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27651.

The nitro alkane which on reaction with nitrous acidforms nitrolic acid is

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`CH_3CH_2NO_2`
`(CH_3)_2CHNO_2`

`(CH_3)_3CNO_2`

SOLUTION :`CH_3-CH_2-NO_2overset(HNO_2)(to)CH_3-UNDERSET("Nitrolic ACID")underset(N-OH)underset(||)(C)-NO_2+H_2O`
27652.

The minimum amount of energy required to remove the most loosely bound electron form an isolated atom in the gaseous state is known as ionization energy or first ionization energy or inonization enthalpy (IE_(1)) of the element. The energy required to remove the second electron from the monovalent cation is called second ionization enthalpy (IE_(2)). Simllarly, we have third, fourth .................... fonization enthalpies. The values of ionization energy depends on a number of factors such as (i) size of the atom (ii) screening effect (iii) nuclear charge (iv) half filled and fuly filled orbitals (v) shape of orbital. In a group, the ionization energy decreases from top to bottom. In a period, the valup of ionisation energy increases from left to right with breaks where atoms have somewhat stable configurations. In a period, the ionization energy is lowest for the

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NOBLE gases
halogens
ALKALINE earth metals
alkali metals

ANSWER :D
27653.

The nitrate which when heated gives off a gas or a mixture of gases which cannot relight a glowing splinter is:

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SODIUM nitrate
Ammonium nitrate
Lead nitrate
Potassium nitrate

Answer :B
27654.

The minimm amount of O_(2)(g) consumed per gram of reactant is for the reaction : (Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)

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`C_(3)H_(8)(g) + 5 O_(2)(g) to 3 CO_(2)(g) + 4 H_(2)O(l)`
`P_(4)(s) + 5 O_(2)(g) to P_(4)O_(10)(s)`
`4 Fe(s) + O_(2)(g) to 2 FeO_(3)(s)`
`2 Mg(s) + O_(2)(g) to 2 MGO(s)`

ANSWER :C
27655.

The nitride ion in lithium nitride is composed of:

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7 PROTONS + 7 electrons
10 protons + 7 electrons
7 protons + 10 electrons
10 protons + 10 electrons

Answer :C
27656.

Thenitration of which of the followingalkaneswill givethemaximum numberof products ?

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PROPANE
NEOPENTANE
ETHANE
METHANE

ANSWER :A
27657.

The minimum amount of energy required for the reacting molecules to undergo reaction is called :

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threshold ENERGY
activation energy
INTERNAL energy
potential energy

Solution :The minimum energy for EFFECTIVE collisions is CALLED threshold energy.
27658.

The nitrate of which metal left globule on heating strongly

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`Pb(NO_(3))_(2)`
`NaNO_(3)`
`AgNO_(3)`
`CU(NO_(3))_(2)`

SOLUTION :AG salts on strong heating from Ag.
27659.

Perovaskite, a mineral containing calcium, oxygen &titanium crystallizes in the given unit cell The oxidation number of titanium in the perovskite is

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`+2`
`+3`
`+4`
`+1`

Answer :C
27660.

The NH_(3) evolved from 0.5 gram of an organic compound in Kjeldahl method,neutralized 20 ml (M)/(2) H_(2)SO_(4). Find the % of nitrogen:-

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`56%`
`28%`
`14%`
`70%`

ANSWER :A
27661.

The mineral of magnesium is :

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Bauxite
Malachite
Carnallite
Haematite

Answer :C
27662.

The NH_(4)^(+) ion is

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SQUARE planar
tetrahedral
a LEWIS base
planar in solution

Answer :B
27663.

The mineral monazite is a rich source of thorium, available in large quantity in kerala. A typical monazite sample contains 9% ThO_(2) and 0.35% U_(3)O_(8). .^(208)Pb and .^(206)Pb are the stable end products in the radioactive decay series of .^(232)Th and .^(238)U respectively. All the lead in monazite is of radiogenic origin. The isotopic ratio of .^(208)Pb//^(232) Th. was found to be 0.104. The half lives of Th and U are 1.41 xx 10^(10) years adn 4.47 xx 10^(9) years respectively. The time elapsed since the formation of monazite sample will be:

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`1.34 xx 10^(9)` years
`2.01 xx 10^(9)` years
`1.41 xx 10^(9)` years
`4.47 xx 10^(9)` years

Solution :`(0.693)/(t_(1//2)Th) = (2.303)/(t_("AGE")) log ((N_(0))/(N))`
`(0.693)/(1.41 xx 10^(10)) = (2.303)/(t_("age")) log (1.104)`
`t_("age") = 2.01 xx 10^(9)` years
27664.

The nitrate of which metal leaves metallic globule on heating strongly :

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`CU(NO_3)_2`
`AgNO_3`
`NaNO_3`
`PB(NO_3)_2`

ANSWER :B
27665.

The mineral monazite is a rich source of thorium, available in large quantity in kerala. A typical monazite sample contains 9% ThO_(2) and 0.35% U_(3)O_(8). .^(208)Pb and .^(206)Pb are the stable end products in the radioactive decay series of .^(232)Th and .^(238)U respectively. All the lead in monazite is of radiogenic origin. The isotopic ratio of .^(208)Pb//^(232) Th. was found to be 0.104. The half lives of Th and U are 1.41 xx 10^(10) years adn 4.47 xx 10^(9) years respectively. Estimated isotopic ratio of .^(206)Pb//^(238) U in the monazite sample will be:

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0.166
0.266
0.366
0.466

Answer :C
27666.

The newman projection formula of 2-3, dimethylbutane is given as X,Y respectively can be:

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`-CH(CH_(3))_(2) " and " H`
`-CH_(3) " and " -C_(2)H_(5)`
`-C_(2)H_(5) " and " -CH_(3)`
`H" and " -CH(CH_(3))_(2)`

Answer :D
27667.

The mineral of copper is:

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Azurite
Malachite
Copper pyrites
All

Answer :D
27668.

The Newman projection shown is the same as

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

Answer :C
27669.

The mineral hawleyite, one form of CdS, crystallizes in one of the cubic lattices, with edge length 5.87 Å. The density of hawleyite is 4.63 g cm^(–3). (i) In which cubic lattice does hawleyite crystallize ? (ii) Find the Schottky defect in g cm^(–3).

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ANSWER :(i)3.90, (II)0.120 g/cc
27670.

Theneutralizationequivalentof an acidis 116. Fusionof sodim salt of theacidwithsodalime givesa hydrocarbon of whichbutanemonobromo substitutionproductis possible .Whatstructural formulamay beassigned to theacid ?

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Solution :The final product is an alkane derviative , thisindicatesthe acidis mono carboxylic, RCOOH.
As neutralequivalentis 116,therefore, formula weightof alky group R is 116-45 (Formulaweight of - COOH) = 71.
Thiscorrespecondsot the pentyl group`(C_(5)H_(11))`. The FORMATIONOF butanederivativesuggestsa branched akyl radical containingfour carbon atomsin thelargestchain. Therefore , R shouldbe .
`CH_3 - CH_2 - underset(CH_(3))underset(|)(C)H- CH_(2)`
`CH_(3)- CH_(2) - underset(CH_(3))underset(|)(C)H- underset("( 3 - methyl pentanoic acid)")(CH_(2) - COOH)`
27671.

The mineral from which potassium permagnate is manufactured

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PYROLUSITE, `MnO_(2)`
BRAUNITE, `Mn_(2)O_(3)`
hausmannite, `Mn_(3)O_(4)`
manganite, `Mn_(2). H_(2)O.`

Answer :A
27672.

The mineral from which is not munufactured by copper :

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GALENA
COPPER Pyrite
Malachite
Chalcopyrite

Answer :D
27673.

The mililitres of 0.2M KMnO_(2) required for the complete oxidation of 0.1 mol Fe^(2+ )in acidic medium is -

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200ml
100ml
400ml
50ml

Answer :B
27674.

The mineral clevelte on heating gives:

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He
Xe
Ar
Ra

Answer :A
27675.

The neutral ligand is :

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CHLORO
Hydroxo
Ammine
Oxalato

Answer :C
27676.

The Miller indices of two parallel planes in a crystal are: (a) same , (b) different

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ANSWER :a
27677.

The milliequivalents in 60 ml of 4.0 M H_(2)SO_(4) is :

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240
48
480
960

Solution :`4.0 M H_(2)SO_(4)` means 4 MOLES of `H_(2)SO_(4)` in 1000ML
or`=(4xx60)/(1000)=0.24` moles in 60 ml
`=0.48` EQUIVALENT in 60 ml
`=480` equivalent in 60 ml
or MOLARITY `=("Moles of" H_(2)SO_(4)xx1000)/("Volume")`
`=("Moles of " H_(2)SO_(4)xx1000)/(60)`
`:.` Moles of `H_(2)SO_(4)=(4xx60)/(1000)=0.24`
Equivalents of `H_(2)SO_(4)=0.48`
Milliequivalents of `H_(2)SO_(4)=0.48xx1000=480`
27678.

The neutral oxide of nitrogen is

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NO
`N_2O`
`NO_2`
Both 1 and 2

Answer :D
27679.

The migration of sol particles under the influence of electric field is called .............

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ELECTRO osmosis
electro dialysis
ELECTROPHORESIS
dialysis

SOLUTION :electrophoresis
27680.

The migratory capacity is

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`H^(-) gt CH_(3)^(-) gt Ph^(-)`
`Ph^(-) gt CH_(3)^(-) gt H^(-)`
`H^(-) gt Ph^(-) gt CH_(3)^(-)`
`CH_(3)^(-) gt Ph^(-) gt H^(-)`

Solution :The migratory capacity is maximum if the negative charge is on a more electronegative atom.
In `Ph^(-)`, the -ve charge is on a `sp^(2)` carbon.
In `OVERSET(-)(C)H(3)`, the -ve charge is on a `sp^(3)` carbon atom.
Out of carbon and hydrogen, carbon is more electronegative. Further an `sp^(2)C` atom is more electronegative than `sp^(3)C` atom.
HENCE migratory capacity of `Ph^(-) gt overset(-)(C)H_(3) gt H^(-)`.
So answer (B) is CORRECT whereas (a), (c) and (d) are incorrect.
27681.

The net redox reaction of a galvanic cell is given below 2Cr(s) + 3Cu_((aq))^(2+) to 2Cr_((aq))^(3+) + 3Cu_((s)) Write the half reactions and describe the cell using cell notation.

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Solution :ANODIC OXIDATION : `2Cr_((s)) to 2Cr_((AQ))^(3+) + 6e^(-)"" ….(1)`
Cathodic reduction : `3Cu_((aq))^(2+) +6C^(-) to 3Cu_((s))"" ….(2)`
Cell Notation is : `Cr_((s)) |Cr_((aq))^(3+)||Cu_((aq))^(2+)|Cu_((s))`.
27682.

The neutral complex, diamminedibromodichloroplatinum(IV) is best represented as

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`[PT(NH_3)_2 Br_2Cl_2]`
`[PtCl_2Br_2(NH_3)_2]`
`PtBr_2Cl_2(NH_3)_2]`
`[Pt(NH_3)_2 Cl_2 Br_2]`

ANSWER :A
27683.

The migration of colloidal solute particles in a colloidal solution , when an electric current is applied to the solution is known as

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Brownian movement
ELECTRO OSMOSIS
electrophoresis
electrodialysis

ANSWER :C
27684.

The net reaction during the discharge of nickel-cadmium battery is :

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`2Ni(OH)_(3)+Cd+H_(2)O to 2Ni(OH)_(2)+Cd(OH)_(2)`
`2Ni_(2)O_(3)+Cd+O_(2) to 2Ni(OH)_(3)+Cd(OH)_(2)`
`2Ni(OH)_(3)+Cd toCdO+2Ni(OH)_(2)+H_(2)O`
`2Ni(OH)_(2)+Cd+O_(2) to 2Ni(OH)_(2)+CDO`

SOLUTION :(a) It is the correct answer.
27685.

The migration of colloidal particles under the influence of an electric field is known as

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ELECTRO osmosis
cataphories
electrodialysis
electrophoresis

Answer :B::D
27686.

The net rate of reaction of the change : [Cu(NH_(3))_(4)]^(2+)+H_(2)O ?" "[Cu(NH_(3))_(3)H_(2)P]^(2+)+NH_(3) is, (dx)/(dt) =2.0xx10^(-4)[Cu(NH_(3))_(4)]^(2+)-3.0xx10^(5)[Cu(NH_(3))_(3)H_(2)O]^(2+)[NH_(3)] Calculate : the direction of reaction in which the abovereaction will be more predominant.

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SOLUTION :More PREDOMINANT REACTION is BACKWARD reaction.
27687.

The migration of positively charged colloidal particles, under an electrical field towards the cathode is called

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Electrophoresis
Electro-osmosis
Tyndall EFFECT
Brownian movement

Answer :A
27688.

The net rate of reaction of the change : [Cu(NH_(3))_(4)]^(2+)+H_(2)O ?" "[Cu(NH_(3))_(3)H_(2)P]^(2+)+NH_(3) is, (dx)/(dt) =2.0xx10^(-4)[Cu(NH_(3))_(4)]^(2+)-3.0xx10^(5)[Cu(NH_(3))_(3)H_(2)O]^(2+)[NH_(3)] Calculate : the ratio of rate constant for forward and backward reactions.

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Solution :Also, `" "K_(F) =2.0xx10^(-4)`
`K_(b) =3.0xx10^(5)`
`THEREFORE" "(K_(f))/(K_(b))=(2.0xx10^(-4))/(3.0xx10^(5))=6.6xx10^(-10)`
27689.

The middle transition element ..... has six different oxidation states.

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SOLUTION :MANGANESE
27690.

The net heat change in a chemical reaction is same whether it is brought about in two or more different ways in one or several steps. It is known as

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HESS’s LAW
Law of conservation of energy
Henry’s law
Joule’s principle

ANSWER :A
27691.

The 'middle oil' fraction of coaltar distillation contains

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Benzene
Anthracene
Naphthalene
Xylene

Answer :C
27692.

The M.F. = C_(3)H_(8)O represents two compounds A and B. Both of them react with sodium to liberate H_(2) and on dehydration results same alkene C". In presence of H_(2)O_(2), the alkene 'C" ads of HBr giving 'D', which on treatment with aqueous KOH gives 'B'. Further A' responds to iodoform test. Identify 'A'.

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`CH_(3)CH_(2)CH_(2)OH`
`CH_(3)CHO`
`CH_(3)-overset(OH)overset(|)CH-CH_(3)`
`CH_(3)-overset(Br)overset(|)CH-CH_(3)`

Solution :`MF=C_(3)H_(2)O` represents ETHER and alcohol. As the compound A and B react with sodium to liberate `H_(2)` gas, so they must be alcohol. Dehydration of A and B results same alkene, HENCE the alkene C is, `C_(3)H_(6)`(propene).
`B= UNDERSET("Propan-1-ol")(CH_(3)-CH_(2)-CH_(2)-OH), A= underset(Propan-2-ol)(CH_(3)-overset(OH)overset(|)CH-CH_(3))`
`underset("(C)")(CH_(3)-CH=CH_(2)) underset(H_(2)O)overset(HBr)to underset("(D)")(CH_(3)-CH_(2)-CH_(2)-Br)`
`CH_(3)-CH_(2)-CH_(2)-Br+KOH_("(aq)")to underset("(B)")(CH_(3)- CH_(2)-CH_(2)-OH+ KBr)`
`underset("(A)")(CH_(3)- underset(OH)underset(|)CH-CH_(3)) overset(I_(2)//NaOH)to underset("iodoform")(CHI_(3))+CH_(3)-COONa+4HI`
27693.

The methods employed for the determination of equilibrium constant are different for different equilibria depending on the nature of the system and the type of reaction. The task becomes comparatively easy when we can measure some physical property such as vapour density, refractive index, optical rotation, electrical conductivity etc. In simple gaseous dissociation such as N_(2)O_(4) rarr 2NO_(2) . 0.1 mole of N_(2)O_(4(g)) is decomposed under atmospheric condition at 25^(@)C. At equilibrium partial pressure of NO_(2) gas is, if K_(p) = 0.14

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0.017atm
0.034 atm
0.28 atm
0.14 atm

Answer :C
27694.

The methods employed for the determination of equilibrium constant are different for different equilibria depending on the nature of the system and the type of reaction. The task becomes comparatively easy when we can measure some physical property such as vapour density, refractive index, optical rotation, electrical conductivity etc. In simple gaseous dissociation such as N_(2)O_(4) rarr 2NO_(2) . 0.1 mole of N_(2)O_(4(g)) is decomposed under atmospheric condition at 25^(@)C. Degree of dissociation for above problem is :

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0.017
0.034
0.29
0.14

Answer :A
27695.

The Nessler.s reagent contains:

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`Hg_2^(2+)`
`HG^(2+)`
`Hgl_2^-`
`Hgl_4^2`

ANSWER :D
27696.

The methodschieflyusedfortheextractionof leadand tinfrom theirores arerespectively

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Selfreduction and Carbonreduction
Self reduction and ELECTROLYTICREDUCTION
Carbonreduction andSelfreduction
Cyanideprocessand Carbonreduction

SOLUTION : Self -reduction forlead`( 2 PbS +3O_ 2to2Pb O+2SO_2 ,PB S +2 PbOto 3P b+SO _ 2 ) and`carbonreduction fortin `(SnO+ CtoSn+CO) `.
27697.

The methods chiefly used for the extraction of lead and tin from their ores are respectively

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self - REDUCTION and CARBON reduction
self - reduction and ELECTROLYTICREDUCTION
carbon reduction and self- reduction
cyanide PROCESS and carbon - reduction

Answer :C
27698.

The methods cheifly used for the extraction of lead and zinc from their ores are respectively

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self-reduction and carbon reduction. 
carbon reduction and self-reduction. 
CYANIDE PROCESS and carbon reduction. 
carbon reduction and cyanide process. 

Solution :`{:(2Pbs + 3O_2 ,to, 2PbO + 2SO_2),(2PbO + PbS , to , 3Pb + SO_2):}}`Self-reduction
`{:(2PbO + PbS ,to , 3Pb + SO_2),(,," "darrc),(,,ZN + CO):}}`Carbon reduction
27699.

The Nernst equation giving dependence of electrode potential on concentration is

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`E = E^(0) + (2.303RT)/(nF) LOG""([M])/([M^(n+)])`
`E = E^(0) + (2.303RT)/(nF) log""([M^(n+)])/([M])`
`E=E^(0) - (2.303RT)/(nF) log""([M^(n+)])/([M])`
`E=E^(0) -(2.303RT)/(nF) log[M^(n+)]`

Answer :B
27700.

Whichmethodof refining is used whena metal of high degree of purity is needed ?

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ZONE refining
Van Arkel's Method
Liquation
Chromatography

Answer :A