl^(37)`
`._(19)K^(39)`

Solution :`._(17)Cl^(37) + ._(1)H^(2) rarr ._(18)Ar^(38) + ._(0)n^(1)`
Discussion
14.

Deuteroammonia (ND_(3)) can be prepared

Answer»

by fractionation of ORDINARY AMMONIA
by action of heavy water on magnesium nitride
by heating a solution of `NH_(4)Cl`in NaOD
none of the above

Solution :`Mg_(3)N_(2)+6D_(2)Orarr 3Mg(OD)_(2)+underset("ammonia")underset("Deutero")(2ND_(3))`
Discussion
15.

Deuterium resembles hydrogen in chemical properties but reacts :

Answer»

SLOWER than HYDROGEN
FASTER than hydrogen
More VIGOROUSLY than hydrogen
Just as hydrogen

Answer :A
Discussion
16.

Deuterium resembels hydrogen in chemical properties but reacts

Answer»

More VIGOROUSLY than HYDROGEN
Faster than hydrogen
Slower than hydrogen
Just as hydrogen

ANSWER :C
Discussion
17.

Deuterium is the isotope of hydrogen of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that C-D bond is slightly (5.0 "KJ" //"mole") stonger than the C-H bond. Reaction rates tend to be slower if a C-D bond as opposed to a C-H bond is broken in a rate limiting step. This effect on the rate is called a kinetic isotope effect.

Answer»

`(K_(H))/(K_(D))=1`
`(K_(H))/(K_(D))~~4`
`(K_(D))/(K_(H))~~4`
`(K_(D))/(K_(H))~~2`

Solution :As deprotonation STEP in case of IODINATION is rate determinining.
Discussion
18.

Deuterium is the isotope of hydrogen of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that C-D bond is slightly (5.0 "KJ" //"mole") stonger than the C-H bond. Reaction rates tend to be slower if a C-D bond as opposed to a C-H bond is broken in a rate limiting step. This effect on the rate is called a kinetic isotope effect. CH_(3)-overset(O) overset(||)(C)-CH_(3)overset(D_(2)O,D^(+))toX

Answer»

`CD_(3)-OVERSET(O) overset(||)(C)-CD_(3)`
`H_(2)C=overset(OD) overset(|)(C)-CH_(3)`
`H_(2)O=overset(OH)overset(|)(C)-CH_(3)`
`CH_(3)-overset(O)overset(||)(C)-CD_(3)`

Solution :
Discussion
19.

Deuterium an isotope of hydrogen is:

Answer»

Radioactive
Non-radioactive
Heaviest
Lightest

Answer :B
Discussion
20.

Dettol is used as :

Answer»

Disinfectant
Antiseptic
Analgesic
Antiallergic

Answer :B
Discussion
21.

Dettol used as an antiseptic is a mixture of .....

Answer»

cholroxylenol and terpineol
potassium PERMANGANATE and furacin 
`2-3%` SOLUTION of ALCOHOL in water
all of the GIVEN 

Solution :cholroxylenol and terpineol
Discussion
22.

Dettol is used as

Answer»

only antisecptic
only disinfectent
antiseptic and disinfectent
antibiotics

ANSWER :A
Discussion
23.

Dettol is the mixture of

Answer»

PHENOL and Iodine
CHLOROXYLENOL and Terpineol
Chloroxylenol and Bithionol
Terpineol and Bithionol

Solution :Dettol (antiseptic) is a MIXTURE of 4.8% chloroxylenol + 9.9% tepineol and ABSOLUTE alcohol.
Discussion
24.

Dettol is a mixture of chloroxylenol and :

Answer»

PHENOL
terpeneol
chloramphenicol
BITHIONAL.

SOLUTION :DETTOL is a mixture of CHLOROXYLENOL and terpeneol.
Discussion
25.

Dettol is a mixture of

Answer»

CHLOROXYLENOL and terpineol
Furacine, and soframicine
Tincture of IODINE and lodoform
Boric ACID and bithional

Answer :A
Discussion
26.

Dettol is a mixture of_____

Answer»

terpineol and BITHIONAL
CHLOROXYLENOL and bithional
chloroxylenol and terpineol
phenol and iodine

Answer :C
Discussion
27.

Dettol is a mixture of :

Answer»

terpineol and bithional
chloroxylenol and bithional
chloroxylenol and terpineol
phenol and iodine.

Answer :C
Discussion
28.

Dettol consists of

Answer»

Xylenol+Terpeneol
Chloroxylenol+terpeneol
Cresol+ethanol
None of these

ANSWER :B
Discussion
29.

Dettol consist of

Answer»

CRESOL and ethanol
XYLENE and terpineol
chloroxylenol and terineol
xylene and salol

Answer :C
Discussion
30.

Dettol, commonly used antiseptic is a mixture of

Answer»




ANSWER :D
Discussion
31.

Determinewhetherthe reactionN_(2) O_(4(g)) to2NO_(2(g)) isspontaneous at 25^(@)C from thefollowingdata : Delta_(f)H_((N_(2)O_(4)))^(@) =9.16 kJ mol^(-1) Delta_(f)H_((NO_(2)))^(@) = 33.2 kJ mol^(-1) S_((N_(2)O_(4)))^(@) = 304.2 JK^(-1)mol^(-1)S_(NO_(2))^(@)= 240JK^(-1) mol^(-1) At whattemperaturewill the reactionbecomespontaneous ?

Answer»

Solution :`(i)Delta G^(@)= +4.85kJ .` REACTIONIS non- spontaneous
(ii) `T= 325 . 6 K`
Discussion
32.

Determinewhetherthe followingreactionis spontaneous understandardconditions. 2H_(2)O_((1))+ O_(2(g))to 2H_(2) O_(2(1))Delta H^(@)= +196 kJ , Delta S^(@)=- 126 JK^(-1) Doesis havea cross- overtermperature?

Answer»

SOLUTION :THEREACTION is NON- SPONTANEOUS . There is nocross- over thetemperaturefor thereaction
Discussion
33.

Determinewhetherthe followingreactionwill bespontaneousor non- spontaneousunderstandard conditions. Zn_((s)) +Cu^(2+) to Zn^(2+) + Cu_((s)) ""DeltaH^(@) =- 219 kJ, DeltaS^(@) =- 21 JK^(-1)

Answer»


SOLUTION :Given : `DeltaH^(@) =- 219 KJ`
`Delta S^(@) =-21 JK^(-1)= 0.021kJK^(-1)`
`Delta G^(@)= ?`
forstandardconditions: Pressure= 1 atm
TEMPERATURE= T = 298 K
`Delta G^(@) = Delta H^(@)= T Delta S^(@)`
`=- 219 - 298 xx (-0.021)`
`=- 219 + 6.258`
`=- 212 .742 kJ`
Since `Delta G lt 0` THEREACTIONIS spontaneous.
Discussion
34.

Determine true statement for Zn-Cu electrochemical cell. [E_(Zn^(+2)|Zn)^(@)=-0.76V,E_(Cu|Cu^(+2))^(@)=-0.34V]

Answer»

Zinc act at cathode and COPPER act as ANODE.
Conduction of electron from copper to zinc electrode.
Zinc act as anode and copper act as cathode.
Standard reduction potential of zinc is more than copper.

Solution :`E_(ZN^(2+)|Zn)^(@)=-0.76V`
`E_(Cu|Cu^(2+))^(@)=-0.34V`
`E_(Cu^(2+)|Cu)^(@)=+0.34V`
So, `E_(Cu^(2+)|Cu)^(@) gt E_(Zn^(2+)|Zn)^(@)`.
So, they form electrochemical cell in which zinc act as anode and copper act as cathode.
Discussion
35.

Determine Vapour pressure of water at 50^(@)C. If enthalpy of vaporisation of water is 40.6 Kj//"mol"

Answer»

10 ATM
0.132 atm
100.15 mm
1.32 atm

ANSWER :B
Discussion
36.

Determine the volume (in ml) of NaOH (aq) needed to prepare 32.8 gm Na_(3)PO_(4) by reaction 3NaOH (aq) + H_(3)PO_(4) (aq) rarr Na_(3)PO_(4) (aq) + 3H_(2)O (l) Sodium hydroxide solution is 20% by mass and have density 1.5 gm//ml. [Divide your answer by 16]

Answer»


Solution :`{:(3NaOH (aq) + H_(3)PO_(4) (aq) rarr,Na_(3)PO_(4) (aq) + 3H_(2) O (l),),(,32.8 gm,):}`
`n_(Na_(3)PO_(4)) = (32.8)/(164) = 0.2` MOLE
` n_(NaOH)` REQUIRED `= 0.2 xx 3 = 0.6` mole
Mass of `NaOH = 0.6 xx 40 = 24 gm`
`20 gm NaOH` in 100 gm solution
`20 gm NaOH` in `(100)/(1.5)` ML solution
For 24 gm `NaOH` volume of solution `= (100)/(1.5) xx (24)/(20) = 80 ml`
`= (80)/(16) = 5 ml`
Discussion
37.

Determine the vlaues of equilibrium constant (K) and DeltaG^(@) for the following reaction: Ni(s)+2Ag^(+)(aq)toNi^(2+)(aq)+2Ag(s),E^(@)=1.05V (1F=96500" C "mol^(-1)).

Answer»


ANSWER :`3.41xx10^(35),-2.02xx10^(5)J`
Discussion
38.

Determine the values of equilibrium constant (K_(c)) and DeltaG^(@) for the reaction Ni(s)+2Ag^(+)(aq)rarrNi^(2+)(aq)+2Ag(s), E^(@)=1.05 V.("Given " 1F=96500"C mol"^(-1))

Answer»

Solution :`DeltaG^(THETA)=-nFE_("cell")^(theta)`
`n=2, E_("cell")^(theta)=1.05V`
`F="96500 C mol"^(-1)`
`DeltaG^(theta)=-2xx1.05xx96500`
`=-202.650kJ`
`DeltaG^(theta)=-RT ln K_(c)`
`lnK_(c)=-(DeltaG^(theta))/(RT)=(-202.650xx10^(3))/(8.314xx298)`
`K_(c)=3.32xx10^(35)`
Discussion
39.

Determine the value of E^0 cell for the following reaction Cu^(+2)+Sn^(+2)toCu+Sn^(+4) Equilibrium constant is 10^6 Cu^(++)+Sn^(++)toCu+Sn^(+4)

Answer»

0.1773
`.01773`
0.2153
1.773

Answer :A
Discussion
40.

Determine the value of Delta E and Delta Hfor the reversible isothermal evaporation of 90.0 g of water at 100^@C . Assume that water vapour behaves as an ideal gas and heat of evaporation of water is 540 cal g^(-1)

Answer»

SOLUTION :44.87 KCAL, 48.6 kcal
Discussion
41.

Determine the units of rate constant for first and zero order reaction.

Answer»

Solution :`K=(MOL)^(1-n) L^(n-1) S^(-1)`
For ZERO ORDER , n=0
So, `K=(mol)^(1-0) L^(0-1) S^(-1)=S^(-1) "mol L"^(-1)`
For FIRST order , n=1
`K=(mol)^(1-n) L^(n-1) S^(-1)`
So, `K=(mol)^(1-1) L^(1-1)S^(-1)`
`=S^(-1)`
Discussion
42.

Determine the type of cubic lattice to which the iron crystal belongs if its unit cell has an edge length of 286 pm and the density of iron crystals is 7.86g cm^(-3).

Answer»


SOLUTION :a=286 pm, d=7.86g `cm^(-3)`, `N_A = 6.023 xx 10^(23)`,
For iron M = 56 g `mol^(-1)`, `d = (ZM)/(N_A xx a^3)` or `Z= (d xx N_A xx a^3)/(M)= (7.86 xx 6.023 xx 10^(23) xx (286 xx 10^(-10))^3)/(56) = 1.98 ~~ 2`
Since Z= 2, iron has bcc structure.
Discussion
43.

Determine the standard enthlpy of the reaction C_(3)H_(8)(g) + H_(2)(g) to C_(2)H_(6)(g) + CH_(4)(g). Using the given enthalpies under standard conditions. Compound H_(2)(g)CH_(4)(g)C_(2)H_(6)(g)C("Graphite") Delta_(c)H^(0)(kJ//mol) –285.8 – 890.0 – 1560. 0 –395.5 The standard enthalpy of formation of C_(3)H_(8)(g) is –103.8 kJ//mol

Answer»

`-55.7 kJ`
`+55.7 kJ`
`-2060.4 kJ`
`+ 2060.4 kJ`

Solution :`C_(3)H_(8)(g) + 5O_(2)(g) to 3CO_(2)(g) + 4H_(2)O(l)`
`Delta_(f)H_(H_(2)O(l)) = Delta_(C)H_(H_2(g))`
`Delta_(C)H_(C_3H_3(g)) = [3 xx Delta H_(CO_2(g)) + 4 xx Delta_(f)H_(H_(2)O(l))]`
`-[Delta_(f)H_(C_(3)H_(3)(g)) + 5 xx Delta_(f)H_(O_2(g))]`
`=[3(-393.5)+4(-285.8)]`
`-(-103.8) = -2219.9 kJ`
`Delta_(f)H_("required") = -[Delta_(C)H_(C_2H_6(g)) + Delta_(C)H_(CH_4(g))]`
`+[Delta_(C)H_(C_3H_8(g)) + Delta_(C)H_(H_2(g))]`
`= -[(-1560.0) + (-890.0)]`
`+[(-2219.9) + (-285.8)] = 55.7 kJ`.
Discussion
44.

Determine thetotalwork done by gas ( inJoule )if system follows an expansion process as shown in figure [Given : 1 barliter =100J]

Answer»

Solution :`P_(B)V_(B)^(2)` =`P_(C)V_(C)^(2)`
`or 50xx(0.4)^(2)=P_(C)xx(0.8)^(2)`
`or P_(C)=(50xx0.4xx0.4)/(0.8xx0.8)=12.5`
`W=W_(AB)+W_(BC)`
`= 50(0.4-0.2)+(P_(C)V_(C)-P_(B)V_(B))/(2+1)`
`=- 50 xx0.2+(12.5xx0.8 -50xx0.4)/(1)`
`= - 10 +(-10)=- 20 "bar LITER " =- 2000 J`
`W_(by)= 2000 J `
Discussion
45.

Determine the product of metalloids and the radio-active elements present in group-13 & group -14

Answer»


ANSWER :3
Discussion
46.

Determine the potential of the following cell: Pt|H_(2)(g,0.1bar)|H^(+)(aq,10^(-3)M)|| MnO_(4)^(-)(aq,0.1M),Mn^(2+)(aq,0.01M),H^(+)(aq.0.01M)|Pt Given :E_(MnO_(4)^(-)|Mn^(2+))^(0)=1.51V

Answer»

`1.54V`
`1.48V`
`1.84V`
`1.51`

SOLUTION :Anode `H_(2(g))rarr2H^(+)(AQ)+2e^(-)`
(i)Cathode `MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O`
(ii)
`(i)XX(5)`
`(ii)xx(2)`
`E=E^(o)-(0.06)/(10)log,([Mn^(2+)]^(2)[H^(+)]_(a)^(10))/([MnO_(4)^(-)][H^(+)]_(ca)^(16)P_(H2)^(5))`
`=1.51-(0.06)/(10)log,((10^(-2))^(2)xx(10^(-3))^(10))/((10^(-1))^(2)xx(10^(-2))^(16)(10^(-1))^(5))`
`=1.51-(0.06)/(10)log10^(5)`
`=1.51-0.03=1.48V`
Discussion
47.

Determine the potential of a Daniell cell, initially containing 1.00 L each of 1.0 M copper (II) ion, after passage of 10^(5) coulomb charge. The E^(@) of the daniell cell is 1.10 volt.

Answer»


SOLUTION :By passage of `10^(5)` COULOMB charge, zinc ion concentration increases by 0.518 MOLE while COPPER ion concentration DECREASES
Discussion
48.

Determine the potential for the cell : Pt|Fe^(2), Fe^(3+)||Cr_(2)O_(7)^(2-), Cr^(3+), H^(+)|Pt in which [Fe^(2+)] and [Fe^(3+)] are 0.5 M and 0.75 M respectively and [Cr_(2)O_(7)^(2-)], [Cr^(3+)] and [H^(+)] are 2M, 4 M and 1M respectively. Given : {:(Fe^(3+)+e^(-) rarr Fe^(2+),,E^(@)=0.770"volt"),(14H^(+)+6e^(-)+Cr_(2)O_(7)^(2-) rarr 2Cr^(3+)+7H_(2)O,,E^(@)=1.35" volt"):}

Answer»


ANSWER :0.56 VOLT
Discussion
49.

Determine the pOH of a solution after 0.1 mole of NaOH is added to 1 litre of solution containing 0.15 M CH_(3)COOH and 0.2 M CH_(3)COONa. Assume no change in volume. [Given: log 3 = 0.48, log 2 = 0.30, pK_(a) "of acetic acid" = 4.74]

Answer»

8.48
5.52
9.66
None

Solution :`CH_(3)COOH + OH^(-) to CH_(3)COO^(-)+H_(2)O`
`{:(0.15,0.1,0.2),(0.05,0,0.3):}`
`pH = 4.74 + log (0.3)/(0.05)`.
Discussion
50.

Determine the pH of the solution that results from the addition of 20.00 mL of 0.01 M Ca(OH)_(2) to 30.00 mL of 0.01 MHCl

Answer»

11.3
10.53
`2.70`
8.35

Solution :Millimoles of `H^(+) = 30 XX 0.01 = 0.3`
Millimoles of `OH^(-) = 20 xx 0.01 xx 2 = 0.4`
Remaining millimoles of `OH^(-) = 0.4 - 0.3 = 0.1`
`[OH^(-)] = (0.1)/(50)` or `2 xx 10^(-3)`
So, `pOH = 2.6999 rArr PH = 14 - 2.6999 = 11.30`.
Discussion