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.

Enthanenitrile on hydrolysis gives________.

Answer»

SOLUTION :ACETIC ACID or ethaoic acid.
Discussion
2.

Enthalpy of vaporization of benzene is +35.3 kJ mol^(-1) at its boiling point, 80^(@)C. The entopy change in the transition of the vapour to liquid at its boiling point [in JK^(-1)mol^(-1)] is….

Answer»

`-441`
`-100`
`+441`
`+100`

SOLUTION :Accordin to Gibb's formula, `DeltaG=DeltaH-TDeltaS`
At equilibrium, `DeltaG=0`
And entropy decreases on going from VAPOUR to liquid, i.e.,
`DeltaH=TDeltaS`
`DELTAS=-(DeltaH)/(T)=-(35300)/(80+273)=-100 JK^(-1)MOL^(-1)`
Discussion
3.

Enthalpy of vaporization of benzene is + 35.3 kJ mol^(-1) at its boiling point, 80^(@)C. The entropy change in the transition of the vapour to liquid at its boilling point [in JK^(-1) mol^(-1)] is ……….

Answer»

`-441`
`-100`
`+441`
`+100`

Solution :ACCORDING to Gibb.s formula
`DeltaG=DeltaH-TDeltaS`
At EUILIBRIUM, `DeltaG=0`
And entropy decreases on going from vapour to liquid. i.e., `DeltaH=TDeltaS`
`DeltaS=-(DeltaH)/(T)=-(35300)/(80+273)=-100"JK"^(-1)"MOL"^(-1)`
Discussion
4.

Enthalpy of vaporisation for water is 186.5kJ mol^-1. The entropy change during vaporisation is____kJ K^-1mol^-1:

Answer»

0.5
1
1.5
2

Answer :A
Discussion
5.

Enthalpy of solution of NaOH (solid) in water is -41.6 kJ mol^(-1). When NaOH is dissolved in water, the temperature of water

Answer»

INCREASES
decreases
does not change
fluctuates indefinitely

Solution :SINCE, process is exothermic, heat is evolved, due to this, TEMPERATURE of WATER increases.
Discussion
6.

Enthalpy of physical adsorption is quite low: Give reason.

Answer»

Solution :Because it INVOLVES WEAK VAN DER WAALS. forces.
Discussion
7.

Enthalpy of neutralization of acetic acid by NaOH is -50.6 kJ//"mol". And the heat of neutralisation of a strong acid with a strong base is -55.9 kJ/mol. What is the value of DeltaH for the ionisation of CH_(3)COOH

Answer»

`+5.3 kJ//mol`
`+6.2 kJ//mol`
`+8.2 kJ//mol`
`+9.3 kJ//mol`

SOLUTION :`DELTAH=DeltaH_("ioniz")+DeltaH_("NEU")=-50.6=DeltaH_("ioniz")+(-55.9)`
`DeltaH_("ioniz")=+5.3 kJ//mol`.
Discussion
8.

Enthalpy of neutralization is defined as the enthalpy change when 1 mole of acid/base is completely neutralized by base/acid in dilute solution. For strong acid and strong base neutralization net chemical change is ""H^(+)(aq) + OH^(-)(aq) rarr H_(2)O(l), "" Delta_(r)H^(@) = -55.84 KJ//mol DeltaH_("ionization")^(@) of aqueous solution of strong acid and strong base is zero. When a dilute solution of a weak acid or base is neutalized, teh enthalpy of neutralization is some what less because of the absorption of heat in the ionization of the weak acid or base, for weak acid/base DeltaH_("neutrilization")^(2) = DeltaH_("ionization")^(@) + Delta_(r)H^(@) "" (H^(+) + OH^(-) rarr H_(2)O) Under the same condition how many mL of 0.1 M NaOHand 0.05 M H_(2)A (strong diprotic acid) solution should be mixed for a total volume of 100 mL produce the highest rise in temperature:

Answer»

`25:75`
`50:50`
`75:25`
`66.66:33.33`

SOLUTION :For max. rise in TEMP. max. neutralization of `H^(+)` and `OH^(-)` REQUIRED .
If we take equal VOLUME , all `H^(+) (5 m-"mole")` will react with all `OH^(-)(5 m-"mole").`
Discussion
9.

Enthalpy of neutralisation of NaOH with H_(2)SO_(4) is -57.3kJ mol^(-1) and with ethanoic acid -55.2 kJ mol^(-1). Which of the following is the best explanation of this difference?

Answer»

Ethanoic ACID is WEAK acid and thus requires less `NaOH` for neutralisation
Ethanoic acid is only partly ionized, neutralisation is THEREFORE incomplete
Ethanoic acid is monobasic while `H_(2)SO_(4)` is dibasic
Some heat is used to ionize the ethanoic acid completely.

Solution :Enthanoic acid is weak acid and it requires so heat for its IONIZATION.
Discussion
10.

Enthalpy of formation of two compounds x and y are - 84 kJ and - 156 kJ respectively. Which of the following statement is correct

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x is more stable than y
x is LESS stable than y
Both x and y are unstable
x and y are ENDOTHERMIC compounds

Solution :The compound Y POSSESS less energy than the X and THUS Y is more stable than X.
Discussion
11.

Enthalpy of formation of HF and HCl are -161 kJ and -92 kJ respectively. Which of the following statements is incorrect?

Answer»

HCL is more STABLE than HF
HF and HCl are exothermic compounds
the affinity of FLUORINE to hydrogen is GREATER than the affinity of chlorine to hydrogen than the affinity of chlorine to hydrogen
HF is more stable than HCl.

Solution :More the energy libertaed during the formation of one mole of a compound, more will be its stability
Or
Higher the enthalpy (-ve value) of formation of a compound greater will be stability of the compound. HF is more stable than HCl.
Discussion
12.

Enthalpy of dissociation is low for

Answer»

`F_(2)` <BR>`Cl_(2)`
`Br_(2)`
`I_(2)`

ANSWER :D
Discussion
13.

Enthalpy of combustion of C_(6) H_(6) (l) is - 3264. 64 kj/ mol . The heat produced by burning 3. 9 g of benzene is

Answer»

`-163.23 KJ`
`` 326 . 4 kJ
` 32.64 kJ`
`-3.254 kJ`

Solution :since , heat evolved when 78 g BENZENE is BURNT
`=-3264.6 kJ`
therefore, heat produced , by buring 3.9 g benzence
`3.9/78xx(-3264.6)=-163.23 kJ`
Discussion
14.

Enthalpy of atomisation is lowest in

Answer»

Sc
Mn
Ni
Zn

Answer :D
Discussion
15.

Enthalpy of a compound is equal to its

Answer»

HEAT of combustion
Heat of formation
Heat of REACTION
Heat of solution

Solution :ENTHALPY of a COMPOUND = Heat of reaction of that compound.
Discussion
16.

Enthalpy is same as:

Answer»

HEAT CONTENT
ENTROPY
Heat
Volume

Answer :A
Discussion
17.

Enthalpy is an ... property.

Answer»

SOLUTION :EXTENSIVE
Discussion
18.

Enthalpy (H) is equal to

Answer»

Internal energy (E)
Product of PRESSURE (P) and volume (V) of gas
Internal energy (E)+PV
Work (W) done by a system

Solution :Enthalpy (H) is DEFINED as the sum of internal energy E+PV, H=E+PV.
Discussion
19.

Enthalpy for the reaction C+O_(2)rarrCO_(2) is

Answer»

Positive
Negative
Zero
None

Solution :It is a COMBUSTION REACTION, `DeltaH=-ve`.
Discussion
20.

Enthalpy for the reaction C+O_2rarrCO_2 is:

Answer»

`=ve`
-ve
zero
none

Answer :B
Discussion
21.

Enthalpy change in the formation of an imaginary lattice Cl^(+)Cl^(-) from Cl_(2)(g) is represented as Delta_(t)H. Given that bond dissociation enthalpy DeltaH_(Ct-ct) =x_(1). Ionisation of chlorine is x_(2), electron gain enthalpy of chlorine is -x_(3) and lattice enthalpy of Cl^(+)Cl^(-) is -x_(4). The value of DeltaH is

Answer»

`x_(1) + x_(2) + x_(3)+x_(4)`
`x_(1) + x_(2)-(x_(3) + x_(4))`
`x_(3) + x_(4) -(x_(1) +x_(2))`
`-(x_(1) + x_(2) + x_(3) +x_(4))`

Answer :B
Discussion
22.

Enthalpy change , when 1 g water is frozen at 0(@)C is (DeltaH_("fus")= 1.435 kcal mol ^(-1) )

Answer»

0.0797 KCAL
`-0.0797 kcal`
1.435 kcal
`- 1.435 kcal`

SOLUTION :`DeltaH=-1.435/18kcal G^(-1)`
`0.0797 kcal g^(-1)`
Discussion
23.

Enthalpy change for the reaction, 4H(g)to2H_(2)(g) is -869.5 kJ. The dissociation energy of H-H bonds is

Answer»

`+217.4kJ`
`-434.8kJ`
`-869.6kJ`
`+434.8kJ`

SOLUTION :The DISSOCIATION ENERGY of H-H BOND is
`=(869.6)/(2)=+434.8kJ`
Discussion
24.

Enthalpy change for the reaction, 4H_((g))rarr2H_(2(g)) is - 869.6 kJ. The dissociation energy of H - H bond is

Answer»

`+ 217.4` KJ
`- 434.8` kJ
`- 869.6` kJ
`+ 434.8` kJ

Solution :The dissociationenergy of H-H BOND is `=(869.6)/(2)=434.8 kJ`.
Discussion
25.

Enthalpy change in the following process is A+ e^(-) to A^(-) Delta H= - X kJ//"mole" Which of the following processes has enthalpy change=X kJ/mole ?

Answer»

`A^(-2) to A^(-)+1E^(-)`
`A+E^(-) to A^(-)`
`A^(-) to A+ e^(-)`
`A^(+)+e^(-) to A`

ANSWER :C
Discussion
26.

Enthalpy change for reaction, (1)/(2)H_(2)+(1)/(2)Cl_(2)rarrHCl, is called

Answer»

ENTHALPY of COMBINATION
Enthalpy of reaction
Enthalpy of FORMATION
Enthalpy of fusion

Solution :Enthalpy of formation of HCL.
Discussion
27.

Enthalpy change for a reaction does not depend upon

Answer»

The physical states of REACTANTS and products
Use of different reactants for the same product
The nature of intermediate reaction STEPS
The DIFFERENCES in initial or final TEMPERATURE of involved substances

Solution :According to Hess law, enthalpy change for a reaction does not depend on the nature of inter mediate reaction steps.
Discussion
28.

Enthalpuy of solution of NaOH (s) in water is -41 .6 KJ mol ^(-1) .when NaOH is dissolved in water , the temperature of water

Answer»

increases
decreases
does not change
fluctuates indefinitely

Solution :since the PROCESS is EXOTHERMIC , heat is evolved DUE to this temperature of water increases.
Discussion
29.

Enthalpies of elements in their standard states are taken as zero. Hence the enthalpy of formation of a compound

Answer»

SHUOLD always be negative
shuold always be positive
will be EQUAL to TWICE the energy of combustion
may be +ve or -ve

Answer :D
Discussion
30.

Enrichment of uranium is made by :

Answer»

distillation
diffusion
evaporation
bleaching

Answer :B
Discussion
31.

Enough moisture in food permits

Answer»

DECREASE the GROWTH of microbs
action of own enzymes
increase the vitamin-C
decrease the growth of BACTERIAL spores

Answer :B
Discussion
32.

Enol content is maximum in:

Answer»

ACETONE
acetophenone
acetic ACID
acetylacetone

Solution :`CH_(3)-OVERSET(O)overset(||)(C)-CH_(2)-overset(O)overset(||)(C)-CH_(3)HARR`
Discussion
33.

enol content will be maximum in which of given compound?

Answer»

`CH_3CHO`


ANSWER :C
Discussion
34.

Enlist the various steps involved in extraction of metals.

Answer»

SOLUTION :
Discussion
35.

Enlist the total number of distinct naturally occurring amino acids obtained by complete acidic hydrolysis of the peptide shown below is :

Answer»

SOLUTION :The PRODUCTS of HYDROLYSIS are SHOWN :
Discussion
36.

Enlist the main points of difference between S_(N)1 and S_(N)2 reactions.

Answer»

SOLUTION :
Discussion
37.

Enlist the condition which facour the formation of an idela solution.

Answer»

SOLUTION :For ANSWER, CONSULT SECTION 6.
Discussion
38.

Enlist the points of difference between fibrous and globular proteins.

Answer»

SOLUTION :The POINTS of DIFFERENCES are :
Discussion
39.

Enlist the main uses of nitrogen.

Answer»

Solution :In manufacturing of ammonia and industrial chemicals such as calcium cyanamide.In iron and steel industry to maintain inert atmosphere.
As inert diluent for reactive chemicals.
Liquid NITROGEN is used as refigerant to preserve BIOLOGICAL MATERIALS, FOOD items and in CRYOSURGERY.
Discussion
40.

Enlist the physical properties of transition elements

Answer»

Solution :The physical properties of transition elements are: Metallic in nature
They are malleable and ductile. All are SOLIDS EXCEPT MERCURY which is liquid.
They have high tensile strength
All the elements except Zn, Cd, Hg have high melting points and have low VOLATILITY. Boiling points are also high.
They produce sound on striking and hence are sonorous.
They have high thermal and electrical conductivity and have metallic lustre.
At normal TEMPERATURES, all the elements except Zn, Cd, Hg and Mn have one or more typical metallic structures.
Discussion
41.

Enlist the characteristics of unit cell.

Answer»

Solution : Its dimensions along three edges a, b, and c. These edges may or may not be MUTUALLY perpendicular.
ANGLES between the edges `alpha`(between b and c),`beta`(between a and c) and ` gamma`(between a and b).
THUS UNIT cell is characterised by six PARAMETERS `a, b, c, alpha, beta` and ` gamma` .
Discussion
42.

Enlist some important vitamins with their sources and the disease caused due to their deficiency in body.

Answer»

SOLUTION :
Discussion
43.

Enlist the characteristic properties of solid state.

Answer»

SOLUTION : Solids have definite mass, volume and shape.
Intermolecular distances are short.
Intermolecular forces are strong.
The constituent particles (atoms, MOLECULES or ions) in solids have fixed positions and can only oscillate about their MEAN positions.
They are rigid and INCOMPRESSIBLE.
Discussion
44.

Enlist the characteristics of crystal lattice.

Answer»

Solution : Each point in a LATTICE is CALLED lattice site or lattice point .
Each point in a crystal lattice REPRESENTS one constituent particles which may be an atom, a molecule (group of atoms) or an ion.
Lattice points are JOINED by straight LINES to bring out geometry of the lattice.
Discussion
45.

E_(ni^(2+)//Ni)^(@)=-0.25 VE_(Au^(3+)//Au)^(@)=1.50 v The emf of the voltaic cell Ni//Ni^(2+)(1.0 M)||Au^(3+)(1.0 M)|Au is

Answer»

1.25 V
`-1.75 V`
1.75V
2.0 V

Solution :EMF of the CELL = `1.50 -(-0.25) = 1.75 V`
Discussion
46.

Energy requried to dissociate 4 g of gaseous hydrogen into free gaseous atoms is 208 kcal at 25^(@)C. The bond energy of H-H will be

Answer»

104 kcal `MOL^(-1)`
52 kcal `mol^(-1)`
10.4 kcal `mol^(-1)`
1040 kcal `mol^(-1)`

Solution :NUMBER of MOLES of hydrogen`=(4g)/(2" g "mol^(-1))=2` moles
Given, `2H_(2)(g)to4H(g),DeltaH=104` kcal `mol^(-1)`. Hence, BOND energy of H-H bond is 104 kcal `mol^(-1)`
Discussion
47.

Enery is stored in our body in the form of ………. .

Answer»

ATP
ADP
FATS
CARBOHYDRATES

ANSWER :A
Discussion
48.

Energy required to separate neutron and proton from the nucleus is called

Answer»

BOND energy
Nuclear energy
Chemical energy
Radiation energy

Answer :B
Discussion
49.

Energy required to ionise 1 model of gaseousHe^+ion present in its ground state is

Answer»

`108.8 N_AeV`
`13.6 `EV
`54.4` eV
`54.4 N_A eV`

ANSWER :D
Discussion
50.

Energy required to dissociate 4 gm of gaseous hydrogen into free gaseous atoms is 208 kcal at 25^(@)C. The bond energy of H-H bond will be

Answer»

104 kcal
10.4 kcal
1040 kcal
104 cal

Solution :4g `H_(2)=2` moles. Bond energy for 1 MOLE of `H_(2)=208//2=104` kcal.
Discussion