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27001.

The number of isomers for the compound with molecular formula C_(2)BrCIFI is:

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3
4
5
6

Answer :D
27002.

The number of isomers exhibited by [Cr(NH_(3))_(3)Cl_(3)] is :

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2
3
4
5

Answer :A
27003.

The only cations present in a slightly acidic solution are Fe^(3+), Zn^(2+) and Cu^(2+) The reagent that when added in excess to this solution would identify and separate Fe^(3+) in one step is

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`2M HCI`
`6 M NH_(3)`
 `6 M NAOH`
`H_(2) S gas`

Answer :B
27004.

The number of isomeric xylenes is:

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3
2
4
5

Answer :A
27005.

The only cations present in slightly acidic solutions are Fe^(3+),Zn^(2+) and Cu^(2+). The reagent that when added in excess to this solution would identify and separate Feo in one step is

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2M HCl
`6M NH_(3)`
6M NaOH
`H_(2)S`

Solution :`NH_(3)` FORMS soluble COMPLEX with `ZN^(2+)` and `Cu^(2+)`m leaving `Fe^(2+)`. Hence, (B) is the correct answer
27006.

The number of isomeric structure for C_3H_9N would be:

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4
3
2
1

Answer :C
27007.

The number of isomeric pentyl alcohols possible is

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Two
Four
Six
Eight

Solution :Draw all possible isomers having all single bonds in them and count their total number.
(i) `CH_(3)CH_(2)CH_(2)CH_(2)CH_(2)OH`
(ii) `{:(CH_(3)CH_(2)CH_(2)CH_(2)CH_(3)),("|"),(""OH):}`
(iii) `{:(CH_(3)CH_(2)CH-CH_(2)CH_(3)),("|"),(""OH):}`
(IV) `{:(CH_(3)CH_(2)CH-CH_(2)OH),("|"),(""CH_(3)):}`(V) `{:(""OH),("|"),(CH_(3)CH_(2)C-CH_(3)),("|"),(""CH_(3)):}`
(vi) `{:(""CH_(2)OH),("|"),(CH_(3)CH_(2)CH-CH_(3)):}`(vii) `{:(""CH_(3) ),("|"),(CH_(3)-C-CH_(2)OH),("|"),(""CH_(3)):}`
(viii) `{:(""CH_(2)-OH),("|"),(CH_(3)-C-CH_(3)),("|"),(""CH_(3)):}`
27008.

The only cations present in a slightly acidic solution are Fe^(3+),Zn^(2+) and Cu^(2+) . The reagent that when added in excess to this solution would identify and separate Fe^(3+) in one step is

Answer»

2 M HCl
`6M NH_(3)`
`6M NaOH`
`H_(2)S` gas

Answer :D
27009.

The number of isomeric pentyl alchohols possible are:

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TWO
four
six
eight

Solution :SEE ART. 9.4 PROBLEM 2
27010.

Which of the following is a tetrabasic acid?

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ORTHO phosphorus ACID
Ortho phosphoric acid
Meta phosphoric acid
PYRO phosphoric acid

Answer :D
27011.

The number of isomeric ketones with formula C_(6)H_(12)O is

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SIX
TWO
FIVE
FOUR

ANSWER :A
27012.

The number of isomeric forms in which [Co(NH_3)_4Cl_2]^+ion can occur is :

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2
3
4
1

Answer :A
27013.

The one which has the highest value of pH is

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Distilled WATER
`NH_(3)` SOLUTION in water
`NH_(3)`
Water saturated with `Cl_(2)`

Solution :In water SOLUTIONS.
`NH_(3) + H_(2)O hArr NH_(4)^(+) + OH^(-)`
Concentration of `OH^(-)` is increased so that solution become more basic and the pH is increased.
27014.

The one which is least basic is

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`NH_3`
`C_6H_5NH_2`
`(C_6H_5)_3N`
(C_6H_5)_2NH`

ANSWER :C
27015.

The number of isomeric compounds of the formula C_(7)H_(7)Br is :

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Three
Four
Five
Six

Solution :
27016.

The one which has lowest ox. No. of Hg?

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`HG(NO_2)_2`
`HgCl_2`
`Hg(NO_3)_2`
Hg_2Cl_2

Answer :D
27017.

The number of isomeric carboxylic acids posible for C_(4)H_(9)COOHare :

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Four
Five
Three
Six.

Solution :`CH_(3)CH_(2)CH_(2)CH_(2)COOH , "" (CH_(3))_(2)CHCH_(2)COOH, "" CH_(3)CH_(2)CH(CH_(3))COOH,(CH_(3))_(3)C COOH`
27018.

The one which has least iodine value is?

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GINGER OIL
Ghee
Groundnut oil
SUNFLOWER oil

ANSWER :B
27019.

The one which has least iodine value is

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ginger OIL
ghee
groundnut oil
sunflower oil

Solution :The degree of unsaturation of a fat or oil is measured by its iodine number which is defined as the number of GRAMS of iodine what would add to carbon carbon double BONDS (C=C) present in 100 grams of the far or oil. Hence, greater the number of doubel bonds in the acid residues of a trglyceride, greatger will be the iodine number. Because a satuaratedfatty acid has no double bonds, it takes up no iodine, this its iodine number is zero.
Among the given options, ghee has the least iodine value as it is SATURATED whereas the other compounds conain some degree of unsaturation.
27020.

The number of isomeric alkyl bromides of molecular formula C_(4)H_(9)Br is :

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Four <BR>Five
Three
Six

Solution :(i) `CH_(3)CH_(2)CH_(2)CH_(2)Br` (ii) `CH_(3)CH_(2)UNDERSET(Br)underset(|)(C)HCH_(3)`
(iii) `CH_(3)-underset(CH_(3))underset(|)OVERSET(CH_(3))overset(|)(C)-Br` (iv) `CH_(3)underset(CH_(3))underset("|")(CHCH_(2)Br)`
27021.

The one which decreases with dilution is

Answer»

conductance(G)
specific conductance( K )
EQUIVALENT conductance `wedge_(eq)`
molar conductance `wedge_(m)`

ANSWER :C
27022.

The one which decreases with dilution is ………… .

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CONDUCTANCE
specific conductance
EQUIVALENT conductance
molar conductance

ANSWER :B
27023.

The one that is synthesized in skin is

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VITAMIN K
Vitamin C
Vitamin D
Vitamin E

Answer :C
27024.

The number of isomeric alcohols of formula C_4H_(10)O is:

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2
4
7
8

Answer :B
27025.

The one that is familiar as ascorbic acid is

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VITAMINE A
Vitamine B
Vitarnine C
Vitamine E

Answer :C
27026.

The number of isomeres possible for the octahedral complex [CoCl_(2)(en)(NH_(3))_(2)]^(+) is

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TWO
three
no ISOMER
FOUR isomers

Solution :
27027.

The one that is a good conductor of electricity in the folowing list of solids is :

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SODIUM chloride
Graphite
Diamond
Sodium carbonate

Answer :B
27028.

Thenumberof isomer(includingstereoisomers) of C_(5)H_(12)O whichcan givepositivehaloformtestis

Answer»


SOLUTION :NA
27029.

The one of the rarest element of group 18 is

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He
Ar
Ne
Xe

Answer :D
27030.

The one of the product formed in the reaction is (AAK_MCP_35_NEET_CHE_E35_005_Q01)

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SUBSTITUTED haloarene
Aromatic hydrocarbon
Organometallic compound
Substituted haloalkane

Answer :B
27031.

The number of ions present in K_(3)[Fe(CN)_(6)] are:

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10
4
3
2

Answer :B
27032.

The one letter code for the amino acid tryptophan is

Answer»

G
V
W
H

Solution :One letter CODE for TRYTOPHAN is 'W'.
27033.

The one electron species having ionisation energy of 54.4 eV is

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`H`
`He^(+)`
`B^(4+)`
`Li^(2+)`

SOLUTION :Ionisation energy = `(13.6Z^(2))/(n^(2))eV=13.6Z^(2)`
for ONE electron species.
`therefore13.6Z^(2)=54.4` or, `Z^(2)=4` or `Z=2`
i.e., `He^(+)`.
27034.

Number of ions present in 2.0 "litre" of a solution of 0.8 M K_(4)Fe(CN)_(6) is:

Answer»

`4.8xx10^(23)`
`4.8xx10^(24)`
`9.6xx10^(24)`
`9.6xx10^(22)`

Solution :Moles of `K_(4)[Fe(CN)_(6)]` present in 2 L of solution
`=(2 L) xx (0.8 mol L^(-1))=1.6` mol
`K_(4)[Fe(CN)_(6)]hArr4K^(+)+[Fe(CN)_(6)]^(-)`
No. of ions = 5
TOTAL ions present = `1.6xx5xxN_(0)`
`=1.6 mol xx 5xx 6.022 xx 10^(23) mol^(-1)`
`=4.8xx10^(24)`
27035.

The olefin which on ozonolysis gives CH_3CH_2CHO and CH_3CHO is

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1-butene
2-butene
1-pentene
2-pentene

Solution :`CH_3CH_2CH=CHCH_3 UNDERSET(Zn//H_2O)OVERSET(O_3)to underset"PROPANAL"(CH_3CH_2CHO)+underset"ETHANAL"(CHOCH_3)`
27036.

The number of ions per mole of the complex CoCl_(3).5NH_(3) in aqueous solution will be :

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3
9
2
4

Answer :A
27037.

The olefin which on ozonalysis gives CH_3CH_2 CHO and CH_3 CHO is

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2-pentene
2-butene
1-pentene
1-butene

Answer :A
27038.

The number of ions given by[Pt(NH_3)_2Cl_4]in aqueous solution is :

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Zero
1
2
4

Answer :A
27039.

The oil used in the froth floatation method for the purification of ores is :

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COCONUT OIL
KEROSENE oil
MUSTARD oil
PINE oil

Answer :D
27040.

The number of ions given by K[Pt(NH_3)Cl_5] in aqueous solution is :

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2
3
4
1

Answer :A
27041.

The OH^- ion concentration of a weak base is :

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C .`K_b`
SQRT(C . K_b)
sqrt(K_b/C)
sqrt(K_b)

ANSWER :B
27042.

The [OH^-] in 100.0mL of 0.016 M-HCl(aq) is :

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`5xx10^12 M`
`3xx10^(-10) M`
`6.25xx10^(-13) M`
`2.0xx10^(-9) M`

Answer :C
27043.

The number' of ions -given by K[Pt(NH_3)Cl_5] in aqueous solution is :

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2
3
4
1

Answer :A
27044.

The -OH group of methyl alcohol cannot be replaced by chlorine by the action of

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chlorine
hydrogen chioride
PHOSPHORUS -TRICHLORIDE
phosphorus pentachloride

Answer :A
27045.

The number of ions given by K_2PtCl_6 in aqueous solution is:、

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2
3
4
Zero

Answer :B
27046.

The - Oh group of an alcohol or COOH group of a carboxylic acid can be replaced by using:

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HYPOCHLOROUS acid
chlorine
hydrochloric acid
phosphorus pentachloride.

Answer :B
27047.

The number of ions given by [Co(NH_3)_4]C1_3 in aqueous solution is:

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2
3
1
4

Answer :D
27048.

The -OHgroup of methyl alcohol cannot be replaced by chlorine by the action of :

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Chlorine
HCI
`PCI_3`
`PCI_5`

ANSWER :A
27049.

The odd decomposition of carbon chlorine bond form

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Two FREE ions
two-carbonium ion
Two carbanion
A cation and an anion

Solution :`R-CH_(2)-Cl overset("Heterolytic bond fission")tounderset("cation")(RCH_(2)^(o+))+underset("anion")(Cl^(ө))`
Cl is more ELECTRONEGATIVE than `C` by which it form anion and hydrocarbon form cation.
27050.

The number of ions given by[CO(NH_(3))_(3)Cl_(3)] in aqueous solution is

Answer»

1
2
3
zero

Answer :D