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 the solubility of an aqueous solution of Mg(OH)_(2), be "X^(@) then its K_(sp) is

Answer»

`4x^(3) `
` 108x^(5) `
` 27x^(4)`
` 9x`

Solution :` MG (OH) _2 hArr Mg^(+ + )+2 underset( 2x) OH^(-) `
` K_(sp)= (S) _x (2S)_(2x) ^(2)=4x^(3) `
Discussion
2.

Let the solubilities of AgCl in H_2 O , 0.01 M CaCl_(2) , 0.01 M NaCl and 0.05 M Ag NO_(3) be S_(1) , S_(2) , S_(3) and S_(4) respectively . What is the correct relationship between these quantities ?

Answer»

`S_(1) gt S_(2) gt S_(3) gt S_(4)`
`S_(1) gt S_(2) = S_(3) gt S_(4)`
`S_(1) gt S_(3) gt S_(2) gt S_(4)`
`S_(4) gt S_(2) gt S_(3) gt S_(1)`

ANSWER :C
Discussion
3.

Let the I.E. of hydrogen like species be 320 eV . Find out the value of quantum number having the energy equal to -20 eV .

Answer»

N = 2
n = 3
n= 4
n = 5

Answer :C
Discussion
4.

Let the colour of the indicator (Hin colourles) will be visible only when its ionised form (pink) is 25 or more in a solution. Suppose Hln (pK_(In) = 9.0) is added to a solution of pH = 9.6. Predict what will happen. (Take log 2 = 0.3)

Answer»

Pink colour will be visible
Pink colour will not be visible
% of ionised form will be less than 25%
% of ionised form will be more than 25%

Solution :` pH =PKA +LOG "" ([In^(-) ])/( [HIn]) , `
` 9. 6 = 9+ log ""([In^(-) ])/( [HIn ]), 0. 6 = log ""([ In^(-)])/( [HIn ]), ([In^(-) ])/( [H In ]) = 4 `
` rArr([In^(-) ])/([In ^(-)]+[HIn ])XX 100 = (4)/(3)xx 100 = 80%`
Discussion
5.

Let r, v and E are the radius of the orbit, speed of the electron and the total energy of the electron respectively. Which of the following quantities are proportional to the quantum number .n. ?

Answer»

`rE`
`ur`
`upsilon/E`
`R/E`

Solution :`upsilon = (2.188xx10^(-8))/(n)` CM/sec
`r= n^2 XX 0.529 xx 10^(-8) cm , E=-(2.18 xx 10^(-18))/(n^2)`
these relations shown that `upsilonr and (upsilon)/(E)` will be proportional to be principal quantum number n.
Discussion
6.

Let Mg TiO_(3) exists in pervoskite structure. In this lattice, all the atoms of one of the face diagonals are removed. Calculate the denstiy of unit cell if the radius of Mg^(2+) is 0.7 Å and the corner ions are touching each other. [Given atomic mass of Mg = 24, Ti = 48]

Answer»


Solution :No. of `Mg^(2+)` PER unit cell = 8 [At corners] `xx (1)/(8) = 1`
No. of Ti per unit cell = 1 [body center ] `xx (1)/(1) = 1`
No of O per unit cell = 6 [Face center] `xx (1)/(2) = 3`
So FORMULA `= MgTiO_(3)`
Atom are removed along face diagonal
No. of `Mg^(2+) = 6`[At CORNER] `xx (1)/(8) = (6)/(8) = (3)/(4)`
No. of Ti per unit cell = 1 [Body center] `xx (1)/(1) = 1`
No. of O per unit cell = 5[Face center] `xx (1)/(2) = (5)/(2)`
So formula of compound `= Mg_((3)/(4)) TiO_((5)/(2))`
Formular MASS `= 24 xx (3)/(4) + 48 + 16 xx (5)/(2) = 18 + 48 + 40 = 106` amu
As corner ion are touching so `= a = 2 r_(Mg^(2+)) = 2 xx 0.7 = 1.4 Å`
`d = ("mass")/("Volume") = (106 xx 1.76 xx 10^(-24))/((1.4)^(3) xx 10^(-24)) g//cm^(3) = 64.5 g//cm^(3) ~~ 65 g//cm^(3)`
Discussion
7.

Let A_(n) be the area enclosed by n^(th) orbit in the H-atoms. The graph ln ((A_(n))/(A_(l))) against ln(n)

Answer»

will PASS through origin
will be STRAIGHTLINE with slope = 6
will be monotonically increasing non-linear CURVE
will be a CIRCLE

Answer :A
Discussion
8.

Lemon juice has a pH = 2.1 . If all the acid in lemon is citric acid (H "Cit." hArr H^(+)+"Cit"^(-1)) and K_(a) for citric acid is 8.4xx10^(-4) mole/litre, what is the concentration of citric acid in lemon juice ?

Answer»


SOLUTION : `pH = 2.1 , i.e., - log [H^(+)]=2.1 or log [H^(+)]= - 2.1 = bar(3). 9 or [H^(+)] = 7.943xx10^(-3)`
`"HCit"HARR H^(+) + "Cit"^(-1)`
`K_(a) = ([H^(+)] [Cit^(-1)])/(["H Cit"])`
`8.4xx10^(-4)=(7.943xx10^(-3))(7.943xx10^(-3))/(["H Cit"]) or ["H Cit" ] = 7.5 xx 10^(-2)M`
Discussion
9.

Leblanc process is employed in the manufacture of

Answer»

Baking soda
Washing soda
Potash
Plaster of paris

Solution :Leblanc PROCESS is USED for the manufacture of `K_(2)CO_(3)` (potash).
Discussion
10.

Least soluble carbonate among alkaline earth metals will be?

Answer»

Solution :`BaCO_3` has high lattice energy and poor HYDRATION energy due to large sized ions. It is the least SOLUBLE CARBONATE among alkaline earth metal CARBONATES.
Discussion
11.

Least contributing resonating structure of nitroethene is :

Answer»




ANSWER :C
Discussion
12.

Least basic among the following is

Answer»

InOH
TlOH
`B(OH)_3`
`Al(OH)_3Al(OH_3)`

SOLUTION :The basic nature of HYDROXIDES INCREASE from `B(OH)_(3) ` to TL `(OH)_3`
Discussion
13.

Least acidic is

Answer»

`F-CH_(2)-CH_(2)-OH`
`Me_(3)N-CH_(2)-CH_(2)-OH`
`H_(2)O`
`CH_(3)OH`

SOLUTION :`CH_3 - OH`
Discussion
14.

What is leaded gasoline?

Answer»


ANSWER :F
Discussion
15.

Lead metal has a density of 11.34 g//cm^(3) and crystallizes in a face-centered lattice. Choose the correct alternatives

Answer»

the volume of one UNIT cell is `1.214 xx 10^(-22) cm^(3)`
the volume of one unit cell is `1.214 xx 10^(-19) cm^(3)`
the ATOMIC radius of lead is 175 pm
the atomic radius of lead is 155.1 pm

Solution :Density `= (Z xx M)/(N_(A) xx "volume")` ltrbgt so, Volume `= (4 xx 207)/(6.02 xx 10^(23) xx 11.34) = 1.213 xx 10^(-22) cm^(3), 4R = asqrt2`
`r = (4.95 xx 10^(-8) xx sqrt2)/(4) = 175 pm`
Volume `= a^(3) = 1.213 xx 10^(-22)` ltrbgt So, `a = (1.213 xx 10^(-24))^(1//3) RARR a = 4.95 xx 10^(-8) cm`
Discussion
16.

Lead (II) sulphide crystal has NaCl structure. What is the distance between Pb^(2+) and S^(2-) in PbS if its density is 12.7 cm^(-3) ? (At mass of Pb = 207)

Answer»


SOLUTION :Distance between `PB^(2+)` and `S^(2-)` IONS =`a/2` (as it has NaCl STRUCTURE)
Discussion
17.

Lead chloride has a solubility productof 1.7xx10^(-5) at 298 K. Calculate itssolubility at this temperature.

Answer»

Solution :The SOLUBILITY equilibrium for `PbCl_(2)` may be represented as :
`PbCl_(2)(s) HARR Pb^(2+) (aq) + 2Cl^(-) (aq)`
Let the solubility of `PbCl_(2) ` be s moles/litre. Then the solution will contain moles of `Pb^(2+)` ions and 2s moles `Cl^(-)` ions respectively per litre. Hence, the solubility product, `K_(sp)` of `PbCl_(2)` WOULD be given by the expression,
`K_(sp) = [ Pb^(2+)][Cl^(-)]^(2) = sxx (2s)^(2)=4S^(3)`
But the value of `K_(sp) = 1.7xx10^(-5) ` (Given)
`4s^(3)=1.7xx10^(-5) = (1.7xx10^(-5))/(4) = 0.425xx10^(-5) or s = (0.425xx10^(-5))^(1//3)=(4.25xx10^(-6))^(1//3)`
Let` x= (4.25)^(1//3) :. log x = 1//3 log 4.25 = 1//3 (0.6284) = 0.2095`
x= Antilog 0.2095 = 1.620
Hence, `s=1.620 xx 10^(-2) "mol L"^(-1)`
Discussion
18.

Le-Chateller's principle is not applicable to

Answer»

`Fe_((s))+S_((s))hArrFeS_((s))`
`H_(2(g))+I_(2(g))hArr2HI_((g))`
`N_(2(g))+O_(2(g))hArr2NO_((g))`
`N_(2(g))+3H_(2(g))hArr2N H_(3(g))`

Solution :Le-Chatelier PRINCIPLE is applicable only for gas-phase EQUILIBRIUM.
Discussion
19.

Le Chatelier's principle is not applicable to

Answer»

`Fe (s) + S(s) hArr F_(e)S(s)`
`H_(2)(g) + I_(2) (g) HARR2 HI (g)`
` N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g)`
` N_(2) (g) + O_(2) (g) hArr 2 NO (g)`

ANSWER :A::B::C::D
Discussion
20.

Le chatelier's principle is applicable to

Answer»

Chemical EQUILIBRIA only
PHYSICAL equilibria only
Both physical and chemical equilibria
Gaseous SYSTEMS only

Solution :Le-chatelier.s is applicable to both physical and chemical EQUILIBRIUM
Discussion
21.

L.C.A.O. Principle is involved in the formationof the molecular orbitals according ot molecular orbital theory. The energy of the bonding molecular orbital is less than that of thecombining atomic orbitals while that of the antibonding molecular orbitals while that of the order (B.O.)=1/2(N_(b)-N_(a)) helps in predicting (i) formation of molecules/molecular ions, bond dossociation energy, stability and bond length. Only the molecules or ions with positive B.O. can be formed. These will be diamagnetic if all molecular orbitals are dilled and paramagnetic if one of more are half filled. The atomic prbitals at the time of overlap must have the same symmetry as well. Which of the followijng combinatinos is not allowed (assume z axis as the internuclear azis) ?

Answer»

2S and 2s
`2p_(x)and 2p_(x)`
2s and `2p_(z)`
`2p_(y)and 2p_(y)`

SOLUTION :`2p_(x)and 2p_(y)` atomic orbitals do not combine due to LACK of SYMMETRY.
Discussion
22.

L.C.A.O. Principle is involved in the formationof the molecular orbitals according ot molecular orbital theory. The energy of the bonding molecular orbital is less than that of thecombining atomic orbitals while that of the antibonding molecular orbitals while that of the order (B.O.)=1/2(N_(b)-N_(a)) helps in predicting (i) formation of molecules/molecular ions, bond dossociation energy, stability and bond length. Only the molecules or ions with positive B.O. can be formed. These will be diamagnetic if all molecular orbitals are dilled and paramagnetic if one of more are half filled. The atomic prbitals at the time of overlap must have the same symmetry as well. In the homonuclear molecule3 which of the following sets of M.O. orbitals are degenerate ?

Answer»

`sigma_(1S)and sigma_(1s)^(**)`
`pi_(2px)and pi_(2py)`
`pi_(2px)and sigma_(2PZ)`
`sigma_(2pz)and pi_(2px)^(**)`

Solution :`pi_(2px) and pi_(2py)` M.O. orbitals are DEGENERATE.
Discussion
23.

L.C.A.O. Principle is involved in the formationof the molecular orbitals according ot molecular orbital theory. The energy of the bonding molecular orbital is less than that of thecombining atomic orbitals while that of the antibonding molecular orbitals while that of the order (B.O.)=1/2(N_(b)-N_(a)) helps in predicting (i) formation of molecules/molecular ions, bond dossociation energy, stability and bond length. Only the molecules or ions with positive B.O. can be formed. These will be diamagnetic if all molecular orbitals are dilled and paramagnetic if one of more are half filled. The atomic prbitals at the time of overlap must have the same symmetry as well. Bond arder is :

Answer»

DIRECTLY RELATED to BOND length
inversely related to bond length
incersely related to bond strength
never fractional.

Solution :is the CORRECT ANSWER.
Discussion
24.

L.C.A.O. Principle is involved in the formationof the molecular orbitals according ot molecular orbital theory. The energy of the bonding molecular orbital is less than that of thecombining atomic orbitals while that of the antibonding molecular orbitals while that of the order (B.O.)=1/2(N_(b)-N_(a)) helps in predicting (i) formation of molecules/molecular ions, bond dossociation energy, stability and bond length. Only the molecules or ions with positive B.O. can be formed. These will be diamagnetic if all molecular orbitals are dilled and paramagnetic if one of more are half filled. The atomic prbitals at the time of overlap must have the same symmetry as well. In the formation of N_(2)^(+) from N_(2), the electron is removed from a

Answer»

`SIGMA`ORBITAL
`PI`orbital
`sigma**`-orbital
`pi**`-prbital.

Solution :The ELECTRON is REMOVED from `sigma` M.O.
Discussion
25.

ldenlify the secondary pollutant among the following.

Answer»

`CH_(4)`
peroxy acetyl nitrate
`SO_(2)`
No

Answer :B
Discussion
26.

Layer test is used to detect the presence of

Answer»

Chlorine
Phosphorous
Bromine
FLUORINE

Answer :C
Discussion
27.

Layer structure is present in

Answer»

GRAPHITE
Coal
Diamond
Coke

Solution :LAYER STRUCTURE is PRESENT in graphite
Discussion
28.

Law of mass action is applicable to

Answer»

HOMOGENEOUS chemical EQUILIBRIUM only
Heterogeneous chemical equilibrium only
Both homogeneous and Heterogeneous chemical equilibria
PHYSICAL equilibrium

Solution :Law of mass action is only APPLICABLE to physical equilibrium
Discussion
29.

Law of mass action is not applicable to C_("(graphite)") harr C_("(diamond)") because

Answer»

It is a physical equilibrium
The PROCESS is not spontaneous
The process spontaneous
Both FORMS are crystalline

Solution :Law of mass ACTION is not APPLICABLE to physical equilibria
Discussion
30.

Law of mass action can not be applied to

Answer»

`2HI_((G)) harr H_(2(g))+I_(2(g))`
`PCI_(5(g)) harr PCI_(3(g))+CI_(2(g))`
`S_("Rhombic") harr S_("Monoclinic")`
`CaCO_(3(s)) harr CaO_((s))+CO_((g))`

Solution :It is not applicable to physical equilibrium
Discussion
31.

Law of conservation of mass does not hold good when we talk of conversion of........into.......

Answer»


ANSWER :MASS, ENERGY
Discussion
32.

Law of combining volumes was proposed by

Answer»

LAVOISIER
GAY Lussac
Avogadro
Dalton

Answer :B
Discussion
33.

Laughing gas is obtained when a mixture of NH_(4)Cl and …………is heated while nitrogen gas is obtained when a mixture of NH_(4)Cl and ………….is heated.

Answer»


ANSWER :`NaNO_(3),NaNO_(2)`
Discussion
34.

Lattice enthalpies of BeF_(2),MgF_(2),CaF_(2) and BaF_(2) are -2906,-2610,-2459 and -2367 kJ mol^(-1) resoectively. Hydration enthalpies of Be^(2+),Ca^(2+),Ba^(2+) and F^(Θ) are -2194,-1921,-1577,-130 and -457 kJ mol^(-1) respectively. Which of the fluorides has the highest solublility in water?

Answer»

Solution :For `BeF_(2)`,
`Delta_(hyd)H^(Θ)(BeF_(2))=Delta_(hyd)H^(Θ)(Be^(2+))+2Delta_(hyd)G^(Θ)(F^(Θ))`
`=[-2494+2(-457)]kJ mol^(-1)`
`=-3408 kJ mol^(-1)`
`Delta_(U)H^(Θ)(BeF_(2))=-2906 kJ mol^(-1)`
`Delta_("Soln")H^(Θ)(BeF_(2))=Delta_(hyd)H^(Θ)(BeF_(2))-Delta_(U)H^(Θ)(BeF_(2))`
`=[-3408-(-2906)]kJ mol^(-1)`
`=-502 kJ mol^(-1)`
For `MgF_(2)`,
`Delta_(hyd)H^(Θ)(MgF_(2))=Delta_(hyd)H^(Θ)(MG^(2+))+2Delta_(hyd)H^(Θ)(F^(Θ))`
`=[-1921+2(-457)]kJ mol^(-1)`
`=-2835 kJ mol^(-1)`
`Delta_(U)H^(Θ)(MgF_(2))=-2610 kJ Mol^(-1)`
`Delta_("soln")H^(Θ)(MgF_(2))=Delta_(hyd)H^(Θ)(MgF_(2))-Delta_(U)H^(Θ)(MgF_(2))`
`=[-2835-(-2610)]kJ mol^(-1)=-225 kJ mol^(-1)`
For `CaF_(2)`,
`Delta_(hyd)H^(Θ)(CaF_(2))=Delta_(hyd)H^(Θ)(Ca^(2+))+2Delta_(hyd)H^(Θ)(F^(Θ))`
`=[-1577+2(-457)] kJ mol^(-1)`
`=-2491 kJ mol^(-1)`
`Delta_(U)H^(Θ)(CaF_(2))=-2459 kJ mol^(-1)`
`Delta_("soln")H^(Θ)(CaF_(2))=Delta_(hyd)H^(Θ) (CaF_(2))-Delta_(U)H^(Θ)(CaF_(2))`
`=[-2491-(-2459)] kJ mol^(-1)`
`=-32 kJ mol^(-1)`
For `BaF_(2)`,
`Delta_(hyd)H^(Θ)(BaF_(2)=Delta_(hyd)H^(Θ)(Ba^(2+))+2Delta_(hyd)H^(Θ)(F^(Θ))`
`=-1305+2(-457)`
`=-2219 kJ mol^(-1)`
`Delta_(U)H^(Θ)(BaF_(2))=-2367 kJ mol^(-1)`
`Delta_("soln")H^(Θ)(BaF_(2))=Delta_(hyd)H^(Θ)(BaF_(2))-Delta_(U)H^(Θ)(BaF_(2))`
`=[-2219-(-2367)] kJ mol^(-1)`
`=148 kJ mol^(-1)`
Since enthalpy of solution `Delta_("soln")H^(Θ)` is maximum for `BeF_(2)`, it has highest SOLUBILITY in water.
Discussion
35.

Lattice energy of NaCl is 'X'. If the ionic size of A^(+2) is equal to that of Na^(+) and B^(-2) is equal to Cl^(-), then lattice energy associated with the crystal AB is

Answer»

X
2X
4X
8X

Answer :C
Discussion
36.

Lattice energy in sodium chloride is x kJ. Assuming the same interionic distance, what will be the lattice energy of magnesium sulphide?

Answer»

Solution :The charge MAGNITUDES of ions in NaCl are1 and 1 respectively.
The produce of `q_(1)` and `q_(2)=1xx1=1`
The charge magnitude of ions in magnesium sulphide are 2 and 2 respevetively.
The product of `q_(1)` and `q_(2)=2xx2=4`
Hence lattice energy of MgS `=4X KJ`.
Discussion
37.

Latic acid on oxidation with Fenton's reagent givesmain product :

Answer»

`CH_(3)COOH`
`H_(2)C_(2)O_(4)`
`CH_(3)CH_(2)COOH`
PYRUVIC acid

Solution :See LACTIC acid
Discussion
38.

Lattice dissociation energy and hydration energy of sodium chloride are respectively 788 kJ mol^(-1) and 784 kJ mol^(-1). Then enthalpy of solution of sodium chloride in water is

Answer»

`+4 KJ MOL^(-1)`
`+1572 kJ mol^(-1)`
`4 kJ mol^(-1)`
`-1572 kJ mol^(-1)`

ANSWER :A
Discussion
39.

Latex is a colloidal suspension of rubber particles, they carry

Answer»

no CHARGE
positive charge only
negative charge only
may be positive or negative charge.

Solution :In LATEX RUBBER particles CARRY negative charge only.
Discussion
40.

Lassaigne's test is not very successful for diazonium salts. Assign reason.

Answer»

Solution :Diazonium salts `(C_(6)H_(5)N_(2)X)` are HIGHLY REACTIVE substances and when heated in a fusion tube, these immediately lose nitrogen `(N_(2))` gas and THEREFORE, do not form sodium CYANIDE with sodium metal.
Discussion
41.

Latent heat of vapourisation of a liquid at 500k and 1atm pressure is 10K. Cal/mole. What is the change in internal energy when 3 moles of the liquid is vapourised at the same temperature

Answer»

`27K.Cal `
`7K.Cal `
`33K.Cal`
`25K.Cal `

ANSWER :A
Discussion
42.

Latent heat of fusion of ice is 6 kJ mol^(-1). Calculate the entropy change in the fusion of ice.

Answer»

Solution :CHANGE in entropy, `DeltaS = q_(rev)//T`
`q_(rev)`= Latent heat of fusion = `6 kJ mol^(-1)`
`= 6000 J mol^(-1)`
T = Freezing point of water = 273K.
`DeltaS = 6000/(273 = 21.98 JK^(-1) mol^(-1)) (or)`
`DeltaS = (6000)/(273 XX 18) = 1.22 JK^(-1)g^(-1)`.
Discussion
43.

Last traces of water is removed is removed from H_(2)O_(2) by

Answer»

electrolysis
crystallisation
condensation
evaporation.

Answer :B
Discussion
44.

Last molecule of H_2Ois removed from H_2O_2by

Answer»

crystallisation
evaporation
DISTILLATION under reduced pressure
electrolysis

Solution :`H_2O_2` solution is CONCENTRATED in 3 steps. In the LAST step its is concentrated by crystallisation
Discussion
45.

Last molecule of H_(2) O is removed from H_(2)O_(2) by

Answer»

CRYSTALLISATION
Crystallisation
Distillation under reduced PRESSURE
ELECTROLYSIS

Answer :A
Discussion
46.

Last molecule of H_(2)O is removed from H_(2)O_(2) by

Answer»

CRYSTALLISATION
EVAPORATION
DISTILLATION under REDUCED pressure
Electrolysis

ANSWER :A
Discussion
47.

Lassigne's test is not used for the detection of :

Answer»

N
S
Cl
O

Answer :D
Discussion
48.

Lassaigne's test (with silver nitrate) is commonly used to detect halogens such as chlorine, bromine and iodine but not useful to detect fluorine because the product AgF formed as

Answer»

volatile
reactive
explosive
SOLUBLE in WATER

Solution :AgF is HIGHLY soluble in `H_(2)O` therefore, NaF formed in the Lassaigne's extract does not produce a ppt. with `AgNO_(3)`.
Discussion
49.

Lassaigne's test (with silver nitrate) is commonly used to detect halogens such as chlorine, bromine and iodine but not useful to detect fluorine because the product AgF formed is (a) volatile (b) reactive (c ) explosive (d) soluble in water (e) a liquid

Answer»


ANSWER :D
Discussion
50.

Lassaigne's test (with silver nitrate) is commonly used to detect halogens such as chlorine, bromine and iodine but not useful to detect fluorine because the product AgF formed as:

Answer»

volatile
reactive
soluble in water
explosive

Answer :C
Discussion