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.

For fixed mass of an ideal gas.

Answer»


Solution :(P) `d = (PM)/(RT)`
`(P)/(d) = (R )/(M) XX T ((P)/(d) = y, T = X)`
(R ) `PV = nRT =` constant
`PV = K`
`P = K (1)/(V) (P = y, (1)/(V) = x)`
`y = mx`
(S) `P = (RT)/(M) xx d`
`Pd = (RT)/(M) xx d^(2) (Pd = y, d = x)`
Discussion
2.

For fixed amount of an ideal gas identify the incorrect graph :

Answer»




ANSWER :C
Discussion
3.

For first order raction, the concentration of ractant

Answer»

is independent of time
varies LINEARLY with time
varies exponentially with time
none

Solution :The expression `[A]= [A] _(0) e^(-kt)` SHOWS that the CONCENTRATION of REACTANT , [A] , varies exponentially with time t .
Discussion
4.

For first order reation the ratio of t_(0.75)" to "t_(0.25) would be

Answer»

`4: 3`
`3 : 2`
`2 :1 `
`1 , 2`

Answer :C
Discussion
5.

For first order reaction, rate constant :

Answer»

is DIRECTLY proportional to concentration of the REACTANT
is proportional to square of concentration of reactant
is dependent on TEMPERATURE
is independent of temperature

ANSWER :C
Discussion
6.

For first order parallel reactions k_(1) and k_(2)are 4and 2min(-1) respectively at 300 K. If the activation energies for the formation of B and C are respectively 30,00 and 38,314 joule/ mol respectively, the temperature at which B and C will be obtained in equimolar ratio is :

Answer»

757.48 k
378.74k
600 k
none of these

Answer :B
Discussion
7.

For first order parallel reactionK_(1) and K_(2) are 4 and 2"min"^(-1) respectively at 300K. If the activation energies for the formation of B and C are respectively 30,000 and 38,314 Joul/mol respectively. The temperature at which B and C will be obtained in equimolar ratio is 47x. Hence x is _____________

Answer»


SOLUTION :`AOVERSET(K)(to)B`
`Aoverset(K)(to)C`
`y_(B)=(K_(1))/(K_(1)+K_(2))=y_(C)=(K_(2))/((K_(1)+K_(2)))`…….1 , `K_(1)=AE^(-epsilon)//RT`…….2
`K_(2)=Ae^(-epsilona_(2)//RT)`…….3, `K=Ae^(-epsilone//RT)`………………..4 If `T_(1)=300K,T_(2)=376K`
Discussion
8.

For [FeF_(6)]^(3-)" and "[CoF_(6)]^(3-), the statement that is correct is :

Answer»

both are COLOURED
both are colourless
`[FeF_(6)]^(3-)` is coloured and `[CoF_(6)]^(3-)` is colourless
`[FeF_(6)]^(3-)` is colourless and `[CoF_(6)]^(3-)` is coloured

Solution :Both have UNPAIRED ELECTRON.
Discussion
9.

For [Fe(CN)_(6)]^(4-).[Ni(CN)_(4)]^(2-)and[Ni(CO)_(4)]. Correct statement

Answer»

all have IDENTICAL geometry
`[Ni(CO)_(4)]` is PARA MAGNETIC
all are dia magnetic
`[FE(CN)_(6)]^(4-)and[Ni(CN)_(4)]^(2-)` are paramagnetic

Solution :All the given compounds are DIAMAGNETIC
Discussion
10.

For extraction of sodium from NaCl, the electrolytic mixture NaCl + KCl + CaCl_2is used. During extraction process, only sodium is deposited on cathode but K and Ca do not because :

Answer»

Na is more REACTIVE than K and Ca
Na is less reactive than K and Ca
NACL is less STABLE than `Na_3AlF_6` and `CaCl_2`
the discharge potential of `Na^(+)`is more than that of `K^(+)`and `Ca^(2+)`ions.

Answer :D
Discussion
11.

For extraction of sodium from NaCI, the electrolytic mixture NaCI + Na_(3)AIF_(6) +CaCI_(2) is used. During extraction process, only sodium is deposited on cathode but K and Ca do not because

Answer»

NA is more reactive than K and Ca
Na is less reative then K and Ca
NaCI is less STABLE than `Na_(3)AIF_(6)` and `CaCI_(2)`
the discharge POTENTIAL of `Na^(+)` is less than that of `K^(+)` and `Ca^(2+)` ions

Solution :Lower discharge potential ions deposited first at cathode.
Discussion
12.

For every 10^@C rise in temperature the rate of reaction increases nearly

Answer»

10 times
2 times
5 times
8 times

Answer :B
Discussion
13.

For estimating ozone in the air, a certain volume of air is passed through an . Alkaline Kl solution when O_(2)is evolved and iodide is oxidized to iodine . When such a solution is acidified , free iodine is evolved which can be titrated with standard Na_(2)S_(2)O_(3) . Solution : in an experiment, 10 L ari at 1 atm and 27^(@) C were passed through an alkaline Kl solution , and at the end, the iodine was entrapped in a solution which on titration as above required 1.5 ml of 0.01 N Na_(2)S_(2)O_(3) solution. Calculate volume percentage of ozone in the sample.

Answer»

Solution :The chemical REACTION is , `H_(2)O +Kl + O_(3) tol_(2) + O_(2) + KOH `
Milliequivalents of iodine = Milliequivalents of Kl = MILLI equivalents of `O_(3)` reacted
Milliequivalnets of `Na_(2)S_(2)O_(3) = 1.5 xx 0.01 = 1.5 xx 10^(-2)`
Millimoles of iodine ` = (1.5 xx 10^(-2))/2 = 7.5 xx 10^(-3)"" [ :." n-factor for iodine " =2]`
Millimoles of ozone ` = 7.5 xx 10^(-3)`
Volume of ozone ` = (nRT)/P = (7.5 xx 10^(-6) xx 0.0821 xx 300)/1 = 184.725 xx 10^(-6) ` litre
Volume PER cent of ozone ` = (184 .725 xx 10^(-6))/10 xx 100 = 1.847 xx 10^(-3)`
Discussion
14.

For equilibrium AB(g) iff A(g) + B(g), K_pis equal to four times the total pressure.Calculate the number of moles of B formed.

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SOLUTION :`2//sqrt5` TIMES INITIAL MOL of AB
Discussion
15.

For emission of alpha-particle from uranium nucleus: ._(92)U^(235) - ._(2)He^(4) rarr ._(90)Th^(231) Shortage of two electrons in thorium is due to

Answer»

CONVERSION of ELECTRON to positron
Adsorption in the nucleus
Annihilation
Combustion with POSITION to EVOLVE energy

Answer :C::D
Discussion
16.

For effusion of equal volume, a gas takes twice the time as taken by methane under similar conditions. The molar mass of the gas is

Answer»

32 g/mol
44 g/mol
64 g/mol
50 g/mol

Answer :A
Discussion
17.

For each value of l, the number of m value is

Answer»

2L
nl
`2l+1`
`2l^(2)`

Solution :For each value of L, there are `(2l+1)`m VALUES.
Discussion
18.

For each of the following pharmaceutical compounds, identify all stereogenic (i.e., all asymmetric carbon atoms) and lable the configuration of each as being either (R ) or (S).

Answer»


SOLUTION :N//A
Discussion
19.

For each of the following processes, tell whether the entropy of the system increases, decreases or remains constant. (a) Melting one mole of ice to water at 0^(@)C (b) Freezing one mole of water of ice at 0^(@)C (c ) Freezing one mole of water to ice at -10^(@)C (d) Freezing one mole of water to ice at 0^(@)C and then cooling it to -10^(@)C

Answer»


ANSWER :[ (a) INCREASES (B) DECREASES (C ) decreases (d) decreases ]
Discussion
20.

For each of the following orders as listed in Column-I pick the correct observation listed in Colomn-II. {:("Column-I","Column-II"),((A)"CgtN",(p)"More favourable (exothermic)electron affinity"),((B)"SegtBr",(q)"The higher first ionization energy "),((C )"MggtK",(r)"The larger size"),((D)"FgtCl",(s)"The higher electronegativity"),(,(t)"The higher number of valence electrons"):}

Answer»


Solution :(A)Nitrogen has less favourable electron affinity because of stable half filled electron configuration `ns^2np^3` Nitrogen has higher first ionisation energy , because of stable half filled electron configuration , `ns^2np^3` Across the period , atomic size decreases , electrons are added in the same valence SHELL but for addition of each successive element the nuclear charge increases by one unit positive charge (i.e. PROTON) Electronegativity of nitrogen (3.0) is greater than that of carbon (2.5) on account of small atomic size.
`C=ns^2 np^2, N=ns^2 np^3`
(B)`Delta_(eg)H, Se= - 198 KJmol^(-1)` , Br = -`325 kJ mol^(-1)`
`Delta_(IE1)`,Se=941 kJ `mol^(-1)` , Br=1142 `kJ mol^(-1)`
Covalent radius/PM, Se=117, Br = 114
Electronegativity, Se=2.48 , Br = 3.0
Se=[Noble gas] `ns^2 np^4` , Br=[Noble gas] `ns^2 np^5`
(C )`Delta_(eg)`H, K=-48 kJ `mol^(-1)`, `MG~~O`(stable configuration `ns^2`)
`Delta_(IE1), K 419 kJ mol^(-1)`, Mn`=737 kJ mol^(-1)`
Metallic radius /pm K=227, Mg=160
Electronegativity , K=0.8 , Mg=1.2
K =[Noble gas ] `ns^1` , Mg=[Noble gas]`ns^2`
(D)`Delta_(IE1)`,F=1680 kJ `mol^(-1)` , Cl = `1256 kJ mol^(-1)`
Covalent radius / pm , F=64 , Cl=99
Electronegativity , F=4.0 , Cl=3.2
Have same number of valence electrons because both belong to same group i.e. halogen
Discussion
21.

For each of the following cells :(a) Write the equation for cell process. (b)Find E^@ for each cell. (c) Expiain the significance of any negative answers in part (b). 1. Fe//Fe(NO_3)_2 (1.0 M) | | Zn^(2+) (1.0 M)/Zn 2. Pt//Cl_2(g)/KCl | | Hg_2Cl_2(s)/Hg 3. Cd//Cd^(2+) (1.0M) | | AgNO_3//Ag E^@ (Fe) = 0.41V , E^@ (Cd) = 0.40 V ,E^@ (Zn) = 0.76 V E^@ (Cl– / Cl_2) = – 1.36 V , E^@ (Ag) = – 0.80V ,E^@ (Hg//Hg_2Cl_2) = – 0.27 V

Answer»


ANSWER :(1) 0.35 V (2) –1.09 V (3) –1.2 V
Discussion
22.

For each of the following complexes, draw a crystal field energy-level diagram, assign the electrons to orbitals, and predict the number of unpaired electrons: (a) [CrF_(6)]^(3-)"" (b) [V(H_(2)O)_(6)]^(3+) "" ( c)[Fe(CN)_(6)]^(3-) (d) [Cu(en)_(3)^(2+) "" ( e) [FeF_(6)]^(3-)

Answer»

SOLUTION :
Discussion
23.

For dry cleaning of clothes instead of tetra chloroethane which is carcinogen in nature, which of the following solvents can be used ?

Answer»

LIQUID `CO_(2)`
`H_(2)O_(2)`
Liquid `O_(3)`
PETROL 

ANSWER :A
Discussion
24.

For drying ether sodium metal can be used, but it cannot be used for drying ethyl alcohol because:

Answer»

NA is very reactive
Ether REACTS easily with Na
Ethyl alcohol reacts with sodium METAL
NONE

Answer :C
Discussion
25.

For dofferemt reactions of H_(2)SO_(4) with(i) HNO_(3) and (ii) HClO_(4), a. Write the equactions for therecactionsand identify the conjugate acids and bases: b. Expain the differentbehaviours of H_(2)SO_(4) II. Nitrobenze can be prepared from benzeneby usinga mixtureof conc. HNO_(3) and conc. H_(2)SO_(4) In the nitratingmixture, HNO_(3) acrs are: a. Base, b. Acid c. Reducingagent, d. Catayst III. Among the following statements on the nitraction of aromatic compounds, the false one is: a. The rateof nitration of benzene is almost the sameas that of hexadeucterobenzene. b. The rate of nitration of touene is grater than that of benzene. c. The rate of nitration of benzene is greater than that of hexadeuterobenze. c. The rate of nitration of benzene is greater than that of hexadeuterobenezene. d. Nitration is an electrophillic subsitution reaction. IV. Select the correct alternatice(s). The following reaction occurs in a mixture of conc. HNO_(3) and cnc. H_(2)SO_(4) as: HNO_(3) + 2H_(2)SO_(4) rarr NO_(2)^(o+) + 2HSO_(4)^(o+) + H_(3)^(o+) Which of the following statements about this reaction is correctgt a. Nitric acid acts as a base. b. Sulphuric acid acts as a base. Sulphureacid acts as a dehydrating agent. d. Addition of H_(2)O will reduce the NO_(2)^(o+) concentration. e. HNO_(3) and NO_(2)^(o+) are conjugate acid-base pair.

Answer»

Solution :a. i.
i. `H_(2)SO_(4)` a STRONGER acid, GIVESA proton to `HNO_(3)`, a weaker acid now acting as a base.
II. `H_(2)SO_(4)` is the weakeracid and acts as the base to accept a proton from `HclO_(4)`
II. a. III. c. IV. a and d.
Discussion
26.

For dissolution of a solid in a liquid, Delta S is generally

Answer»

`+ve`
`+ve`
zero
Both (A) and (B)

Solution :Dissolution of a solid in a LIQUID always lead to an increases in randomness i.e, `DELTAS ` is +ve
Discussion
27.

For dissolution of an ionic solid in water Delta_("sol") H^((Theta)) =

Answer»

`Delta_("lattice")H^(Theta)+Delta_(f)H^(Theta)`
`Delta_(f)H^(Theta)+Delta_(HYD)H^(Theta)`
`Delta_(a)H^(Theta)+Delta_(f)H^(Theta)`
none of these

Solution :`Delta_("sol") H^(Theta) =Delta_("lattice")H^(Theta)+Delta_(hyd)H^(Theta)`
(For dissolution of an IONIC solid in WATER)
Discussion
28.

For dilution of H_2SO_4, why water should not be added to concentrated H_2SO_4 ?

Answer»

Solution :On adding water to `H_2SO_4` a LARGE amount of HEAT is evolved. This LEADS to spontaneous change of water into steam. This may lead to splashing of the LIQUID CAUSING burns, etc
Discussion
29.

For dibasic acid correct order is

Answer»

`K_(a1) lt K_(a2)`
`K_(a1) gt K_(a2)`
`K_(a1)=K_(a2)`
not certain

Solution :In DIBASIC acids the loss of second proton occurs much less readily than the FIRST. Usually the `K_(a)` values for successive loss of protons from these acids DIFFER by at least a factor of `10^(-3) i.e., K_(a_(1)) gt K_(a_(2))`
`H_(2)X hArr H^(**) ** HX^(**) K_(3_(1))`
`HX^(**) hArr H^(**) ** X^(@**) K_(a_(2))`
Discussion
30.

For diamond, state the element present at the lattice sites, the number of nearest neighbours for each atom and the type of cell. State the hybridization of the carbon atom in diamond.

Answer»

Solution :Carbon is PRESENT at the lattice SITES, one carbon atom is linked with four other carbon atoms forming a network crystal. It has TETRAHEDRAL UNITS and `sp^3` hybridization.
Discussion
31.

For detection of sulphur in an arganic compound, sodium nitroprusside is added to the sodium extract. A violet colour is obtained due to the formation of:

Answer»

`FE(CN)_2`
`K_3Fe(CN)_5NS`
`Na_4[Fe(CN)_5NOS]`
`Na_4Fe(CN)_6`

Answer :C
Discussion
32.

For denaturation of ethanol

Answer»

Copper SULPHATE is ADDED to give colour and pyridine for FOUL smell
Nickel sulphate is added to give colour and aniline for foul smell
Copper sulphate is added to give colour and aniline for foul smell
Nickel sulphate is added to give colour and pyridine for foul smell

Answer :A
Discussion
33.

For dehyetohalogenation, the order of reactivity of alkyl halides considering E1 mechanism is

Answer»

`1° GT 2° gt 3°`
`2° gt 1° gt 3°`
`2° gt 3° gt 1°`
`3° gt 2° gt 1°`

ANSWER :D
Discussion
34.

For dehydration and also purification of gases like CO_(2), N_(2), Cl_(2), O_(2)and He, ........... and ........ are employed.

Answer»

SOLUTION :ALUMINA , SILICA
Discussion
35.

For decolouration of 1 mole of KMnO_4, the moles of H_2 O_2 required is

Answer»

`1//2`
`3//2`
`5//2`
`7//2`

SOLUTION :`5//2`
Discussion
36.

For cyclohexane, which of the following factors does not make the boat conformation less stable than the chair conformation

Answer»

1,3-diaxial interactions
flag POLE interactions
angle strain
torsional strain

Answer :A
Discussion
37.

For [CrCl_(3) xNH_(3)] , elevation in boiling point of one molal solution is double of one molal urea solution . Hence , the value of x (assuming complete dissociation ) is ____

Answer»


ANSWER :D
Discussion
38.

For d-electrons the orbital angular momentum is

Answer»

`sqrt(6)h/(2PI)`
`sqrt(2)h/(2pi)`
`h/(2pi)`
`2 h/(2pi)`

SOLUTION :ORBITAL angular MOMENTUM `=sqrt(l(l+1))h/(2pi)`
For d l=2
`:.` Orbital angular momentum
`=sqrt(2(2+1))h/(2pi)=sqrt(6)h/(2pi)`
Discussion
39.

For Cyclooctatetraene following is correct ?

Answer»

There are two TYPES of C-C bond
Structure is non planar and resonance is not OBSERVED in molecule
Extensive resonance is found with in the molecule and all bonds are of same type
its heat of HYDROGENATION is EQUAL to that of `4pi` bonds hydrogenated

Solution :Structure of cyclooctatetrane is
(non planar) so there is no resonance . Hence , 4 bonds are C-C (SINGLE ) and 4 bonds are C=C (double bond)
Discussion
40.

For CuSO_(4)*5H_(2)O, which is the correct mole relationship?

Answer»

9 `XX` mole of CU = mole of O
5 `xx` mole of Cu = mole of O
9 `xx` mole of Cu = mole of `O_(2)`
Mole of Cu = 5 `xx` mole of O

Answer :A
Discussion
41.

For Cr_(2)O_(7)^(2-)(aq)+14H^(+)(aq)+6e^(-) to 2Cr^(3+)(aq)+7H_(2)O(aq) E^(@)=1.33" V". At [Cr_(2)O_(7)^(2-)]=4.5" millimole" , [Cr^(3+)]=15" millimole", E is1.067" V". The pH of the solution is :

Answer»

SOLUTION :`-E_(cell)=-((772xx1000" CV "mol^(-1)))/((4)xx(96500" C " mol^(-1)))=2V` ltbr? For the given reaction :
`Cr_(2)O_(7)^(2-)(aq)+14H^(+)(aq)+6e^(-) to 2Cr^(3+)(aq)+7H_(2)O(aq)`
`1.067=1.33-(0.0591)/(6)"log"([Cr^(3+)]^(2)[H_(2)O]^(7))/([Cr_(2)O_(7)^(2-)][H^(+)]^(14))`
`1.067=1.33-9.85xx10^(-3)"log"((15xx10^(-3))^(2)xx(1)^(7))/((4.5xx10^(-3))[H^(+)]^(14))`
`2 log[15xx10^(-3)]-log[4.5xx10^(-3)]-14" log "[H^(+)]=(0.263)/(9.85xx10^(-3))`
`2xx(-1.82)+2.34+14 pH=26.7`B
`14 pH=26.7+1.3=28`
`pH=(28)/(14)=2`.
Discussion
42.

For Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)to2Cr^(3+)+7H_(2)O,E^(@)=1.33V At [Cr_(2)O_(7)^(2-)]=4.5 millimole, [Cr^(3+)]=15 millimole, E is 1.067V. Calculate the pH of the solution.

Answer»

Solution :For the GIVEN reaction,
`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-)to2Cr^(3+)+7H_(2)O`
`E=E^(@)=-(0.0591)/(6)"log"([Cr^(3+)]^(2)[H_(2)O]^(7))/([Cr_(2)O_(7)^(2-)][H^(+)]^(14))`
`1.067=1.33-9.85xx10^(-3)"log"([15xx10^(-3)]^(2)[1]^(7))/([4.5xx10^(-3)][H^(+)]^(14))`
`(-0.263)/(-9.85xx10^(-3))=2LOG(15xx10^(-3))-log(4.5xx10^(-3))-14log(H^(+))`
`26.7=2(-1.82)+2.34+14pH` or `pH=(28)/(14)=2`.
Discussion
43.

For Cr_2O_7^(2-) + 14 H^(+) + 6e^(-) to 2Cr^(+3) + 7H_2O, E^(@) = 1.33 V " At" [ Cr_2O_7^(2-)]=4.5 millimoles , [Cr^(+3)]=15 millimole , E is 1.067V. The pH of the solution is nearly equal to

Answer»

2
3
5
4

Solution :For the GIVEN reaction,
`Cr_(2)O_(7)^(2-)+14H^(+)+6e^(-) rarr 2Cr^(3+)+7H_(2)O`
`E=E^(@)-(2.303RT)/(nF)log. ("[Products]")/("[Reactants]")`
`1.067=1.33-(0.0591)/(6)log. ([Cr^(3+)]^(2)[H_(2)O]^(7))/([Cr_(2)O_(7)^(2-)][H^(+)]^(14))`
`1.067=1.33-9.85xx10^(-3)log. ([15xx10^(-3)]^(2)[1]^(7))/([4.5xx10^(-3)][H^(+)]^(14))`
`(-0.263)/(-9.85xx10^(-3))=2 log[15xx10^(-3)]-log[4.5xx10^(-3)]-14log[H^(+)]`
`26.7=2xx-1.82+2.34+14`pH
`26.7= - 3.64+ 2.34+14 ` pH
`14pH=26.7+3.64-2.34`
`pH=(28)/(14)=2`
Discussion
44.

For copper half cell, the graph between reduction potential (Y-axis) and log[Cu^(2+)]is a straight line with the Y -intercept + 0.34V. Reduction potential of copper electrode with 0.01 M CuSO_4solution is

Answer»

+0.40V
-0.28V
+ 0.28 V
-0.40 V

ANSWER :C
Discussion
45.

For converting aniline into chlorobenzene which of the following reagents is not used?

Answer»

`Cl_2`
`HCL`
`HNO_2`
`CUCL`

ANSWER :A
Discussion
46.

For converting aniline into chlorobenzene which of the following reagents is not used ?

Answer»

`Cl_(2)`
`HCl`
`HNO_(2)`
`CuCl`

ANSWER :A
Discussion
47.

For complex ion/compound formation reactions (I) Co^(3+) (aq) + EDTA^(4-) to P (II) Ni^(2+) (aq) + "dmg (excess)" overset(NH_4 OH)to Q (III) Zn^(2+) (aq) + "gly (excess)" to R (IV) Pt^(4+)aq + en (excess to S Which of the following complex ion/compound does not exhibit optical activity ?

Answer»

P
Q
R
S

Solution :Chlorodiaquatriamminecobalt (III) CHLORIDE is `[COCL(NH_3)_3 (H_2 O)_2]Cl_2`.
Discussion
48.

Forcomplete oxidationof 4 litres of CO at NTP , therequiredvolume of O_(2)at NTPis

Answer»

4 LITRES
8 litres
2 litres
1 litres

Answer :C
Discussion
49.

For converting a solution if 100 ml KCl of 0.4 M concentration into a solution of KCl 0.05 M concentration. The quantity of water added is

Answer»

900 ML
700 ml
500 ml
300 ml

Solution :We know that, `M_(1)V_(1)= M_(2)V_(2)`
`0.05xx V_(1)=0.4xx100`
`therefore V_(1)=(0.4xx100)/(0.05)=800`
`V_(2)-V_(1)=800-100 RARR 700 ml`.
Discussion
50.

For conversion C(graphite) rarr C(diamond) the DeltaS is

Answer»

Zero
Positive
NEGATIVE
Unknown

Solution :CONVERSION of graphite into DIAMOND is an endothermic reaction. So, HEAT of diamond is higher than that of graphite. But `DeltaS` WOULD be negative for the conversion of graphite into diamond.
Discussion