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87701.

A regular hexagonal crystalline lattice of ice is mostly formed by :

Answer»

Ionic BOND
HYDROGEN bond
COVALENT bond
Metallic bond

Answer :B
87702.

A regular copolymer of ethylene and vinyl chloride contains alternate monomers of each type . What is the weight precent of ethylene in this copolymer.

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Solution :The structre of the copolymer is

Mol.WT. of ETHYLENE `=28`
Mol.wt.of vinyl CHLORIDE `=62.5`
E.F.wt. of copolymer `=28+62.5=90.5`
Wt. percent of ethylene is the copolymer `=(28)/(90.5)xx100=30.94`
87703.

A refrigerator is used to remove heat from enclosure at 0^(@)C at the rate of 600 watt. If the surroundings temperature is 30^(@)C, calculate the power needed:

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303 watt
11000 watt
65.9 watt
110 watt

Answer :C
87704.

(A) Reduction of a metal oxide is easier if the metal formed is in liquidstate at the temperature of reduction . (R)Delta G^(@)for the net reaction, reduction of metal oxide with the reductant, is more-ve when the metal formed is in molten state

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Both (A) and (R) are TRUE and (R) is the CORRECT EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanationof (A)
(A) is true but (R) is false
Both (A) and (R) are false

Answer :A
87705.

(A) Reducing sugars undergo mutarotion. (R) During mutarotation, one pure anomer is converted into an equilibrium mixture of two anomers

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Both A & R are true and R is the correct EXPLANATION of A 
Both A & R are true, but R is not the correct explanation of A 
A is true, R is false 
A is false, R is true 

Answer :A
87706.

A reducing agent in a redox reaction undergoes

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a decrease in oxidation number
an INCREASE in oxidation number
loss of ELECTRONS
gain of electrons

Solution :A REDUCING AGENT is the ONE which loses electrons.
87707.

Areddish substance on heating gives of a capour whichcondenses on the sides of the test tube and the substance turn blue If on cooling water is added to the residue it turm to itsoriginal colour .The substance is

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IODINE crystals
Copper sulphate crystals
Cobalt CHLORIDE crystals
ZINE oxide

Solution :It is a preperty of `CoCI_(2)`
87708.

A reddish-pink substance on heating gives off a vapour which condences on the sides of the test tube and the substance turns blue. Il on cooling water is added to the residue it turns to its original colour. The substance is :

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IODINE sulphate CRYSTALS
Copper sulphate crystals
Cobalt CHLORIDE crystals
Zinc oxide

Answer :C
87709.

A reddish brown sol ("containing" Fe^(3+)) is obtained by:

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the addition of small amount of `FeCl_(3)` solution to FRESHLY prepared `Fe(OH)_(3)` precipitate
the addition of `Fe(OH)_(3)` to freshly prepared `FeCl_(3)` solution
the addition of `NH_(4)OH` to `FeCl_(3)` solution dropwise
the addition of `NaOH to `FeCl_(3)` solution dropwise

Solution :Reddish BROWN sol is prepared by adding `FeCl_(3)` in `Fe(OH)_(3)` precipitate.
87710.

A reddish brown residue in the charcoal cavity test is given by

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`Pb^(2+)`
`Bi^(3+)`
`Cd^(2+)`
`Zn^(2+)`

Solution :CDO is the residue. It is reddish BROWN in colour.
87711.

A red solid is insoluble in water.However , it becomes solubleif some KI is added to water.Heating the red solidin a test tube results in liberation of someviolet coloured fumesand droplets of a metal appear on the cooler partsof the test tube.The red solid is

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`HgI_(2)`
`HgO`
`Pb_(3)O_(4)`
`(NH_(4))_(2)Cr_(2)O_(7)`

SOLUTION :`UNDERSET("RED ppt.")(HgI_(2)) +2KI rarrunderset("Soluble")(K_(2)[HgI_(4)])`
87712.

A red solid is insoluble in water.However it becomes soluble if some KI is added to water.Heating the red solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube.The red solid is :

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`(NH_(4))_(2)Cr_(2)O_(7)`
`HgI_(2)`
`HgO`
`Pb_(3)O_(4)`

Solution :`Hgl_(2) darr + 2l^(-) to [Hgl_(4)]^(2-)` (colourless SOLUBLE COMPLEX)
`Hgl_(2) overset(DELTA)to Hg+l_(2) uarr` (violet colour)
87713.

A red solid is insoluble in water. However it becomes soluble if some KI is added to water. Heating the red solid in a test tube results in liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is

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`(NH_(4))_(2)Cr_(2)O_(7)`
`HgI_(2)`
`HGO`
`Pb_(3)O_(4)`

SOLUTION :`UNDERSET("Red solid")(HgI_(2)+)2KItoK_(2)(HgI_(4))` soluble
`HgI_(2)toHg+I_(2)` violet fumes.
87714.

A red coloured solid is insoluble in water. However, it becomes soluble if some KI is added to water. Heating red solid in a test tube results in the liberation of some violet coloured fumes and droplets of a metal appear on the cooler parts of the test tube. The red solid is

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`(NH_(4))_(2)Cr_(2)O_(7)`
`HgI_(2)`
HgO
`Pb_(3)O_(4)`

Solution :The RED coloured solid in `Hgl_(2)`
`Hgl_(2)+2Kl to underset("Soluble")(K_(2)Hgl_(4))`
`Hgl_(2) overset("heat")toH_(g)+underset("Violet fumes")(I_(2))`
87715.

A red coloured mixed oxide (X) on treatment with conc. HNO_(3) gives a compound (Y). (Y) with HCl produces a chloride (Z) which is insoluble in cold water but soluble in hot water, (Z) can also be formed by treating (X) with conc. HCl. Compounds (X), (Y) and (Z) are :

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`Pb_(3)O_(4), PbNO_(3), PbCl_(2)`
`Mn_(3)O_(4), MnO_(2), MnCl_(2)`
`Fe_(3)O_(4), Fe_(2)O_(3),FeCl_(3)`
`Fe_(2)O_(4), FEO, FeCl_(2)`

Answer :A
87716.

A recemic mixtureof (+-) 2-phenylpropanoic acidon esterification with (+) 2-butanol gives two ester. Mention the stereochemistry of the two esters produced.

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ANSWER :`(##RES_CHM_SIM_E03_002_A01##)`
87717.

A recent investigation of the complexation of SCN– with Fe3^+ represented by constant K_1, K_2 and K_3 as 130, 16, and 1.0 respectively. What is the overall formation constant of Fe(SCN)_3 from its component ions, and what is the dissociation constant of Fe(SCN)_3 into its simplest ions on the basis of these data ?

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Answer :`K_(d)=1/(K_(F))=4.8 xx 10^(-4)`
87718.

A red blood corpuscles (RBC) is composed of heme group which............. Complex play an important role in carrying oxygen from lungs to tissues.

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ANSWER :`FE^(2+)` PORPHYRIN
87719.

A reaserch scholar get a mixture of three product during an experimant with ammonia. In product I only one H of ammonia is replaced by ethyl group and in II two H atoms of ammonia are replaced by ethyl groups and in III alll the H-atoms are replaced by ethyl groups. Which test he should use to distinguish or separate the products :

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Carbyl AMINE test
Iodoform test
Fehling solution test
Hinsberg test

Solution :`1^(@), 2^(@)` and `3^(@)` amine MIXTURE can be separated by Hinsberg REAGENT.
87720.

A recemic substance is composed of

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25% D- FORM and 75% L - form
50% D-form and 25% L - form
75% d-form and 25% 1 - form
50% d- form and 50% 1 - form

ANSWER :D
87721.

A real gas tends to behave more ideally at

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LOW TEMPERATURE and low PRESSURE
low temperatureand high pressure
high temperature and low pressure
high temperature and high pressure.

Answer :C
87722.

A real gas obeying van der Waals' equation :( P + ( an^(2))/( V^(2))) ( V - b ) = nRTwill closely resemble an ideal gas if

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the CONSTANTS a and B are LARGE
a is large and b is amll
a is SMALL and b is large
a and b are both small.

Answer :D
87723.

A real gas most closely approaches the behaviour of an ideal gas at :

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15 atmosphere and 200 K
1 atmosphere and 273 K
0.5 atmosphere and 500 K
15 atmosphere and 500 K

Answer :C
87724.

A real gas at high pressure occupies under identical conditions :

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More volume than that of an ideal GAS
LESS volume than that of an ideal gas
Same volume as that of an ideal gas
More or less volume than that of an ideal gas depending upon the NATURE of the gas

Answer :B
87725.

A reagent which reacts differently with CH_(2)O, C_(2)H_(4)O,C_(3)H_(6)O is

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`NH_(2)OH`
`C_(6)H_(5)NHNH_(2)`
NCN
`NH_(3)`

SOLUTION :The reagents a , b , c reacts with the CARBONYL COMPOUNDS in a same manner and only `NH_(3)` react differently. Formaldehyde is reacted with ammonia give urotropine. Acetaldehyde reacts ammonia give acetaldehyde AMINE andacetone reacts ammonia give diacetone amine.
87726.

A reagent used to test for unsaturation of alkene is

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Conc. `H_2SO_4`
Ammoniacal `Cu_2Cl_2`
Ammoniacal `AgNO_3`
Solution of `Br_2` in `C Cl_4`

Solution :Solution of BROMINE in CARBON TETRACHLORIDE is used to test for unsaturation of ALKENE. Red colour of bromine disappears due to the FORMATION of colourless dibromo ethane `(C_2H_4Br_2)`
87727.

A reagent used for identifying nickel ion is:

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POTASSIUM ferrocyanide
Phenolphthalein
Dimethyl glyoxime
EDTA

Answer :C
87728.

A reagent that can separate Fe from Zn is,

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NAOH
HCL
`H_3S`
`NaNO_2`

ANSWER :A
87729.

A reagent that can detect any of Cu^(2+),Fe^(3+),Zn^(2+) and Cd^(2+) is ______(a)______.

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ANSWER :`K_(4)[FE(CN)_(4)]`
87730.

A reagent suitable for the determination of N-terminal residue of a peptide is

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p-toluene SULPHONYL chloride
2,4-dinitrophenyl hydrazine
carboxypeptidase
2,4-dinitrofluorobenzene

Solution :2,4-dinitrofluorobenzene is called Sanger's reagent. When this reagent reacts with amino group of PEPTIDE chain, it form 2,4-dinitrophenyl DERIVATIVES which on hydrolysis form DNP derivatives of amino ACIDS.
87731.

A' reacts by following two parallel reactions to give B& C If half of 'A' goes into reaction I and other half goes to reaction-II . Then , select the correct statement(s) A+N overset(I) to B +L A+N overset(II)to (1)/(2) B+(1)/(2)(C)+L

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B will be always GREATER than C
If 2 mole of C are FORMED then total 2 mole of B are also formed
If 2 mole of C are formed then total 4 mole of B are also formed
If 2 mole of C are formed then total 6 mole of B are also formed

Answer :A::D
87732.

'A' reacts with C_2H_5Igiving 'B' and Nal. Here 'A' and 'B' respectively are

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`CH_3COONa, CH_2OCH_3`
`C_2H_5OC_2H_5, C_2H_5COOC_2H_5`
`C_2H_5ONa, C_2H_5OC_2H_5`
`C_2H_5OH, C_2H_5OC_2H_5`

ANSWER :C
87733.

A reacts to form P , A plot of the reciprocal of the concentration of A vs time is a straight line . When the initial concentration of A is 1.0 xx 10^(-2) M , its half-life is found to be 20 min . When initial concentration of A is 3.0 xx 10^(-3) M , the half -life will be

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20 MIN
40 min
56 min
67 min

Solution :`A to P`
A plot of 1/[A] vs time is a STRAIGHT line of second order .
For second order , `t_(1//2) = (1)/(k[A]_(0))`
where , k =rate constant
`[A]_(0) ` = initial concentration of A.
when , `[A]_(0) = 1 xx 10^(-2)` M , then `t_(1//2) = 20` min
`20 = (1)/(k[1 xx 10^(-2)]) implies k = (1)/(20) [ 1 xx 10^(-2)] "" ... (i)`
when , `[A]_(0) = 3 xx 10^(-3)` M , then `t_(1//2) = `?
`t_(1//2) = (1)/(k[3 xx 10^(-3)])`
substituting value of k from equation (i)
`t_(1//2) = (1 xx 20 [1 xx 10^(-2)])/([3 xx 10^(-3)]) = 6.66 xx 10` min = 67 min .
87734.

A reacton is first order in A and second order in B. (i)Wite the differential rate equation. (ii)How is the rate affected on increasing the concentration of B three times? (iii)How Is the rate affected when the concentration of both A and B are doubled?

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Solution :Reaction:`A+BtoP`
Order with respect to A=1
Order with respect to B=2
Total order of reaction =1+2=3
(i)Rate law of REACTON is as follow:
`(d[R])/(DT)=k[A]^(1)[B]^(2)`
In short r=`k[A]^(1)[B]^(2)`
(ii)Effect on rate if the concentration increase by three times:
`r_(2)=k[A]^(1)[3B]^(2)` `r_(2)=9 r_(1)`
Rate increases by 9 times.
(iii)If concentration A and B both double then the rate:
Concentration of A=2A
Concentration of B=2B
So,`r_(3)=k[2A]^(1)[2B]^(2)`
`=k2[A]^(2)xx4[B]^(2)` `therefore` The rate becomes 8 times
(Note:If only concentration of B is double then rate become double)
87735.

(A) Reactivity of primary alcohol is more than secondary alcohols towards sodium metal(R) Primary alcohol is more acidic than secondary alcohol

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Both (A) and (R) are true and (R) is the correct EXPLANATION of (A)
Both (A) and (R) are true and (R) is not the correct explanation of (A)
(A) is true but (R) is FALSE
Both (A) and (R) are false

Answer :A
87736.

A reaction Xrarr product, completed 50% in 25 min. If concentration of 'X' is doubled, 50% completed in 50 min, the order of the reaction is

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0
1
2
3

Answer :A
87737.

(A) Reactions of higher order are rare (R ) The chances of multimolecular collisions are extremely less.

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Both (A) and (R ) are TRUE and (R ) is the correct EXPLANATION of (A)
Both (A) and (R ) are true and (R ) is not the correct explanation of (A)
(A) is true but (R ) is false
Both (A) and (R ) are false

Answer :A
87738.

A reaction which is of first order w.r.t reactant A has a rate constnat is 6"min"^(-1). If we start with [A]=0.5"mol"L^(-1) when would [A] reach the value of 0.05 "mol"L^(-1)

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0.384 min
15 min
20 min
3.84 min

Answer :A
87739.

A reaction which is first order with respect to A has rate constant 6 "min"^(-1). If we start with [A] = 0.5 "mol L"^(-1), when would [A] reach the value of 0.05 ML^(-1) ?

Answer»

SOLUTION :`k=(2.303)/t "log" ([A_0])/([A])`
`k=6 "MIN"^(-1), [A]_0 =0.5, [A]=0.05`, t=?
`t=(2.303)/6 "log" 0.5/0.05=2.303/6 "log"10` = 0.3838 min
87740.

A reaction was found to be second order with respect to the concentration of carbon monoxide . If the concentration of carbon monoxide is doubled with everything else kept the same the rate of reaction will :

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triple
INCREASE by a factor of 4
double
reamain unchanged

Solution :(B) `-(DX)/(dt) prop [CO]^(2) ` DOUBLING the conc .of CO rate will become four TIMES .
87741.

A reaction was found to be of second order with respect to concentration of carbon monoxide . If the concentration of carbon monoxide is doubled, the rate of reaction will :

Answer»

triple
increase by a FACTOR of 4
double
remain unchanged

Answer :B
87742.

A reaction that takes place with the absorption of energy is

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BURNING of a candle
RUSTING of iron
Electrolysis of water
Digestion of food

Solution :`H_(2)O_((L))OVERSET("electrolysis")rarrH_(2(g))+(1)/(2)O_(2(g))`.
87743.

A reaction that takes place with the absorption of energy is __________ .

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BURNING of a candle
Rusting of iron
electrolysis of water
respiration

Answer :C
87744.

A reaction taking palce with absorption of energy is:

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BURNING of a candle
Electrolysis of water
Digestion of food
Rusting or iron

Answer :B
87745.

A reaction takes place in various steps. The rate constant for first, second, third and fifth steps are k1, k2, k3 and k5 respectively. The overall rate constant is is given by: k = k_(2)/k_(1) (k_(1)/k_(2))^(1/2) activation energy are 40,60, 50 and 10 kJ/mol respectively, find the overall energy of activation (kJ/mol).

Answer»


SOLUTION :`k = A.e^(-E_(a)//(RT))`
`therefore` Effective overall energy of activation
`E_(a) = E_(a)(2) -E_(a)(3) + 1/2 E_(a)(1) -1/2 E_(a)(5)`
`=60-50 + 1/2 xx 40 - 1/2 xx 10 = 25` kJ/mol
87746.

A reaction takes place in various steps. The rate constatn for first, second, third and fifthsteps are k_(1),k_(2),k_(3) and k_(5) respectively The overall rate constant is given by k=(k_(2))/(k_(3))(k_(1)/(k_(5)))^(1//2) If activation energy are 40, 60, 50, and 10 kJ/mol respectively, the overall energy of activation (kJ/mol) is :

Answer»

10
20
25
none of these

Answer :C
87747.

A reaction takes place in a several sequential steps A,B,C and D. the vlaue of enthalpy change for sequential steps are p,q,r and s respectively. If A and C have equal values of enthalpy change with total enthalpy change is t, then choose the incorrect statement among the following.

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`q=r`
`q+r=t-(s+p)`
`2q=t-(s+p)`
`q+r=p+s`

SOLUTION :According to the question, the diagramatic representation of given reaction can be drawn as

As given, q=r[Hence, (a) is correct]
USING concept of Hess's law
`t=p+q+r+s`
`therefore q+r=t-(s+p)`[hence (b) is correct]
`2q=t-(s+p)`[hence (C) is correct]
`[becauseq=r]`
but q+r=p+s is INCORRECT because EXACT amount of enthalpy change for each step is not given.
87748.

A reaction takes place in 3 steps, the rate constant are K_(1),K_(2),K_(3) and energies of activation are 40, 30 and 20 Kj reaspectively. If overall rate constant K=(K_(1),K_(3))/(K_(2), the overall energy of activation is:

Answer»

10
15
30
60

Answer :C
87749.

A reaction S_8 (g) hArr 4S_2 (g) is carried out by taking 2 moles ofS_8(g)and 0.2 mol of S_2(g) in a reaction vessel of 1 litre. Which is (are) correct if K_c=6.30xx10^(-6)

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`K_p =2.55 atm^3`
REACTION PROCEEDS in backward direction
Reaction proceeds in forward direction
Reaction quotient is `8xx 10^(-4)`

Answer :A::B::D
87750.

A reaction rate is found to depend upon the two concentration terms. The order of the reaction is

Answer»

1
3
2
can't be predicted

Answer :D