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.

KIO_(3)+5KI+6HCl to 3I_(2)+6KCl+3H_(2)O Correct statement of this reaction is

Answer»

`I^(-)` is REDUCED to `I_(2)`
`IO_(3)^(-)` is oxidised to `I_(2)`
`IO_(3)^(-)` is reduced to `I_(2)`
OXIDATION number of I decreases from 01 (in KI) to zero `("in "I_(2))`

Answer :C
Discussion
2.

Kinetic theory of gases is a generalization offered by Maxwell. Boltzmann, Clausius, etc. to explain the behaviour of ideal gases. This theory assumes that ideal gas molecules neither attract nor repel each other. Average kinetic energy of gasmolecules is directly proportional to the absolute temperature. A gas equation called kinetic gas equation was derived on the basis of kinetic theory PV = (1)/(3) mv^(2) {:("Gas","Density",),(A,0.82 g L^(-1),),(B,0.26 g L^(-1),),(C,0.51 g L^(-1),):} Pick up the correct statement/statements 1. Gas A will tend to lie at the bottom 2. The number of atoms fo various gases A, B and C are same. 3. The gases will diffuse to form homogeneous mixture. 4. Average kinetic energy of each gas is same

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2, 4
1, 4
1
3, 4

Answer :D
Discussion
3.

Kinetic theory of gases proves:

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only BOYLE's law
only CHARLE's law
only Avogardo's law
all of these

ANSWER :d
Discussion
4.

Kinetic energy of photoelectrons is indepen dent on --------- of incident radiation.

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WAVELENGTH 
WAVE number
Frequency 
Intensity

Answer :D
Discussion
5.

Kinetic energy of molecules is least in

Answer»

Gases
Solids
Liquids
Solutions

ANSWER :B
Discussion
6.

Kinetic energy of electron in a mono electronic species is +1313 lJ/mole. Then which of the following statements are correct (I) The electron is present in the 2^(nd) orbit of He^(+) ions (II) The electron is present in the 2^(nd) orbit of H atom (III) The electron is present in the 3^(rd) orbit of Li^(+2) ion (IV) The electron is present in the 4^(th) orbit of He^(+) ion

Answer»

I and II
II and III
I and III
III and IV

Solution :K.E. = `+ (Z^2)/(n^2) (13.6 eV)`/ATOM,
K.E. `= + (Z^2)/(n^2) (1312 KJ)`/MOLE
Discussion
7.

Kinetic energy of 0.30 moles of He gas in a container of maximum capacity of 4 litres at 5 atmosphere, must be (R = 0.0821 atm litre "mole"^(-1) K^(-1) )________xx 10^(1)atm.lit

Answer»


SOLUTION :`kE = 3/2 PV = 3/L (4)(5)= 30 ` LIT atm.
Discussion
8.

KI combines with I_(2) and forms polyiodide . The number of hybrid orbitals on the central iodine atom is "_____" .

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<P>

ANSWER :A-p,R,s , B - p,Q,r , C-p,r , D-p,q,s
Discussion
9.

Kharasch effect operates in Which of the following?

Answer»

`CH_3CH_2CH = CH_2 + HCl`
`CH_3CH_2 - CH = CH_2 + HBr`
`CH_3CH= CH- CH_3 + HBr`
`CH_3 CH_2 CH = CH_2 + HI`

Solution :KHARASH effect operates in ADDITION of HBr to ASYMMETRIC alkenes
`CH_2 = CH - CH_2 - CH_3 to CH_3CH_2 - CH_2 - CH_2Br`
Discussion
10.

KF has NaCl structure. If the distance between K^+ and F^- is 269 pm, find the density of KF (N_A=6.02 xx 10^23 mol^(-1), atomic masses K=39, F = 19 amu)

Answer»


ANSWER :`2.48 " G CM"^(-3)`
Discussion
11.

KF has ccp structure. Calculate the radius of unit cell if the side of the cube or edge length is 400 pm. How many F^- ions and octahedral voids are there in this unitcell ?

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Solution :ccp is same as fcc, Hence, `r=a/(2sqrt2)`=0.3535 a=0.3535 X 400 nm=141.4 PM
In fcc structure , there are `4F^-` ions in the packing and hence 4 OCTAHEDRAL voids.
Discussion
12.

KFhas ccp structure. Calculate the radius of unit cell if the side of the cube or edge length is 400 pm.How manyFe^(-)ions and octahedral voids are there in the unitcell ?

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Solution :ccpis same as fcc, hence,`r= a/ (2sqrt2) = 0.3535 a = 0.3535 xx 400 "pm" = 141.4 ` pm.
In fcc, struture, THER are ` 4 F^(-)`ions in the PACKING and hence 4 octahedral VOIDS.
Discussion
13.

KF combines with HF to form KHF_(2) . The compound contains the species

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` K_(+).F^(-) and H^(+)`
`K^(+), F^(-) and HF`
`K^(+) and [HF_(2)]^(-)`
`[KHF]^(+) and F_(2)` .

ANSWER :C
Discussion
14.

KF combines with HF to form KHF_(2). The compound contains the species

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`K^(+) ,F^(-)and H^(+)`
`K^(+) ,F^(-)and HF `
`K^(+) and[HF_(2)]^(-1)`
`[KHF]^(+) and F^(-)`

Answer :C
Discussion
15.

KF combines with HF to form KHF_(2) The compound contains the species .

Answer»

`K^(o+),F^(Θ)` and `H^(o+)`
`K^(o+),F^(Θ)` and `H^(o+)`
`K^(o+)` and `[HF_(2)]^(Θ)`
`[KHF]^(o+)` and `F_(2)`

Solution :`KHF_(2) RARR K^(o+) + HF_(2)^(Θ)`
`HF_(2)^(Θ)` ion is STABILISED by H-bonding .
Discussion
16.

Ketone is formed by the reaction

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(I)(i)(Q)
(IV)(III)(R)
(III)(iv)(Q)
(II)(iii)(P)

ANSWER :A
Discussion
17.

keto-enol tautomerism is observed in

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SOLUTION :
Discussion
18.

Keto and enol forms of a compound are related to each other as

Answer»

RESONANCE structures
Conformations
Configurational isomers
Constitutional isomers

Answer :D
Discussion
19.

Kerosene is a mixture of

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AROMATIC hydrocarbons
aliphatic hydrocarbons
unsaturated hydrocarbons
saturated hydrocarbons.

Answer :B
Discussion
20.

K.E. of one mole of helium at 273 K is:

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819 cal
81.9cal
8.19cal
none of these

ANSWER :a
Discussion
21.

Keiserite is an ore of ...... and has the composition.

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SOLUTION :MAGNESIUM, `MgSO_4 . H_2O`
Discussion
22.

KCl in water deviates from ideal behaviour - why?

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Solution :A solution of potassium chloride in water DEVIATES from ideal behavior because the solute dissociates to give `K^(+)` and `Cl^(-)` ion which FORM strong ion-dipole interaction with water MOLECULES.
Discussion
23.

KCl crystallizes in the same type of lattice as does NaCl. Given that r_(Na+)//r_(Cl-)=0.55 and r_(K+)//r_(Cl-)=0.74. Calculate the ratio of the side of the unit cell of KCl to that of NaCl.

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1.123
0.891
1.414
0.414

Solution :Our aimis `(r_(K^+)+r_(Cl^-))//(r_(Na^+)+r_(Cl^-))`
`"Given"(r_(Na^+))/(r_(Cl^-))=0.55`
`and(r_(K^+))/(r_(Cl^-))=0.74`
`therefore(r_(Na^+))/(r_(Cl^-))+1=1.55`
and`(r_(K^+))/(r_(Cl^-))+1=1.74`
`i.e.,(r_(Na^+)+r_(Cl^-))/(r_(Cl^-))=1.55""....(i)`
and`(r_(K^+)+r_(Cl^-))/(r_(Cl^-))=1.74"".......(II)`
`"Dividing"(ii)"by"(i)`,
`(r_(K^+)+r_(Cl^+))/(r_(Na^+)+r_(Cl^-))=(1.74)/(1.55)=1.125`
Discussion
24.

KCl in water deviates from ideal behaviour -why ?

Answer»

SOLUTION :A solution of POTASSIUM chloride in water deviates from ideal behaviour because the SOLUTE DISSOCIATES to GIVE ` K^(+)andCl^(-) ` ion which form strong ion-dipole interaction with water molecules.
Discussion
25.

KCl crystallises in the same type of lattice as does NaCl. Given that r_(Na^+)//r_(Cl^-)=0.55 and r_(K^+)// r_(Cl^-)=0.74 . Calculate the ratio of the side of the unit cell of KCl to that of NaCl

Answer»

1.123
0.891
1.414
0.414

Solution :We AIM at :`(r_(K^+)+r_(Cl^-))//(r_(Na^+)+r_(Cl^-))`
Given `r_(Na^+)/r_(Cl^-)=0.55`and `r_(K^+)/r_(Cl^-)`=0.74
`r_(Na^+)/r_(Cl^-)+1=1.55` and `r_(K^+)/r_(Cl^-)+1=1.74`
i.e., `(r_(Na^+)+r_(Cl^-))/(r_(Cl^(-)))=1.55`
`(r_(K^+)+r_(Cl^-))/(r_(Cl^(-)))=1.74`
Dividing (II) by (i) , `(r_(K^+)+r_(Cl^-))/(r_(NA^+)+r_(Cl^-))=1.74/1.55`=1.123
Discussion
26.

KCl contains 52% of potassium and KI contains 23.6% potassium by weight. What is the weight percentage of iodine in iodine monochloride?

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ANSWER :78.2
Discussion
27.

KCl cannot be used as a salt bridge for cell Cu(s)|CuSO_(4)(aq)| |AgNO_(3)(aq)| Ag(s) because

Answer»

`CuCl_(2)` gets precipited
`Cl_(2)` GAS is evolved
AgCl GEST precipitated
None of the above

Solution :`CuCl_(2)` gets precipitated.
Discussion
28.

KCI in water deviates fromideal behaviour-why?

Answer»


Answer :A solution of potassium CHLORIDE in water deviates from IDEAL behavior because the solute dissociates to give `K^(+) and CL^(-)` ION which form strong ion-dipole interaction with water MOLECULES.
Discussion
29.

K_(C) value of a gaseous reaction is 5mole/lit. If pressure is increased

Answer»

Forward reaction is favoured
Backward reaction is favoured
Reaction is uneffected
`K_c` VALUE increases

Solution :Units of `K_(C) = ("mole" // "LIT") implies Delta n_(g) =1`
`:.` if pressure is increased backward reaction is favoured
Discussion
30.

K_C/K_P for the reaction,N_2(g) + 3H_2(g) hArr 2NH_3(g) is ……………………. .

Answer»

`1/(RT)`
`SQRT(RT)`
`RT`
`(RT)^2`

SOLUTION :`N_2(G) + 3H_2(g) hArr2NH_3(g), Deltan_g = 2-4 = -2`
`:. K_p = K_C(RT)^(-2)`
`rArr K_C/K_P = (RT)^2`
Discussion
31.

(K_(c))/(K_(p)) for the reaction, N_(2)(g)+3H_(2)(g)hArr2NH_(3(g)) is

Answer»

`(1)/(RT)`
`sqrt(RT)`
RT
`(RT)^(2)`

SOLUTION :`K_(p)` and `K_(C)` is the relationship between `K_(p)` is EQUAL to `K_(c)`. (RT) `Deltan_(g)`.
Discussion
32.

K_C for the reaction 2X leftrightarrow Y+Z is 2 times 10^-3 at a given time the composition of reaction mixture [X]=[Y]=[Z]=2.8 times 10^-4 M. IN what direction, the reaction will proceed?

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SOLUTION :BACKWARD
Discussion
33.

K_c is 6.3 times 10^14 for the reaction NO+O_3 leftrightarrow NO_2+O_2 at 1000K, calculate K_c for the reverse reaction.

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SOLUTION :`1.587 TIMES 10^-15`
Discussion
34.

The equilibrium constant K_(c) for the SO_(2(g)) + NO_(2(g)) hArr SO_(3(g)) + NO_((g)) reaction is 16. if 1 mole of eachof all the four gases is taken in ldm^(3)vessel, the equilibrium concentration of NO would be

Answer»

SOLUTION :`1.6 MOL L^-1`
Discussion
35.

K_c for the reaction H_2+I_2 leftrightarrow 2HI at 500^@C is 45.9. calculate the equilibrium concentration of HI.

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SOLUTION :`4.4 MOL L^-1`
Discussion
36.

K_c for the reaction Ag(CN)_2^(-) leftrightarrow Ag^(+) +2CN^(-) at 298K is 4 times 10^-19 M^2. Starting with 0.1M KCN and 0.03 AgNO_3. Calculate the equilibrium concentration of Ag^+

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SOLUTION :`7.5 TIMES 10^-18 M`
Discussion
37.

K_(c) "for" CO(g) + H_(2) O (g) hArr CO_(2) (g) + H_(2) (g) " at " 986 ^(@)C " is " 0*63 . " A mixture of 1 mole of" H_(2)O (g) and" 3 moles " CO (g) " is allowed to react to an equilibrium . The equilibrium pressure is " 2*0atm .

Answer»

Solution :`{:(,CO(G),+,H_(2)O,hArr,CO_(2)(g),+,H_(2)(g)),("Intial moles",3,,1,,0,,0),(" Moles at eqm.",3-x,,1-x,,x,,x):}`
Total no. of moles at EQUILIBRIUM `= 3-x x+1 - x + x + x = 4 `
` K_(c) = (x xxx)/((3-x)(1-x)) , i.e.,0*63= (x^(2))/(3 + x^(2) - 4x ) `
On solving, it gives `x = 0*681 "" ( x =-(bpm sqrt(b^(2) - 4ac))/(2A) )`
` :." Moles of " H_(2)" present at eqm . " = 0* 681 "mole" `
Total PRESSUREAT eqm. = 2 atm ( Given )
Totalmolesat eqm . = 4
` :. p_(CO)= ( 3- 0*681 )/4 xx 2 = 1*16"atm " , p_(H_(2)O) = (1-0*681)/4 xx 2 = 1*16 " atm " , p_(CO_(2)) = p_(H_(2)) = (0* 681)/4 xx 2 = 0* 34 "atm" `
Discussion
38.

Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium : 2BrCl(g)hArrBr_(2)(g)+Cl_(2)(g) for which K_(c)=32 at 500K. If initially pure BrCl is present at a concentration of 3.3xx10^(-3)molL^(-1), what is its molar concentration in the mixture at equilibrium?

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SOLUTION :`3 TIMES 10^-4 MOL L^-1`
Discussion
39.

K_C for N_2+O_2 harr 2NO is 'X', then for NO harr 1//2N_2 +1//2 O_2, it is

Answer»

`X^2`
`sqrt(X)`
`1/(sqrt(X))`
`1/X^2`

Solution :`K_(C)=(1)/(sqrt(x))`
Discussion
40.

K_(c)" for "CS_(2)(g)+4H_(2)(g)hArrCH_(4)(g)+2H_(2)S(g) is 0.28 at 900K. Calculate K_(p). (R = 8.314 JK^(-1)mol^(-1)).

Answer»

SOLUTION :`CS_(2)(g)+4H_(2)(g)hArrCH_(4)(g)+2H_(2)S(g)`
`Deltan=3-5=-2`
`K_(p)=K_(c)(RT)^(Deltan)" becomes "K_(p)=K_(c)(RT)^(-2)=(0.28)/((8.314xx900)^(2))=5XX10^(-9)`.
Discussion
41.

K_c for N_2+O_2 harr 2NO at certain temprature is 1.6 xx 10^(-3), then K_p for NO harr 1//2N_2+1//2O_2 at same temperature will be

Answer»

25
25 atm
5 atm
`1.6 xx 10^(-3) "atm"^(-1)`

Solution :`KP=KC(RT)^(Delta n_(G)), Delta n_(g)=0, K_(P)=1.6 xx 10^(-3)` for `NO harr (1)/(2) N_(2)+(1)/(2)O_(2)`
`K_(P)^(1)=(1)/(sqrt(K_(P)))=(1)/(sqrt(1.6 xx 10^(-3)))=25`
Discussion
42.

K_(C) for 2 SO_(2) + O_(2) hArr 2 SO_(3) in a 10 lit flask at certain T is 100 lit -mol^(-1) . Now , if equilibrium pressures of SO_2 and SO_3 are equal , then mass of O_(2) present at equilibrium is

Answer»

`6.4` G
12.8 g
3.2 g
16 g

ANSWER :C
Discussion
43.

K_c and K_p for heterogeneous equilibrium: Write the equilibrium constant expressions K_c and K_p, if applicable, for each of the following heterogeneous system: (i) S(s)+H_2SO_3(aq.)hArr H_2SO_3(aq.) (ii) 2NH_3(g)+H_2SO_4(l)hArr (NH_4)_2SO_4(s) (iii) P_4(s)+6Cl_2(g) hArr 4PCl_3(l) (iv) AgCl(s)hArr Ag^(+)(aq.)+Cl(aq.) (v) (NH_4)Se(s) hArr 2NH_3(g)+H_2Se(g)

Answer»

Solution :Strategy: Apply the definitions of `K_c` and `K_p` to each reaction. Remember that in HETEROGENEOUS equilibrium, pure solids or pure LIQUIDS MUST be present (HOWEVER, small it may be) for the equilibrium to exist, but their active masses do not appear in the expression of the equilibrium constant because they are in incoroprated into the value of equilibrium constant.
Solution:
(i) `K_c=(C_(H_2S_2O_3))/(C_(H_2SO_3))`, `K_p` cannot be defined as no gases are involved
(ii) `K_c=(1)/(C_(NH_3))`, `K_p=(1)/(P_(NH_3))`
(iii) `K_c=(1)/(C_(Cl_2)^6)`, `K_p=(1)/(P_(Cl_2)^6)`
(iv) `K_c=C_(Ag^(+))C_(C1^(-))`, `K_p` undefined, no gases involved
(V) `K_c=C_(NH_3)^2C_(H_2Se)`, `K_p=P_(NH_3)^1P_(H_2)Se`
Discussion
44.

KBr (potassium bromide) contains 32.9% by weight of potassium. If 6.40 g of bromine react with 3.60 g of potassium, calculate the number of moles of potassium which combine with bromine to form KBr.

Answer»

Solution :The AMOUNT of potassium in KBr = 32.9%
This MEANS that 67.1 parts of bromine combine with 32.9 parts of potassium to form KBr.
`therefore`The amount of potassium that reacts with 6.40 g of bromine `=(32.9)/(67.1) XX 6.40 = 3.14 g`
`therefore` The number of moles of potassium PRESENT in this amount = `3.14/39.10 = 8.03 xx 10^(-2)`
Hence, `8.03 xx 10^(-2)`moles of potassium combine to form KBr.
Discussion
45.

KBr has fcc structure. The density of KBr is 2.75 "g cm"^(-3). Find the distance between K^+ and Br^-. (At mass of Br = 80.0)

Answer»


ANSWER :330 PM
Discussion
46.

K_b of a monoacidic base BOH is 2 xx 10^(-5). When BOH is neutralised with HCI, what is the equilibrium constant?

Answer»

SOLUTION :`5 XX 10^(10)`
Discussion
47.

K_(a_(1)), K_(a_(2)) and K_(a_(3)) are the respective ionisation constants for the following reactions. H_(2)S hArr H^(+) + HS^(-) HS^(-) hArr H^(+)+S^(2-) H_(2)S hArr 2H^(+)+S^(2-) The correct relationship between K_(a_(1)) , K_(a_(2)) and K_(a_(3))is

Answer»

`K_(a_(3))=K_(a_(1))xxK_(a_(2))`
`K_(a_(3))=K_(a_(1)) + K_(a_(2))`
`K_(a_(3))=K_(a_(1)) - K_(a_(2))`
`K_(a_(3))=K_(a_(1)) // K_(a_(2))`

Solution :`K_(a_(1))=([H^(+)][HS^(-)])/([H_(2)S]),K_(a_(2))=([H^(+)][S^(2-)])/([HS^(-)]), K_(a_(3))=([H^(+)]^(2)[S^(2-)])/([H_(2)S])`
Obviously, `K_(a_(3))=K_(a_(1))xxK_(a_(2))`.
Discussion
48.

K_a=1.4xx10^(-5) of propanoic acid. Calculate its pH of 0.1 M solution.

Answer»

SOLUTION :`[H_3O^+]=1.2xx10^(-3)` , pH=2.92
Discussion
49.

K_(a_1), K_(a_2) and K_(a_3), are the respective ionisation constants for the following reactions. H_2S hArr H^(+) + HS^(-) HS^(-) hArr H^(+) + S^(2-) H_2S hArr 2H^(+) + S^(2-) The correct relationship between K_(a_1),K_(a_2),K_(a_3) is

Answer»

`K_(a_3) = K_(a_1) xx K_(a_2)`
`K_(a_3) = K_(a_1) + K_(a_2)`
`K_(a_3)=K_(a_1) - K_(a_2)`
`K_(a_3)=K_(a_1) // K_(a_2)`

Solution :For the reaction, `H_2S HARR H^(+)+ HS^(-)`
`K_(a_1)=([H^+][HS^-])/([H_2S])`
For this reaction, `HS^(-) hArr H^(+) + S^(2-)`
`K_(a_2) = ([H^+][HS^(2-)])/([HS^-])`
As, the above two reactions are added, their EQUILIBRIUM constants are MULTIPLIED, As
`K_(a_3)=([H^+]^2[S^(2-)])/([H_2S]) K_(a_1)xxK_(a_2)`
So `K_(a_3)= K_(a_3) xx K_(a_2)`
Discussion
50.

The Ksp values of MA , MB , MC , MD are 1.8 xx 10^(-10) , 4 xx 10^(-3) , 4 xx 10^(-8) & 6 xx 10^(-5) respectively . If a 0.01 M solution of MX is added dropwise to a mixture containing A , B , C , D solution the one to be precipitated first will be :

Answer»

`(PH)_(N_BA) LT (pH) _(NaB) `
`(pH) _(NAD) lt(pH) _(NaB) `
` (pH)_(NaA) lt (pH) _(NaD) `
`(pH) _(NaB) =7`

SOLUTION :` pH= 7+ ( (pKa+log c)/( 2)) , pH uparrow if pKa uparrow `
Discussion