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.

{:("Column-I ","Column-II "),("A) Atm L mol"^(-1),"P)" a//Rb),("B) " sqrt((2RT)/(M)), "Q)" a//b),("C) Boyle temperature ","R)" V_m),("D) Volume of one mole","S) "u_(MP)):}

Answer»

Solution :`T_b = a/(Rb) , V_m` = molar volume
`U_(MP) = sqrt((2RT)/(M)) , a/b = ("atm - LIT"^2)/("mol"^2 XX ("lit")/("mol")) = ("atm - lit")/("mol")` .
2.

Column IColumn II

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SOLUTION : FOLLOW cannizaroreactionmechanism .
3.

{:("List -I Salt","List -II Kg"),((A) AgCI,P.27S^(4)),((B)PbI_(2),Q.108(S)^(5)),((C )AS_(2)S_(3),R.4S^(3)),((D)Ag_(3)PO_(4),S.S^(2)):}

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Solution :` AgCl hArr Ag^(+) +CL ,AgCl hArr UNDERSET(s) Ag^(+)+underset( S) Cl^(+) `
` Ksp = S^(2) `
` AgCl hArr Ag^(+) +Cl ,AgCl hArr underset(s) Ag^(+)+underset( S) Cl^(+) `
` Ksp = S^(2) `
` AS_2 S_3 hArr 2 underset(2S) AS^(+3)+underset( 3S) 3S^(-2) `
` K_(sp)=[As^(+3) ] ^(2)[S^(-2) ]^(3)=(2S)^(2)(3s)^(3)=108 S^(5)`
` Ag_3 PO_4 hArr 3underset( 3s) Ag^(+)+underset(S) POunderset( 4) overset(-3) `
`K_(sp)=(3S)^(3)(S)=27 S^(4) `
4.

Property to be determined Method used for determination{:("(P) Estimation of carbon and hydrogen in an organic compound","(i) Lassaigne's test"),("(Q) Estimation of nitrogen in aniline","(ii) Carius method"),("(R ) Estimation of chlorine in carbon tetrachloride","(iii) Liebig's method"),("(S) detection of nitrogen, sulphur and halogens","(iv) Kjeldahl's method"):}

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SOLUTION :ESTIMATION of ORGANIC COMPOUNDS.
5.

Column -I and Column -II contain four entries each. Entries of column-I are to be matched with some entries of column-II. One or more than one entries of column-I may have the matching with the same entries of column-II.

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<P>

Answer :A `RARR`P;B`to`Q,R;C`to`P;D`to`R, S
6.

{:("Column-"," Column-II "),("A) Boyle's law ","P)" P_("obs") = P_("atm") +P_("water vapour")),("B) Avagadro's law ","Q)" V_1 = n_1 ((V_2)/(n_2))),("C) Charles' law ","R)" V_t = V_0 (1 + t/273)),("D) Dalton's law ","S)" V_1 = P_2((V_2)/(P_1))):}

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SOLUTION :GAS LAWS.
7.

Column-1 and column-II contains four entries each. Entries of column-I are to be matched with some entries of column-II .One or more than one entries of column-I may have the matching with the same entries of column-II

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ANSWER :A::B::C::D
8.

Colour of colloids depend on which of the following factors

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Size
Mass
Charge
Nature

Solution :Colour of colloidal solution depends upon particle size. As the PARTICLES size increases, the colour of gold sol changes from red to blue and FINALLY to GOLDEN.
9.

Coloumn I & Column II contain data on Schrondinger Wave-Mechanical model, where symbols have their usual meanings. Match the column :-

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SOLUTION :s-orbital is SPHERICAL HENCE it is non-directional.
10.

Colors of super oxides of alkali metals are ..... or ......

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SOLUTION :YELLOW, ORANGE
11.

Colloids and crystalloids differ w.r.t.

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PHYSICAL STATE
SIZE of the particle
PROTECTIVE agent
viscosity

Solution :Size of the crystalloid particles in solution is less than colloidal particles.
12.

Colloidion is a colloidal solution of

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SOURCE in water
cellulose in water
cellulose NITRATE in water
cellulose nitrate in ETHYL alcohol

Answer :D
13.

Colloidal sulphur is used as

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a disinfectant
a germ KILLER
a TREATING EYE diseases
a to raise vitality of human system

Solution :Germ killer
14.

Colloidal solutions of gold prepared by different methods are of different colours because of

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variable valency of GOLD
different CONCENTRATION of gold particles
impurities produce by different methods
different diameters of colloidal gold particles.

Solution :Colour of a colloidal solutio, in general, depends UPON the size of the colloidal particles and POSITION of the viewer.
15.

colloidal solutions of gold prepared by different methods are of different colours because of :

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variable valenct of GOLD
impurities produced by DIFFERENT methods
different DIAMETERS of colloidal gold PARTICLES
different concentration of gold particles

ANSWER :C
16.

Colloidal solution cannot be obtained from two such substances which are

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INSOLUBLE in each other
In same PHYSICAL state
In DIFFERENT physical state
None of these

Answer :D
17.

Colloidal solutioin of gold cannot be prepared by

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BREDIG's are METHOD
MECHANICAL dispersion
Reduction of GOLD chloride
Echange of solvents

Answer :D
18.

Colloidal sol found effectivein treating eye diseases is

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COLLOIDAL sulphur
colloidal antimony
colloidal gold
colloidal silver

Answer :D
19.

Colloidal particles of soap sol in water is

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NEGATIVELY charged
neutral
Positively charged
Unpredictable

Answer :A
20.

Colloidal gold is prepared by

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MECHANICAL dispersion
Peptisation
BREDIG's ARC METHOD
Hydrolysis

Solution :By Bredig's arc method
21.

Colloidal gold is given by injection

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to act as a disinfectant
to act as a GERM killer
for treating EYE diseases
to raise VITALITY of human system

Solution :To raise vitality of human beings
22.

Colloid of which one of the following can prepared by electrical dispersion method as well as reduction method ?

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Sulphur
Ferric hydroxide
Arrhenius sulphide
Gold

Solution :Gold SOL by Bredig's METHOD .(Dispersion method) and by REDUCTION method `(AuCl_(3)+` Tannic ACID `to` Gold sol).
23.

Colloid of which one of the following can be prepared by electrical dispersion as well as reduction method?

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Sulphur
Ferric hydroxide
Arsenious sulphide
Gold.

Solution :Gold COLLOID can be prepared by BREDIG arc method as WELL as by reduction of `AuCl_(3)` with `SnCl_(2)`
24.

Collodian is a 4% solution of which one of the following alcohol ether mixture

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nitroglycerine
cellulose acetate
glycol dinitrite
NITROCELLULOSE

SOLUTION :Collodian is a solution of nitrocellulose in alcohol ETHER mixture.
25.

Collision frequency (Z) of a gas at a particular pressure

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decreases with the rise in temperature
increases with the rise in temperature
decreases initially and thereafter increases
unpreductable

Solution :Collision FREQUENCY = Collisions which all MOLECULES undergo in a unit volume per second
`= (1)/(2)` collision no. `xx N//V`
or `Z = (1)/(2) sqrt2 PI bar(v) sigma^(2) N//V xx N//V`
`= (1)/(sqrt2) pi bar(v) sigma^(2) ((N^(2))/(V^(2)))`
or `Z prop T` SINCE `bar(v) prop T`
26.

Collision diameter is least in case of (1) H_2 , (2) He: (3) CO_2 : (4) N_2 Indicate with the number in brackets.

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SOLUTION :He has LEAST `SIGMA.`
27.

Collision cross-section is an area of an imaginary sphere of radius sigma around the molecule within which the centre of another molecule cannot penetrate. The volume swept by a single molecule in unit time is V=(pisigma^(2))overline(u)where overline(u) is the average speed If N^(**) is the number of molecules per unit volume, then the number of molecules within the volume V is N=VN^(**)=(pisigma^(2)overline(u))N^(**) Hence, the number of collision made by a single molecule in unit time will be Z=N=(pi sigma^(2)overline(u))N^(**) In order to account for the movements of all molecules, we must consider the average velocity along the line of centres of two coliding molecules instead of the average velocity of a single molecule . If it is assumed that, on an average, molecules collide while approaching each other perpendicularly, then the average velocity along their centres is sqrt(2)overline(u) as shown below. Number of collision made by a single molecule with other molecule per unit time is given by Z_(1)=pisigma^(2)(overline(u)_("rel"))N^(**)=sqrt(2) pisigma^(2)overline(u)N^(**) The total number of bimolecular collisions Z_(11) per unit volume per unit time is given by Z_(11)=(1)/(2)(Z_(1)N^(**))"or" Z_(11)=(1)/(2)(sqrt(2)pisigma^(2)overline(u)N^(**))N^(**)=(1)/(sqrt(2))pisigma^(2)overline(u)N^(**2) If the collsion involve two unlike molecules then the numberof collisions Z_(12) per unit volume per unit time is given as Z_(12)= pisigma _(12)^(2)(sqrt((8kT)/(pimu)))N_(1)N_(2) where N_(1) and N_(2) are the number of molecules per unit volume of the two types of molecules, sigma_(12) is the average diameter of the two molecules and mu is the reduced mass. The mean free path is the average distance travelled by a molecule between two successivecollisions. We can express it as follows : lambda=("Average distance travelled per unit time")/("NO. of collisions made by a single molecule per unit time")=(overline(u))/(Z_(1)) "or"lambda=(overline(u))/(sqrt(2)pisigma^(2)overline(u)N^(**))implies(1)/(sqrt(2)pisigma^(2)overline(u)N^(**)) For a given gas the mean free path at a particular pressure is :

Answer»

independent to TEMPERATURE
DECREASES with rise in temperature
increases with rise in temperature
directly proportional to `T^(2)`

Answer :C
28.

Colligative properties of a solution depend in ………………..present in it.

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SOLUTION :NUMBER of SOLUTE PARTICLES
29.

Colemanite + Na_(2)CO_(3)overset(" Fused")rarr (A)+(B)+CO_(2)underset("solution)((A))+CO_(2)rarr (B)+Na_(2)CO_(3)(B)+conc.HCl rarr NaCl+underset("Acid")((C )) underset("Acid")((C ))+H_(2)O rarr underset("Acid")((D))(D)overset("strong heating")rarr (E )(E )+CuSO_(4)overset("Heated")underset("in flame")rarr underset("Blue coloured compound")((F))Compound (D) is

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`H_(2)B_(4)O_(7)`
`HBO_(2)`
`H_(3)BO_(3)`
`H_(2)O`

Solution :`D-H_(3)BO_(3), E-B_(2)O_(3) , F-Cu(BO_(2))_(2)`
30.

Colemanite + Na_(2)CO_(3)overset(" Fused")rarr (A)+(B)+CO_(2)underset("solution)((A))+CO_(2)rarr (B)+Na_(2)CO_(3)(B)+conc.HCl rarr NaCl+underset("Acid")((C )) underset("Acid")((C ))+H_(2)O rarr underset("Acid")((D))(D)overset("strong heating")rarr (E )(E )+CuSO_(4)overset("Heated")underset("in flame")rarr underset("Blue coloured compound")((F))Compound (B) is

Answer»

`Na_(2)B_(4)O_(7)`
`NaBO_(2)`
`Na_(3)BO_(3)`
NaOH

Solution :`B-Na_(2)B_(4)O_(7), C-H_(2)B_(4)O_(7)`
31.

Colemanite + Na_(2)CO_(3)overset(" Fused")rarr (A)+(B)+CO_(2)underset("solution)((A))+CO_(2)rarr (B)+Na_(2)CO_(3)(B)+conc.HCl rarr NaCl+underset("Acid")((C )) underset("Acid")((C ))+H_(2)O rarr underset("Acid")((D))(D)overset("strong heating")rarr (E )(E )+CuSO_(4)overset("Heated")underset("in flame")rarr underset("Blue coloured compound")((F))Colemanite is

Answer»

`Ca_(2)B_(2)O_(11)`
`Ca_(2)B_(6)O_(11)`
`Ca_(4)B_(4)O_(11)`
`Ca_(6)B_(6)O_(11)`

SOLUTION :COLEMANITE `- Ca_(2)B_(6)O_(11) , A - NaBO_(2)`
32.

[Co(H_2O)_6]_((aq))^(3+) + 4Cl_((aq))^(-) hArr [CoCl_4]_((aq))^(2-) + 6H_2O this reaction is endothermic and if blue colour is due to CoCl_(4(aq))^(2-). If this mixture kept in ice than what happen ?

Answer»

Solution :This reaction is endothermic. If placed in ICE so temperature DECREASES than heat release and reaction proceed in REVERSE direction and blue colour of `CoCl_4^(2-)` decreases and PINK colour of `[Co(H_2O)_6]^(3+)`increases.
33.

[Co(H_2O)_6]_((aq))^(3+) + 4Cl_((aq))^(-) hArr [CoCl_4]_((aq))^(2-) + 6H_2O this reaction is endothermic and if blue colour is due to CoCl_(4(aq))^(2-). If this mixture kept in ice than What happened when reaction vessel ofis kept in 80^@C containing water ?

Answer»

Solution :Thus, temperature increase i.e., ENDOTHERMIC reaction So, reaction will be forward. The CONCENTRATION of BLUE colour product `[CoCl_4]^(2+)`is increases and blue colour BECOME dark.
34.

[Co(H_(2)O)_(6)]^(2+)(aq)("pink")+4Cl^(-)(aq)hArr[CoCl_(4)]^(2+)(aq)("blue")+6H_(2)O(l) In the above reaction at equilibrium, the reaction mixture is blue in colour at room temperature. On cooling this mixture, it becomes pink in colour On the basis of this information, which one the following is true ?

Answer»

`DeltaHgt0` for the FORWARD REACTION
`DeltaH=0` for the REVERSE reaction
`DELTAHLT0` for the forward reaction
SIGN of the `DeltaH` cannot be predicted bassed on this information.

Answer :A
35.

[Co(H_2O)_6]^(2+) (aq) " (pink) " + 4Cl^(-) (aq) hArr [COCl_4]^(2-) (aq) "(blue)"+6H_2O (l) In the above reaction at equilibrium,the reaction mixture is blue in colour at room temperature. On cooling this mixture,it becomes pink in colour. On the basis of this information, which one of the following is true ?

Answer»

`Delta gt 0` for the forward reaction
`Delta H = 0` for the reverse reaction
`Delta lt 0 ` for the forward reaction
Sign of the `Delta H` cannot be PREDICTED based on this information.

Solution :On COOLING, reverse reaction predominates and the solution is pink in colour.
`:.` Decrease in temperature, favours the reverse reaction i.e., reverse reaction is exothermic `(DELTAH LTO)` and for the forward reaction is endothermic `(DeltaH gt 0)`.
36.

CO_((g)) + 2H_(2(g)) hArr CH_3OH_((g)) for this reaction is K_c is 0.5. If the concentration of CO and H_2 at equilibrium 0.18M and 0.22 M respectively what is the concentration of CH_3OH ?

Answer»

SOLUTION :`4.56xx10^(-3)` M
37.

Coefficients of MnO_(4)^(-), C_(2)O_(4)^(2-) and H^(+) in the balanced reaction, MnO_(4)^(-)+C_(2)O_(4)^(2-)+H^(+) to Mn^(2+)+CO_(2)+H_(2)O

Answer»

2,3 and 8
2,5 and 8
2,3 and 16
2,5 and 16

Answer :D
38.

COD of a water sample is 8ppm the weight of acidified K_(2)Cr_(2) O_(7) required to oxidise the organie matter present in one litre of water sample is

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98 MG
49 mg
196 mg
98 g

Answer :B
39.

Cod Liver oil is

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FAT DISPERSED in water
Water dispersed in fat
Water dispersed in oil
Fat dispersed in fat.

Answer :C
40.

COD is a measure of

Answer»

ORGANIC substances in water
Oxides of S, P and N in air
Inorganic pollutants in water
SALINITY of water

Answer :A
41.

Coconut oil on alkaline hydrolysis gives

Answer»

Glycol
Alcohol
GLYCEROL
Glucose

Solution :Glycerol (by SAPONIFICATION)
42.

Coal tar is the main source of

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AROMATIC compounds
alicyclic compounds
aliphatic compounds
nitro compounds.

Answer :A
43.

Coal miners may suffer from …………. .

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TUBERCULOSIS
BLACK LUNG DISEASE
White lung disease
Malaria

Solution :Black lung disease
44.

Coal miners may suffer from...........

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SOLUTION :BLACK LUNG DISEASE
45.

Coal gas

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BURNS with a smoky flame
burns with non-smoky flame
is a GOOD fuel
is not USED for lighting purpose

Solution :Coal gas is a good fuel and burns with non-smoky flame
46.

CO_(3)O_(4) is an oxide of Ca (III) and Cu (II) .If its formula is CO_(x)(II)CO_(y)(II)O_(4) , then what is the value of x and y ?

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SOLUTION :The SUM of oxidaiton number of the elements in a compound is equal to ZERO
`CO_(x)(II)CO_(y)(II)O_(4)`,2x+3y -4xx2=0
or, 2x+3y=9
The only solution for this EQUATION is x=1 and y=2.
47.

CO_(2)andC_(2)H_(2) (acetylene) are non-polar.

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SOLUTION :Because both have LINEAR STRUCTURE.
48.

CO_2 is used for extinguishing fire because

Answer»

It has RELATIVELY high CRITICAL temperature
In SOLID state, it is called dry ICE
It is neither combustible nor a supporter of combustion
It is colourless GAS

Answer :C
49.

CO_(2) is not a poisnous gas but there is increase in concentration of CO_(2) in the atmosphere due to during of fossil fuels and decomposition of limestone. The increase in concentration of CO_(2) may lead to

Answer»

increase in photosynthesis in plants
higher concentration of `CO_(2)` in water
increase in greenhouse effect, thus raising the temperature
increase in FORMATION of metal carbonates

Solution :Increase in concentration of `CO_(2)` will RESULT in increase in greenjhouse effect thus raising the temperature which MIGHT have serious CONSEQUENCES.
50.

CO_(2) is linear whereas SO_(2) is bend-shaped. Give reason? Why does H_(2)O have bent structure?

Answer»

Solution :`CO_(2)` does not have LONE PAIRS. `SO_(2)` has two lone pairs