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.

Let M_(1) and M_(2) be the masses of the nuclei ""_(1)H^(2) and ""_(2)H^(4) respectively. Aslo let m_(p) and m_(n) be the masses of proton and neutron respectively.

Answer»

`-87`
`-170`
`+170`
ZERO

ANSWER :B
Discussion
2.

Delta G^(@)(HI, g) ~= + 1.7 kJ. What is the equilibrium constant at 25^(@)C for 2HI(g)hArrH_(2)(g) + I_(2) (g) ?

Answer»

`24.0`
`3.9`
`2.0`
`0.5`

SOLUTION :`Delta G^(@) = - 2.303xx8.314xx10^(-3)xx298 LOG K_(p)`
`1.7 = - 2.303xx8.314xx10^(-3)xx298xxlog K_(p)`
`K_(p) = 0.5`
Discussion
3.

Delta G^(0) for the reaction x + y hArr z is -4.606 kcal. The value of equilibrium constant of the reaction at 227^(@)C

Answer»

100
10
2
0.01

Solution :`K = e^(-(Delta G^(0))/(RT)) = 100`
Discussion
4.

The standard free energy of Cu_((aq))^(+)is 50 kJ mol^(-1) and that of Cu_((aq))^(2+) is 66 kJ. mol^(-1). Calculate the change in free energy for the reaction, Cu_((aq))^(+) to Cu_((aq))^(2+) + e^(-).

Answer»

`-16`
`+116`
`-116`
`+16`

ANSWER :D
Discussion
5.

Delta G_(sys) is equal to

Answer»

`T Delta S _(sys)`
`T Delta S _("total")`
`-T Delta S _("total")`
`Delta S _("sys") Delta S _("SURROUNDING")`

ANSWER :C
Discussion
6.

Which property is common in diamond and graphite ?

Answer»

`-2.9`
`+2.9`
`+5.8`
`-5.8`

ANSWER :C
Discussion
7.

Delta G^(@) of a reaction is 120 kJ mol^(-1). Calculate the K_(p) at 20^(@)C.

Answer»

Solution :`Delta G^(@) = -2.303 RT log_(10) K_(p)`
`-120,000 = -2.303 xx 8.314 xx 293 xx log_(10)K_(p)`
`log_(10)K_(p) = (-120,000)/(-2.303 xx 8.314 xx 293)`
`log_(10)K_(p) = 21.39`
`K_(p) = "ANTILOG" 21.39`
`- 2.455 xx 10^(21)`
Discussion
8.

Delta _(C) U^( Theta ) of formation of CH_(4(g)) at certain temperature is -393 "kJ mol"^(-1). The value of Delta_(f) H^( Theta ) is

Answer»

<P>zero
`LT Delta_(f) U^( Theta )`
`gt Delta_(f) U^( Theta )`
equal to `Delta_(f) U^( Theta )`

Solution :For the reaction,
`CH_(4(g)) + 2O_(2(g)) to CO_(2(g)) + 2H_(2) O_((l))`
`Deltan_(g) = (n_(p) - n_(r) )_(g) = 1-3 = -2`
`Delta_(f) H^( - Theta )= Delta_(r) U^( Theta ) + Deltan_(g) "RT"`
As `Deltan_(g) = -2`
`therefore Delta_(f) H^( Theta ) lt Delta_(f) U^( Theta )`
Discussion
9.

The kind of delocalisation involving sigma bond orbitals is called…….

Answer»

Mesomeric EFFECT
Tautomeric effect
ELECTROMERIC effect
Hyperconjugative effect

ANSWER :D
Discussion
10.

Delicate materials like silk, wool and ivory are bleached by

Answer»

`H_(2)O_(2)`
`CaOCl_(2)`
`Cl_(2)`
None

Answer :A
Discussion
11.

Deionised water is prepared by the following method

Answer»

Clark's
Ion exchange
Permutit
Calgon

ANSWER :B
Discussion
12.

Dehydrohalogenation reaction is also known as _____.

Answer»

ADDITION
`ALPHA`-DEHYDRATION
`BETA`-ELIMINATION
Substitution

Solution :`beta`-elimination
Discussion
13.

Dehydrohalogenation involves removal of the halogenn atom together with a hydrogen atom from carbon adjacent to the one with halogen atom. Alcoholic KOH is used for dehydrohalogenation. According to saytzeff's rule, when two alkenes may be formed, the alkene which is most substituted is the major product. Q. Arrange the following alkyl halides in decreasing order of the rate of beta-elimination reaction with alcoholik KOH. underset((i))(CH_(3)CH_(2)CH_(2)Br) underset((ii))(CH_(3)-underset(CH_(3))underset(|)(C)H-CH_(2)Br) underset((iii))(CH_(3)-CH_(2)-underset(Br)underset(|)(C)H-CH_(3))

Answer»

iigtiiigti
iiigtiigti
igtiigtiii
iigtigtiii

Answer :B
Discussion
14.

Dehydrohalogenation of alkyl halide is

Answer»

Addition reaction
Elimination reaction
REARRANGEMENT reaction
Substitution reaction

Solution :Involves removal of HYDROGEN and HALOGEN ATOMS.
Discussion
15.

Dehydrohalogenation involves removal of the halogenn atom together with a hydrogen atom from carbon adjacent to the one with halogen atom. Alcoholic KOH is used for dehydrohalogenation. According to saytzeff's rule, when two alkenes may be formed, the alkene which is most substituted is the major product. Q. What are the products of dehydrohalogenation of 2-iodopentane?

Answer»

2-Pentene(major),1-Pentene (minor)
1-Pentene (major), 2-Pentene (minor)
2-Pentene (50%), 1-Pentene (50%)
NONE of these

Solution :`CH_(3)CH_(2)CH_(2)-UNDERSET(I)underset(|)(C)H-CH_(3) overset("ALC. KOH")to`
`underset("2-Pentene (major)")(CH_(3)CH_(2)-CH=CH-CH_(3))+underset("1-Pentene (minor)")(CH_(3)CH_(2)CH_(2)-CH=CH_(2))`
Discussion
16.

Dehydrogenation of isobutyl alcohol yields

Answer»

Butanone
Butanal
2-Methylpropanol
2-Methylpropene

Solution :`underset("Isobutyl alcohol")(CH_(2)-underset(CH_(3))underset(|)(CH)-CH_(2)OH) underset(-H_(2))OVERSET(Cu//575K)RARR underset("2-Methylpropanal")(CH_(3)-underset(CH_(3))underset(|)(CH)-CHO)`
Discussion
17.

Dehydration of which of the following alcohols produce alkenes exhibiting cis-trans-isomerism ?

Answer»

ISOPROPYL alcohol
Tertiary butyl alcohol
Secondary butyl alcohol
3-Pentanol

Solution :Both `2^@` butyl alcohol and 3-pentanol on DEHYDRATION GIVE alkenes which show cis-trans isomerism.
`UNDERSET(2^@ "Butyl alcohol")(CH_3-CHOH-CH_2CH_3)underset((-H_2O))overset"Dehydration"to underset"2-Butene"(CH_3-CH=CH-CH_3)`
`underset"3-Pentanol"(CH_3CH_2-CHOH-CH_2CH_3)underset((-H_2O))overset"Dehydration"to underset"2-Pentene"(CH_3CH_2-CH=CHCH_3)`
Discussion
18.

Dehydration of hydrates of halides of calcium, barium and strontium i.e., CaCl_(2)*6H_(2)O, BaCl_(2)*2H_(2), SrCl_(2)*2H_(2)Ocan be achieved by heating. These become wet on keeping in air. Which of the following statements is correct about these halides?

Answer»

Act as dehydrating agent
Can absorb moisture from air
Tendency to FORM HYDRATE decreases from CALCIUM to barium
All of the above

Solution :CHLORIDES of alkaline earth metals are hydrated. These are hygroscopic in nature.
Discussion
19.

Dehydration of hydrates of halides of calcium , barium and strontium i.e, CaCl_(2) . 6 H_(2)O , BaCl_(2) . 2H_(2)O , SrCl_(2) . 2H_(2)O , can be achieved by heating . These become wet on keeping in air . Whichof the following statements is correct about these halides ?

Answer»

ACT as dehydrating agent
Can ABSORB moisture from air
TENDENCY to form hydrate DECREASES from calcium to barium
All of the above

Answer :D
Discussion
20.

Dehydration of glycerol gives

Answer»

Propane
Propene
ACROLEIN
Benzene

Solution :Acrolein `(CH_(2)=CH-CHO)`
Discussion
21.

Dehydration of formic acid gives "…………..".

Answer»


ANSWER :`CO`
Discussion
22.

Dehydration of alcohols to alkenes is always carried out with conc. H_2SO_4 and not with HCl or HNO_3. Explain .

Answer»

Solution :Dehydration of ALCOHOLS to alkenes occurs through CARBOCATION intermediate. If HCl is USED , the chloride `(Cl^-)` ion being a good nucleophile brings about substitution reaction to give an alkyl halide.
`{:(underset"ALCOHOL"(R-CH_2-CH_2-OH)underset(-H_2O)overset(H^+)to,underset"Carbocation"(R -CH_2-overset+CH_2),underset(-H^+)to underset"(Elimination product)"underset"Alkene"(R-CH=CH_2)),(,underset"(Substitution product)"underset"Alkyl chloride"overset(darr +Cl^-)(R-CH_2-CH_2-Cl),):}`
If `H_2SO_4` is used, then the bisulphate `(HSO_4^-)` ion being non-necleophilic cannot bring out the substitution reaction. INSTEAD , the carbocation loses a proton to form an alkene. Conc. `HNO_3` is not used because it being a strong oxidising agent, preferentially brings about the oxidation of alcohols first to aldehydes or ketones and then to carboxylic acids. Thus, dehydration of alcohols to form alkenes is always carried out with such strong acids (e.g., `H_2SO_4 , H_3PO_4,H_3PO_4, KHSO_4`) which have non-nucleophilic anion)
Discussion
23.

Dehydration of 2,2,3,4,4-pentamethy 1-3-pentanol gave two alkenes A and B. The ozonolysis product of A and B are

Answer»




NONE of these

SOLUTION :
Discussion
24.

Dehydration of 1-butanol or 2-butanol with conc. H_2SO_4 always gives the same mixture of 2-butene (80 %) and 1-butene (20 %) but dehydrohalogenation of 1-bromobutane with alc. KOH gives 1-butene as the major product while that of 2-bromobutane gives 2-butene as the major product . Explain why ?

Answer»

Solution :During dehydration of alcohols , carbocations are the intermediates. In CASE of 1 butanol , first `1^@` carbocation is FORMED which being less stable rearranges to the more stable `2^@` carbocation before losing a proton to form a mixture of 2-butene (80 %) and 1-butene (20%) .

Since 1-butanol and 2-butanol give the same carbocation intermediate, therefore , on losing a proton, both give the same mixture of 2-butene (80%) and 1-butene (20%) in accordance with the Saytzeff's rule. In case of dehydrohalogenation of alkyl halides with alc. KOH , carbocations are not the intermediates. Instead the reaction occurs by a concerted mechanism involving a transition state . Further , since in 1-bromobutane, there is `beta`-H only on left side of `alpha`-carbon carrying the Br atom, therefore , only 1-butene is formed as the major product. however, in 2-bromobutane there are `beta`-H ONE either side of the `alpha`-carbon carrying the Br atom, therefore, elimination of a proton can occur from either `beta`-H leading to the formation of a mixtures of ALKENES . Since according to Saytzeff rule, 2-butene is more stable than 1-butene, therefore, dehydrohalogenation of 2-bromobutane gives 2-butene as the major product.

Discussion
25.

Dehydration means

Answer»

REMOVAL of HYDROGEN molecule from ADJACENT carbons
Removal of hydrogen molecule from the same carbon
Removal of WATER moelcule from the same carbon as H and OH
Removal of H and OH from the adjacent carbons as water molecule

Answer :D
Discussion
26.

Degree of ionization depends on which factor.

Answer»

SOLUTION :It depends on (i) STRENGTH of BOND (II) solution of IONS produced.
Discussion
27.

Degree of liness of water is expressed in terms of ppm of

Answer»

CAO
`CaCl_2`
`CaCO_3`
`CA(HCO_3)_2`

ANSWER :C
Discussion
28.

Degree of hardness of a water sample is 100ppm. If that sample contains hardness causing salt Mg(HCO_3)_2,calculate weight of Ca(OH)_2required for the treatment of 10kg of water sample.

Answer»

Solution : Degree of hardness is 100ppm. Thus `10^(6)` g of HARD WATER contains 100g of `CaCO_(3)` which is equivalent to 146g of `MG(HCO_(3))_(2)`
Weight of `Mg(HCO_(3))_(2)` present in 10kg of water `=(146xx10^(4))/(10^(6))=1.46g`
`Mg (HCO_(3))_(2)+2CA(OH)_(2)to2CaCO_(3)+Mg(OH)_(2)+2H_(2)O`
146g of Mg`(148zxx1.46)/(146)=1.48g`
Discussion
29.

Degensrate atomic orbitals have :

Answer»

EQUAL energy in absemce of magnetic FIELD
NEARLY equal energy
different energy
equal in presence of magnetic field

Solution :N/a
Discussion
30.

Deflection back of a few particles on hitting thin foil of gold shows that

Answer»

NUCLEUS is heavy
nucleus is small
Both a and b
eletrons CREATE hindrance in the movement of `alpha-particles`

ANSWER :C
Discussion
31.

Define'molarheat capacity'

Answer»

Solution :MOLAR heat capacity of a SUBSTANCE is DEFINED as the amount of heat required to raise the TEMPERATURE of one mole of the substance through the `1^(@)C`.
Discussion
32.

Define Zeroth law of thermosynamics (or Law of thermal equilibrium.

Answer»

Solution :ZEROTH LAW of thermodynamics states that .If two systems at different temperatures are separately in thermal equilibrium with a third ONE then they TEND to be thermal equilibrium with themselves. .
Discussion
33.

Define Wave number.

Answer»

Solution :It is the NUMBER of waves present PER meter and is EQUAL to the reciprocal of wavelength. `(1/(LAMDA))=barv`
Discussion
34.

State Zerothlaw of thermodynamic.

Answer»

Solution :Zeroth law of thermodynamics states that .If TWO SYSTEMS at DIFFERENT temperatures are SEPARATELY in thermal equilibrium with a third one, then they tend to be in thermal equilibrium with themselves
Discussion
35.

Define Walden inversion.

Answer»

SOLUTION :The attach of nucleophile OCCURS from the back side (i.e., opposite to the side in which the HALOGEN is attacked). The carbon at which substitution occurs has inverted configuration during the course of reaction just as an UMBRELLA has tendency to invert in a WIND storm. this inversion of configuration is called walden inversion.
Discussion
36.

DefineVolumepercentage .

Answer»

Solution :Volume PERCENTAGE is defined as the ratio of volume of SOLUTE in ML to the volume of solution in ml multiplied by 100.
Volume percentage = `("Volume of solute (in mL)")/("Volume of solution (in mL)") XX 100`
Discussion
37.

Define visocity of a liquid.

Answer»

Solution :The INTERNAL resistance to the FLOW of the liquids which one LAYER offers to another layer trying to pass over is called its VISCOCITY. It depends upon the nature of the LIQUID and temperature.
Discussion
38.

Define Van't Hoff factor.

Answer»

SOLUTION :van.tHoff FACTOR (i) is defined as the ratio of the actual molar MASS to the ABNORMAL molar mass of the solute.
`I = ("Normal (actual) molar mass")/("Abnormal (observed) molar mass")`
`I = ("Observed colligative property")/("Calculate colligative property")`
Discussion
39.

Define vander waals radius.

Answer»

Solution :It is half the distance between the NUCLEI of NON bonded ATOMS of neighbouring molecules in a solid.
Discussion
40.

Define valency . How is it determined ?

Answer»

Solution :The valency of an element MAY be defined as the combining capacities of elements. The electrons present in the outermost SHELL are CALLED valence electrons and these electrons determine the valency of the ATOM.
Discussion
41.

Define- Valency.

Answer»

Solution :Define VALENCY of an ELEMENT MAY be defined as the combining capacities of elements. The electrons present in the outermost shell are CALLED valence electrons and these electrons DETERMINE the valency of the atom .
Discussion
42.

Define - Valency.

Answer»

SOLUTION :The VALENCY of an element may be defined as the combining capacities of elements. The electrons PRESENT in the outermost SHELL are called valence electrons and these electrons determine the valency of the ATOM.
Discussion
43.

Define torsional strain.

Answer»

Solution :To CHECK the increased repulsive forces, molecule will have to possess more energy and thus has lesser STABILITY. The repulsive interaction between the electron clouds, which affects stability of a conformation is called torsional STRAIN.
Discussion
44.

Define third law of thermodynamic.

Answer»

SOLUTION :The entropy of a perfectly CRYSTALLINE solid may be TAKEN as zero at the absolute zeroof TEMPERATURE.
Discussion
45.

Define the terms, Bond length

Answer»

SOLUTION :BOND length: It is the EQUILIBRIUM DISTANCE between nuclei of two bonded atoms in a molecule.
Discussion
46.

Define the term 'radiation'.

Answer»

SOLUTION :RADIATION is the emission and TRANSMISSION of ENERGY through SPACE in the form of waves.
Discussion
47.

Define the term osmotic pressure. Describe how the molecular mass of a substance can be determined by a method based on measurement of osmotic pressure ?

Answer»

Solution :`PI = CRT`
`pi = (n)/(V)RT`
`or pi V = n RT`
` or pi V = (W _(B))/(M _(B)) RT`
`or M _(B) = (W_(B) RT)/(piV)`
OSMOTIC pressure is INVERSELY proportional to the MOLECULAR mass of the solute.
Discussion
48.

Define the term isotonic solution.

Answer»

Solution :(i) Two solutions having same osmotic pressure at a given tempreture are called ISOTONIC solutions.
(II) When such solutions are seprated by a semipermeable MEMBRANE, sovlent flow between one to the other on either direction is same , the net solvent flow between two isotonic solutions is zero.
Discussion
49.

What are isotonic solutions?

Answer»

Solution :(i) Two solution having same osmotic pressure at a GIVEN TEMPERATURE are called istonic solutions.
(ii) When such solutions are separated by a semipermeable membrane, solvent FLOW between one to the other on either direction is same. i.e., the net solvent flow between two ISOTONIC solutions is zero.
Discussion
50.

Define the term 'isotonic solution'.

Answer»

Solution :(i) TWO solutions having same osmotic PRESSURE at a given temperature are calledisotonic solutions.
(ii) When such solutions are separated by a semipermeable membrane SOLVENT flow between one to the other on either DIRECTION is same i.e. the net solvent flow between the two ISOTONIC solutions is zero.
Discussion