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.

If in a solvent, n simple molecules of solute combine of form an associated molecule, alpha is the degree of association, then Van't Hoff factor is equal to

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`(1)/(1- N alpha)`
`(1-alpha + n alpha)/(1)`
`(1-alpha+ alpha//n)/(1)`
`(alpha/n - 1+ alpha)/(1)`

Solution :`{:(" nX",HARR," X"_(n)),(1-alpha,,alpha//5):}`
Total moles after ASSOCIATION `=1-alpha+ alpha//n`
Van.t Hoff factor, `i=(1-alpha+alpha//n)/(1)`
2.

If in 3160 years a radioactive substance becomes one fourth of the original amount . Find its half life periods.

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SOLUTION :`t_(1//2)=1580` YEARS
3.

If in a certain element contain one electron in sub shell of 5d-orbital, then the element is..

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LA, GA and Lu
Tb, ND and Ho
Ce, PR and Sm
Tm, YB and Dy

Answer :A
4.

If I.E. of Na, Mg and Si are respectively 496, 737 and 786 kJ "mol"^(-)The I.E. of Al in KJ "mol"^(-) is :-

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575
760
390
1120

Solution :`underset(3S^(-1))(Na) lt underset(3s^(2))(Mg) gt underset(3s^(2)3p^(1))(Al) lt underset(3s^(2)3p^(2))(Si)`
Sequential ORDER`to underset(786)(Si ) gt underset(737) (Mg) gt underset (575)(Al) gt underset(496)(Na)`
5.

If I_(2) is dissolved in aqueous KI, the intense yellow species, I_(3)^(-) is formed. The structure of I_(3)^(-) ion is

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square PYRAMIDAL
trigonal bipyramidal
octahedral
PENTAGONAL bipyramidal

Solution :Trigonal bipyramid `(SP^(3)d)` with three equatiorial positions OCCUPIED by lone PAIRS.
6.

If (i)C+O_(2)rarrCO_(2), (ii) C+1//2O_(2)rarrCO, "(iii)"CO+1//2O_(2)rarrCO_(2), the heats of reaction are Q, -12,-10 respectively. Then Q=

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`-2`
2
`-22`
`-16`

Solution :`C+O_(2)rarrCO_(2),Delta=q`
`C+1//2O_(2)rarrCO,DeltaH=-12"....(i)"`
`CO+1//2O_(2)rarrCO_(2), DeltaH=-10"....(ii)"`
ADDING equation (i) and (ii) we can get
`DeltaH=-12+(-10)=-22`.
7.

If 'I' is the intensity of absorbed light and 'C' is the concentration of AB for the photochemical process AB+hvtoAB^(**), the rate of formation of AB^(**) is directly proportional to

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C
I
`I^(2)`
C.I

Answer :B
8.

If 'I' is the intensity of absorbed light and C is the concentration of AB for the photochemical process , AB + hv to AB^(**) . The rate of formation of AB^(**) is directly proportional to

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C
I
`I^(2)`
C.I.

Solution :The rate of photochemical process VARIES with the intensity of ABSORPTION since greater the intensity of absorbed LIGHT more photons will fall at a point , and further each photon causes one molecule to undergo REACTION.
9.

If hydrogen atoms (in the ground state) are passed through an homogeneous magnetic field, the beam is split into two parts.This interaction with the magnetic field shows that the atoms must have magnetic moment.However, the moment cannot be due to the orbital angular momentum since l=0. Hence one must assume existance of intrinsic angular momentum, which as the experiment shows, has only two permitted orientations. Spin of the electron produces angular momentum equal to s=sqrt(s(s+1))h/(2pi) where s=+1/2. Total spin of an atom =+n/2 or n/2 where n is the number of unpaired electron. The substance which contains species with unpaired electrons in their orbitals behave as paramagnetic substances.The paramagnetism is expected in terms of magnetic moment. The magnetic moment of an atom mu=sqrt(s(s+1))(eh)/(2pimc)=sqrt(n/2(n/2+1))(eh)/(2pimc)s=n/2 rArrmu_s=sqrt(n(n+2))B.M. n=number of unpaired electrons 1. B.M.(Bohr magneton)=(eh)/(2pimc) If magnetic moment is zero the substance is diamagnetic. Which of the following is a paramagnetic substance

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`MG^(2+)`
`CU^(+)`
`Mn^(+7)`
`TI^(+2)`

SOLUTION :There is two unpaired electron in `Ti^(2+)`
10.

If hydrolysis of ester is carried out inpresence of an acid , usig water containing redioactive oxygen atom, the product most likely to be radioactive is ,

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alcohol
ester
acid
both 'a' and 'c'

Solution :`RCOOR' + H_(2)O underset(DELTA) overset(conc. H_(2)SO_(4)) HARR RCOOH + R' OH`
Theradioactive OXYGEN must be in acid. This technique is known as TRACER technique.
11.

If hydrogen electrode dipped I 2 solution of pH=3 and pH=6 and salt bridge is connected the e.m.f. of resulting cell is

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0.177V
0.3V
0.052V
0.104V

Solution :`E_(CELL)=-0.059log((10^(-6))/(10^(-3)))=-0.059log10^(-3)`
`=0.059xx(-3)=0.177V`.
12.

If HgI_(2) is stoichiometrically dissolved in a KI solution to get K_(2)[HgI_(4)] complex the osmotic pressure is decreased by5x% . What is x ?

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ANSWER :5
13.

If heat of neutralisation is -13.7 k cal and H_f^(@)H_2O=-68 k cal, then enthalpy of OH^- would be:

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54.3 K CAL
`-54.3 k cal`
71.3 k cal
None

Answer :B
14.

If heavy water is taken as solvent of normal water while performing cannizzaro reaction, the products of the reaction are

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`RCOO^(-)+RCH_(2)OH`
`RCOO^(-)+RCH_(2)OD`
`RCOOD+RCD_(2)OD`
`RCOO^(-)+RCD_(2)OD`

Solution :
option (d) is correct.
15.

If helium and methane are allowed to difuse out of the container under the similar conditions of temperature and pressure, then the ratio of rate of diffusion of helium to methane is

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2
1
0.5
4

Answer :A
16.

If heat of hydrogenationof 1,3 - pentadiene and 1,4 - pentadiene are X and Y kcal respectively , heat of hydrogenation of 2,3 - pentadienewill be

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between X and Y
less thanX as well as Y
greater than X as well as Y
it can'tbe PREDICTED

Solution :Heat of HYDROGENATION is a measure of stability ,more stablean alkene, less is its heatof hydrogenation andvice- versa . Stability of alkadiene followsthe ORDER .
Conjugated diene `gt` Isolated diene ` gt` Cummulativediene (1,3 -pentadiene ) (1,4 - pentadiene) (2,3 - pentadiene )
17.

If heatof formation of C Cl_4 is 316 kcal mol^-1 the dissociation energy of C-Cl is :

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`79 kcal mol^-1`
`316 kcal mol^-1`
`97 kcal mol^-1`
`158 kcal mol^-1`

ANSWER :A
18.

If heat of combustion of methane is -800kJ, then heat of combustion of 4 xx 10^-4kg of methane is ________ .

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`-3.2 XX 10^4`KJ
`-800`kJ
`-280`kJ
`-20`kJ

Answer :D
19.

If He and CH_(4) are allowed to diffuse out of the container under similar conditions of temperature and pressure then the ratio of the rate of diffusion of He to CH_(4) is

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`2.0`
`1.0`
`0.5`
`4.0`

ANSWER :A
20.

If halfe-life of a substance is 5 yrs, then the total amount of substance left after 15 years, when initial amount is 64 grams is

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16 GRAMS
2 grams
32 grams
8 grams

Solution :`t_(1//2) = 5 yrs, t = 15 yrs`
`:. n = (t)/(t_(1//2)) = (15)/(5) = 3`
Now `N = (N_(0))/(2^(n)) = (N_(0))/(2^(3)) = (1)/(8) N_(0) = (1)/(8) XX 64 = 8` grams,
21.

If half lives of a radioactive element, undergoing parallel path alpha-"decay" and alpha-"decay" are 4 years and 12 years respectively, then percentage of element that remains after 12 years will be:

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0.5
0.125
0.0625
0.25

Answer :C
22.

If H_(2)SO_(4) ionises as H_(2)SO_(4) + 2H_(2)O to 2H_(3)O^(+) + SO_(4)^(-2). Then total number of ions produced by 0*1 M of H_(2)SO_(4) will be

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`9*03xx10^(21)`
`3*01xx10^(22)`
`6*02xx10^(22)`
`1*8xx10^(23)`

Solution :1 M produces IONS `= 3` moles
`:. 0*1` M `H_(2)SO_(4)` produces ions `=0*3` mol
No. of ions `=0*3xx6*02xx10^(23)=1*8xx10^(23)`
23.

If half-life of a certain radioactive nucleus is 1000s. the disintegration constant is

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`6.93 xx 10^(2) s^(-1)`
`6.93 xx 10^(-4)s`
`6.93 xx 10^(-4) s^(-1)`
`6.93 xx 10^(3) s`

SOLUTION :`K = (0.693)/(t_(1//2)) = (0.693)/(1000S) = 0.000693 = 6.93 xx 10^(-4) s^(-1)`
24.

If H_2SO_4 ionises as : H_2SO_4 + 2H_2O to 2H_3O^(+) + SO_4^(2) Then total number of ions produced by 0.3 M H_2SO_4 will be :

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`9.03 XX 10^21`
`3.01 xx 10^22`
`6.02 xx 10^22`
`5.40 xx 10^23`

ANSWER :D
25.

If H_2S is passed through an acidified K_2Cr_2O_7solution, the colour of the solution :

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Will remain unchanged
Will change to DEEP red
Will change to DARK GREEN
Will change to dark brown

Answer :C
26.

If H_(2) molecule is acting as ligand via oxidate addition : [Ir^(I)(CO)ClL_(2)] ("Vaska's complex")+H_(2) to [Ir^(III)(CO)Cl(H_(2))L_(2)],L=PPh_(3) then, choose the correct statement about the H_(2) molecule fate before and after reaction .

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`{:("Before","Trasition State","After"),(B.O=1,1 sigma"-DONOR,"sigma(**)-"ACCEPTOR",B.O.=0):}`
`{:("Before","Trasition State","After"),(B.O=1,1 sigma"-donor,"sigma(**)-"acceptor",B.O.=1):}`
`{:("Before","Trasition State","After"),(B.O=1,1 sigma"-donor,"pi(**)-"acceptor",B.O.=0):}`
`{:("Before","Trasition State","After"),(B.O=1,1 sigma^(**)"-donor,"sigma-"acceptor",B.O.=0):}`

Solution :
27.

If H_2(g)=2H(g), triangleH=104 cal, then heat of atomisation of hydrogen is:

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52 kcal
104 kcal
208 kcal
Nove of these

Answer :A
28.

If H_2(g)+Cl_2(g)→2HCl, ^@=-44 kcal 2Na(s)+2HCl(g→)2NaCl(s)+H_2(g), △H =-152 kcalthen, Na(s)+0.5Cl_2(g)→NaCl(s),△H =?

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108 KCAL
196 kcal
`-98 kcal`
54 kcal

Answer :B
29.

If {:(H_(2)+1//2O_(2)rarrH_(2)O",",,,,DeltaH= -68 kcal),(K+H_(2)OrarrKOH(aq)+1//2H_(2)",",,,,DeltaH= -48 kcal),(KOH+"water"rarrKOH(aq)",",,,,DeltaH= -14 kcal):} Find the heat of formation of KOH. Find the heat of formation of KOH.

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SOLUTION :N//A
30.

If H^++OHrarrH_2O+13.7 kcal then the heat of neutralisation for complete neutralisation of one mole of H_2SO_4 by a base will be:

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`13.7 KCAL`
`27.4 kcal`
`6.85 kcal`
3.425 kcal

Answer :B
31.

If H^(+)+OH^(-)rarrH_(2)O+13.7kcal, then the heat of neutralization for complete neutralization of one mole of H_(2)SO_(4) by base will be

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13.7 KCAL
27.4 kcal
6.85 kcal
3.425 kcal

Solution :If acid or BASE or both are STRONG, heat of neutralization = 13.7 kcal `xx` 2 = 27.4 kcal.
32.

If H is considered as the function of P and T, then which of thefollowing relations is/are corret?

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`DH=((dH)/(DT))_(P)dT=((dH)/(DP))_(T)dP`
`dH=C_(P)dT+((dH)/(dP))_(T)dP`
`((dH)/(dP))_(T)=-V`
All of the above

Answer :D
33.

If H^(+) ion concentration of a solution is increased by 10 times its pH will be

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Increase by ONE
Remains unchanged
Decrease by one
Increase by 10

Solution :When concentration of `[H^(+)]` INCREASED then the value of PH is DECREASES.
`pH = log.(1)/([H^(+)])`
34.

If group IV the ppt of Zn(OH)_(2) dissolve in excessof NaOHdue to theformation of

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`Na_(2)ZnO_(2)`
`NaZnO_(2)`
`Zn`
`Na_(3)ZnO_(2)`

ANSWER :a
35.

If graph log Kto(1)/(T) is plotted ,a straight line is obtained then what will be value of slope ?

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`(E_(a))/(2.303R)`
`(-E)/(3.203R)`
`(-2.303R)/(E_(a))`
`(-E_(a))/(2.303R)`

ANSWER :D
36.

If glycerene C_(3)H_(5)(OH), and methyl alcohol, CH_(3)OH are sold at the same price per kg, which would be cheaper for preparing an antifreeze solution for the radiator of an automobile ?

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ANSWER :`CH_(3)OH`
37.

If gold number of A, B, C and D are 0.005, 0.05, 0.5 and 5 respectively, then which of the following will have the greatest protective value?

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A
B
C
D

Answer :A
38.

If given are following standard electrodes potential, then find out standard potential of cell. [Zn^(+2)|Zn=-0.763V,Ag^(+)|Ag=+0.799V].

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0.562 V
1.562V
`2.560V`
`1.560V`

SOLUTION :`E_(CELL)^(@)=E_(CATHODE)^(@)-E_(ANODE)^(@)`
`=0.799-(-0.763)`
`=0.799+0.763=1.562V`
39.

If galvanic cell is formed by standard hydrogen electrode with by E_(Cu^(2+)|Cu)^(Theta)=+0.34Vin standard condition, so what is Cu^(2+)|Cu electrode ?

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ANODE electrode, negative electrode
CATHODE electrode, positive electrode
None of above
Anode positive electrode

Solution :(a) `E_(cell)^(Theta)=E_("CATHOD")^(Theta)-E_("Anode")^(Theta)`
`=E_(H^(+)|H_(2))^(Theta)-E_(Cu^(2+)|Cu)^(Theta)`
`=0.0-(0.34)=0.34V(A)` is wrong
(b) `E_(cell)^(Theta)=E_("Cathod")^(Theta)-E_("Anode")^(Theta)`
`=E_(Cu^(2+)|Cu)^(Theta)-E_(H^(+)|H_(2))^(Theta)`
`=0.34-0.0=0.34V` So, (B) is true.
MEANS, `Cu^(2+)|Cu` Cathode (Positive) electrode and `H^(+)|H_(2)` anode.
40.

If formic acid, acetic acid, propanoic acid and benzoic acid is mixed with phosphorus and bromine then how many product are formed

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Solution :Formic acid and BENZOIC acid do not CONTAIN `alpha-H` atoms so they cannot give HVZ reaction `CH_(3)- COOH overset(P+Br_(2))to underset(Br)underset(|)(C)H_(2)-COOH`
`CH_(3)-CH_(2)-COOH overset(P+Br_(2))to underset((d+l))(CH_(3)-underset(Br)underset(|)overset(**)(C)H-COOH`.
41.

If freezing point of 5% w/w aqueous solution of sucrose has 271 K and pure water has freezing point of 273.15 K then calculate freezing point of 5% w/w aqueous solution of glucose.

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ANSWER :269.07 K
42.

If formic acid , acetic acid, propanic acid and benzoic acid is mixed with Phosphorus and Bromine then how many product are formed.

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Solution :FORMIC ACID and BENZOIC acid do not CONTAIN `alpha`-H atoms so they cannot give HVZ reaction
43.

If formaldehyde and potassium hydroxide are heated, then we get

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ACETYLENE.
methane
methyl ALCOHOL
ETHYL formate.

ANSWER :C
44.

If formaldehyde and KOHare heated then we get

Answer»

METHANE
acetylene
ETHYL formate
methyl alcohol

SOLUTION :We KNOW that,
`2HCHO + KOH OVERSET(Delta) to HCOOK + CH_(3)OH`.
Thus in this reaction, methyl alcohol `(CH_(3)OH)` is produced . This reaction is known as Cannizaro's reaction .
45.

If for the raction at 300 K, 2Mg(g)+O_(2)(g)to2MgO(s), Delta_(r)H=-1202kJ"mol"^(-1) and Delta_(r)S=-217.0JK^(-1)"mol"^(-1) The total entropy change (DeltaS)_(T) and gibbs energy change during the course of reaction (Delta_(r)G) are respectively.

Answer»

`3.79xx10^(3)JK^(-1)"mol"^(-1),-1136.9kJ`
`3.79xx10^(3)Jk^(-1)"mol"^(-1)+1000J`
`+1000JK^(-1)"mol"^(-1),3.79xx10^(3)J`
`-1136.9kJ, 3.79xx10^(3)J`

Solution :`DeltaG=DeltaH-TDeltaS`
`=-1202-(-217xx10^(-3)xx300)=-1136.9kJ`
Heat released will be ABSORBED by the SURROUNDING to increase the entropy of surroundings.
`DeltaS_("surr")=+(1202xx10^(3))/300=+4.01xx10^(3)JK^(-1)"mol"^(-1)`
`DeltaS_("TOTAL")=-217+4.01xx10^(3)=+3793JK^(-1)"mol"^(-1)`
46.

If for two gases of molecular weights M_(A) and M_(B) at temperature T_(A) and T_(B), T_(A)M_(B)=T_(B)M_(A), then which property has the same magnitude for both the gases.

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Density
Pressure
KE PER mol
`V_(rms)`

Solution :(i) density of a gas `(RHO) =(PM)/(RT)`
Since `(M_(B))/(T_(B))=(M_(A))/(T_(A)) therefore` at the same pressure `rho_(A) =rho_(B)`
But f pressure is DIFFERENT then `rho_(A) NE rho_(B)`
(ii) Pressure of the gases would be equal if their densities are equal other wise not.
KE per mol `=(3)/(2)RT`
`therefore` It will be different for the two gases.
(iii) `V_(rms) =sqrt((3RT)/(M))," since "(T_(A))/(M_(A))=(T_(B))/(M_(B)), V_(rms)" of "A=V_(rms)" of "B`
Hence, (D) is the correct answer.
47.

If for the half reactions Cu^(2+)+e^- to Cu^+""E^0=0.15V Cu^(2+)+2e^- to Cu""E^0=0.34V Calculate E^0 of the half cell reaction Cu^+ +e^- to Cu also predict whether Cu^+ undergoes also predict whether Cu^+ undergoes disproportionation or not

Answer»


ANSWER :A::C::D
48.

If for two gases of molecular weight M_Aand M_B at temperature T_A and T_B,T_AM_B= T_BM_A , then which property has the same magnitude for both the gases :

Answer»

Density
Pressure
KE PER mole
rms speed

Answer :D
49.

If for H_(2(g))+1/2S_(2(s))hArrH_(2)S_((g))and H_(2(g))+Br_(2(g))hArr2HBr_((g))The equlibrium constants are K_(1) and K_(2) respectively, the reaction Br_(2(g))+H_(2)S_((g))hArr2HBr_((g))+1/2S_(2(g)) would have equlibrium constnt

Answer»

`K_(1)xxK_(2)`
`K_(1)//K_(2)`
`K_(2)//K_(1)`
`K_(2)^(2)//K_(1)`

SOLUTION :`K_(1)=([H_(2)S])/([H_(2)][S_(2)]^(1//2))K_(2)=([HBr]^(2))/([H_(2)][Br_(2)])`
`K_(3)=([HBr]^(2)XX[S_(2)]^(1//2))/([Br_(2)]xx[H_(2)S]),(K_(2))/(K_(1))=K_(3)`
50.

If H_(2)(g)=2H(g),DeltaH=104cal , then heat of atomisation of hydrogen is

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`52 KCAL`
`104 kcal`
`208 kcal`
NONE of these

Answer :A