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88951.

A liquid hydrocarbon can be converted to gaseous hydrocarbon by

Answer»

Cracking
Hydrolysis
Oxidation
Distillation under reduced pressure

Solution :`UNDERSET"HEXANE LIQUID"(C_6H_14)oversetDeltatounderset"Butane"(C_4H_10)+underset"ETHENE Gas"(C_2H_2)`
Discussion
88952.

A hydrocation (A) (C_(8) H_(10)) (E) is a stream volatile compound and on nitrationgives two mononitro derivatives. (A) gives the followingreactions. The name of (H) is:

Answer»

Dibenzocylobutane
Dibenzocyclobutadiene
Dibenzocylobutane
Benzocyclobutane

Solution :`D.U`. in`(A) = ((2n_(C) + 2) - n_(H))/(2)=((8xx2+2)-10)/(2)= 4^(@)`
`4 D.U`. and `C:H = 1:1` suggst that `(A)` contains benzene RING with two extra `C` atom [i.e., two `(Me)` groups]. Since compound`(A)` is steam volatie and on NITRATION GIVES two nitro-derivaties, so `(A)` is ortho-xylene.
Discussion
88953.

(A) Liquid helium is used in NMR spectrometers and MRI systems(R) Liquid helium can be used in making powerful super conducting magnets

Answer»

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
Discussion
88954.

(A) Liquid helium can be used as a cryogenic liquid(R) Liquid helium can produce a very low temperatures of around 4.2k

Answer»

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
Discussion
88955.

A hydrocarbon (X) with vapour density = 27 containing C = 88.88% decolourizes KMnO_(4) and Br_(2) water without evolving HBr. It gave no ppt. with ethereal ammonical AgNO_(3) or Cu_(2)CI_(2) When reacted with dil. H_(2)SO_(4) in the presence of mercuric sulphate, methyl ethyl ketone (Y) is formed. The compound is

Answer»

`C_(8)H_(12)`
`C_(4)H_(6)`
`C_(2)H_(3)`
`C_(5)H_(10)`

SOLUTION :Based on the percentages (88.88% C and 11,12% H) its emphirical formula `C_(2)H_(3)` `MW = 2 XX 27 = 54 , MF = C_(4)H_(6)`
Discussion
88956.

A liquid has vapour pressure 35xx10^(3)Nm^(-2) at 298 K. If the solution contains 0.2 mole fraction of a solute, the vapour pressure of the solution will be

Answer»

`2.8xx10^(3)NM^(-2)`
`7XX10^(3)Nm^(-2)`
`7.0xx10^(4)Nm^(-2)`
`28xx10^(3)Nm^(-2)`

Answer :A::B::C::D
Discussion
88957.

A hydrocarbon (X) reacts with Cl_2 giving the substitution product (Y). By the Wurtz reaction 'Y' yields a gaseous hydrocarbon, 'Z' having vapour density = 14. Identify X.

Answer»

`CH_4`
`CH_3 CH_3`
`CH_3 OH`
`CH_3Cl`

Solution :Vapour density of Z = 14.
`implies` MOLECULAR mass of the hydrocarbon (Z) = `2 xx 14 = 28` .
Hence the hydrocarbon is ethane `(C_2 H_6)` .
It can be obtained from METHYL chloride by Wurtz reaction , so the substitution product .Y. is `CH_3 - Cl` and the hydrocarbon .X. is `CH_4` (methane).
`underset("(X)")(CH_(4)) overset(Cl_(2)) (to) CH_(3) underset("(Y)") (C)l overset("Na/ether")("Wurtz reaction") (to) CH_(3) underset("(Z)")(-)CH_3`
Discussion
88958.

A liquid freezes at 300K and boils at 400K. If Kf and Kb values for the liquid are 5 and 2.5 ^(@)C/molalrespectively, then the ratio of latent heat of vaporisation to latent heat of fusion is :-

Answer»

`1:1`
`3:1`
`32:9`
`9:32`

SOLUTION :`Kb=(RTb^(2))/(1000 L_(V )) K_(f )=(RT_(f ) ^(2))/(1000 L_(f)`
Discussion
88959.

A hydrocarbon (X) of the formula C_6H_12 does not react withbromine water but reacts with bromine in presence of light,forming compound (Y).Compound (Y) on treatment with Alc. KOH gives compound (Z) which on ozonolysis gives (T) reduces Tollens reagent which gives compound (W) after acidification . (W) gives iodoform test and after acidification produce compound (U) which when heated with P_2O_5 forms a cyclic anhydride (V). Compound 'x' is

Answer»




SOLUTION :
Discussion
88960.

A hydrocarbon (X) of the formula C_6H_12 does not react withbromine water but reacts with bromine in presence of light,forming compound (Y).Compound (Y) on treatment with Alc. KOH gives compound (Z) which on ozonolysis gives (T) reduces Tollens reagent which gives compound (W) after acidification . (W) gives iodoform test and after acidification produce compound (U) which when heated with P_2O_5 forms a cyclic anhydride (V). Compound W is

Answer»

`COOH-(CH_2)_2-COOH`


ANSWER :C
Discussion
88961.

A hydrocarbon (X) of the formula C_6H_12 does not react withbromine water but reacts with bromine in presence of light,forming compound (Y).Compound (Y) on treatment with Alc. KOH gives compound (Z) which on ozonolysis gives (T) reduces Tollens reagent which gives compound (W) after acidification . (W) gives iodoform test and after acidification produce compound (U) which when heated with P_2O_5 forms a cyclic anhydride (V). CompoundV is

Answer»



`CH_3-oversetOoverset(||)C-CH_2-CH_2-CH=O`
`CHO-CH=CH-CHO`

ANSWER :B
Discussion
88962.

A liquid aerosol is a colloidal system of:

Answer»

A LIQUID dispersedin a solid
A liquid dispersed INA gas
A gas dispersed in a liquid
A solid dispersed in a gas

ANSWER :B
Discussion
88963.

A hydrocarbon X adds on one mole of hydrogen to give another hydrocarbon and decolourises bromine water. X reacts with KMnO_4in presence of acid to give two moles of the same carboxylic acid. The structure of Xis

Answer»

`CH_2=CH-CH_2CH_2CH_3`
`CH_3CH_2CH_2-CH=CHCH_3`
`CH_3CH_2CH=CHCH_2CH_3`
`CH_3CH=CHCH_2CH_2CH_3`

ANSWER :C
Discussion
88964.

(A) Liquid ammonia is used for refrigeration (R ) The moisture in ammonia is removed by using CaO

Answer»

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 :2
Discussion
88965.

A hydrocarbon X decolourises bromine water and with HI it forms isopropyl iodide X is :

Answer»

Propyne
Propylene
Ethylene
Isobutylene

Solution :`underset(("Decolourises bromine water"))(CH_(3) CH = CH_(2)) OVERSET(HI)RARR underset("Isopropyl iodide")(CH_(3) - underset(I)underset(|)(CH_(2)) - CH_(3))`
Discussion
88966.

A hydrocarbon with molecular formula C_5H_12on mono chlorination in presence of light gives four compounds A,B,C,D. A is optically inactive and dehydrohalgenation gives E( major) which on ozonolysis gives acetone and acetaldehyde. B is optically active gives on dehydrohalogenation E also gives E as the major product. Further C is optically active while D is optically inactive. Allcompounds A,B,C,D on reduction give 2-methyl butane.The structure B is

Answer»

`(CH_3)_2CHCHClCH_3`
`(CH_3)_2C(CL)CH_2CH_3`
`ClCH_2CH(CH_3)CH_2CH_3`
`(CH_3)_2CHCH_2CH_2Cl`

SOLUTION :
Discussion
88967.

A hydrocarbon with molecular formula C_(8)H_(18) gives only one monochloro derivative. It can be :

Answer»

n-octane
2-methyl HEPTANE
2, 2, 3, 3-tetramethyl butane
2, 2, 4-trimethyl pentane

ANSWER :C
Discussion
88968.

A hydrocarbon with molecular formula C_5H_12on mono chlorination in presence of light gives four compounds A,B,C,D. A is optically inactive and dehydrohalgenation gives E( major) which on ozonolysis gives acetone and acetaldehyde. B is optically active gives on dehydrohalogenation E also gives E as the major product. Further C is optically active while D is optically inactive. Allcompounds A,B,C,D on reduction give 2-methyl butane.Which of the following is likely structure of C

Answer»

`ClCH_2CH(CH_3)CH_2CH_3`
`(CH_3)_2CHCHClCH_3`
`(CH_3)_2CHCH_2CH_2Cl`
`CH_3CH_2CH(CL)CH_2CH_3`

SOLUTION :
Discussion
88969.

A hydrocarbon with molecular formula C_5H_12on mono chlorination in presence of light gives four compounds A,B,C,D. A is optically inactive and dehydrohalgenation gives E( major) which on ozonolysis gives acetone and acetaldehyde. B is optically active gives on dehydrohalogenation E also gives E as the major product. Further C is optically active while D is optically inactive. Allcompounds A,B,C,D on reduction give 2-methyl butane.Which of the following is the correct structure of A

Answer»

`ClCH_2CH(CH_3)CH_2CH_3`
`(CH_3)_2C(CL)CH_2CH_3`
`(CH_3)_2CHCH(Cl)CH_3`
`(CH_3)_2CHCH_2CH_2Cl`

SOLUTION :
Discussion
88970.

AhydrocarbonwithformulaC_(8)H_(12)givesonemonochloro derivative, thehydrocarbonis:

Answer»

N- OCTANE
2- methylheptane
2,2,4- trimethypentane
2,2,3,3- tetramehtybutane

ANSWER :D
Discussion
88971.

A hydrocarbon reacts with hypochlorous acid to give 2-chloroethanol. The hydrocarbon is:

Answer»

Methane
Ethylene
Acetylene
Ethane

Answer :A
Discussion
88972.

A hydrocarbon reacts with hypochlorous acid to give 1-chloro-2-hydroxyethane. The hydrocarbon is

Answer»

Ethylene
Methane
Ethane
Acetylene

Solution :`CH_2=CH_2+Hoverset-OCoverset+l to UNDERSET"1-Choro-2-hydroxy ethane "(undersetunderset(OH)(|)OVERSET(2)CH_2-undersetunderset(CL)(|)overset1CH_2)`
Discussion
88973.

A hydrocarbon reacts with hypochlorous acid to give 1-chlor-2-hydroxy ethane. The hydrocabon is

Answer»

METHANE
Acetylene
Ethylene
Ethane

Answer :C
Discussion
88974.

A hydrocarbon (R) has six membered ring in which there is no unsaturation.Two alkyl group are attached to the ring adjacent to each other.One group has 3 carbon atoms with branching at 1^(st) carbon and another has 4 carbon atoms.The larger alkyl group has main chain of three carbon atoms of which second carbon is substituted. Correct IUPAC name of compound (R) is

Answer»

1-(1-Methylethyl)-2-(1-methylpropyl)cyclohexane
1-(2-Methylethyl)-2-(1-methylpropyl)cyclohexane
1-(1-Methylethyl)-2-(2-methylpropyl)cyclohexane
1-(1-Methylethyl)-2-butylclohexane

Solution :
Discussion
88975.

A hydrocarbon 'P' having molecular formula C_(8)H_(10) can form only three monochloroderivative. 'P' on oxidation with KMnO_(4) yields a compound, which on heating with aq. NH_(3) yields 'Q'. 'Q' on treatment with NaOH followed by NaOCl yields a salt which on acidification yields with (NaNO_(2)+HCl) and then with base to give dipolar ion 'S' which readily losses two gases two give an intermediate 'T' is finally converted into hydrocarbon 'U' on dimerisation. degree of unsaturation in the hydrocarbon 'U' will be:

Answer»


ANSWER :9
Discussion
88976.

A liquid (A) is treated with Na_(2) CO_(3) solution. A mixture of two salts (B ) and ( C ) are product in the solution. The mixture on acidification with sulphuric acid and distillation produces the liquid (A ) again. Which of the following Halogen molecule can have Reddish brown colour ?

Answer»

`F_(2)`
`Br_(2)`
`Cl_(2)`
`I_(2)`

Solution :Halogen molecule can have reddish brown COLOUR is `Br_(2)`.
`underset("(X)")(CaOCl_(2))+ 2KI + 2HCl RARR CaCl_(2) + 2KCl + H_(2) O +I_(2)`
`CaOCl_(2) + 2KI + 2HCl rarr CaCl_(2) + 2KCl + H_(2) O + I_(2) , CaOCl_(2) + CO_(2) rarr CaCO_(3) + Cl_(2)`
Discussion
88977.

A hydrocarbon of molecular mass of 72 g mol^(-1) gives a single monochloro derivative and two dichloro derivatives on photochlorination. Give the structures of the hydrocarbon.

Answer»

Solution :The hydrocarbon having molecular mass of 72 g `mol^(-1)` must be `C_(5)H_(12)`. Since on photochlorination, it gives a single monochloro derivative, THEREFORE, all the twelve HYDROGENS in this hydrocarbon are equivalent. Thus, the hydrocarbon must be neopentane or 2,2-dimethylpropane, it forms only one monochloro derivative and two dichloro derivatives as shown below:
`underset("2,2-Dimethylpropane (All the 12 H's are equivalent)")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-CH_(3))""underset("1-Chloro-2,2-dimethylpropane (Monochloro derivative)")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-CH_(2)Cl)""underset("1,3-Dichloro-2,2-dimethylpropane (Dichloro derivative)")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(2)Cl)overset(|)(C)-CH_(2)Cl)""underset("1,1-Dichloro-2,2-dimethylpropane (Dichloro derivative)")(CH_(3)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-CHCl_(2))`.
Discussion
88978.

A liquid (A) is treated with Na_(2) CO_(3) solution. A mixture of two salts (B ) and ( C ) are product in the solution. The mixture on acidification with sulphuric acid and distillation produces the liquid (A ) again.The mixture of salt's are

Answer»

NABR, NaCl
NaBr`, NaBrO_(3)`
both a and b
NONE of these

Solution :MIXTURE salts are `NaBr, NaBrO_(3)`
Discussion
88979.

A liquid (A) is treated with Na_(2) CO_(3) solution. A mixture of two salts (B ) and ( C ) are product in the solution. The mixture on acidification with sulphuric acid and distillation produces the liquid (A ) again.The liquid compound 'A' is

Answer»

`H_(2)O`
`Br_(2)`
`I_(2)`
NONE of these

SOLUTION :LIQUID COMPOUND A is `Br_(2)`
Discussion
88980.

A hydrocarbon of molecular mass 72 g mol^(-1) gives a single monochloro derivative and two dichloro derivatives on photo chlorination. Give the structure of the hydrocarbon.

Answer»

Solution :`C_(5)H_(12)`, Pentane has MOLECULAR mass 72 g `MOL^(-1)`. Neo-pentane YIELD monochlorinated product because all 1,2-hydrogen are equivalent.
`CH_(3)-overset(CH_(3))overset("|")underset(CH_(3))underset("|")("C")-CH_(3) overset(Cl_(2))underset(" Light ")rarr CH_(3)-overset(CH_(3))overset("|")underset(CH_(3))underset("|")("C")-CH_(2)Cl`
Dichloro derivatives :
`underset((i))(CH_(3)-overset(CH_(3))overset("|")underset(CH_(3))underset("|")("C ")-CHCl_(2))`
`CH_(3)-overset(CH_(2)-Cl"")overset("|")underset(CH_(2)-Cl" ")underset("|")(C-CH_(3))`
Discussion
88981.

A hydrocarbon of formula C_6H_10 absorbs only one molecule of H_2 upon catalytic hydrogenation. Upon ozonolysis, the hydrocarbon yields O=oversetoversetH|C-CH_2-CH_2-CH_2-CH_2-oversetoversetH|C=O The hydrocarbon is

Answer»

CYCLOHEXANE
Benzene
Cyclohexene
Cyclobutane

Solution :
Discussion
88982.

A hydrocarbon of formula C_(6)H_(8) absorbs two mole of H_(2) upon hydrogenation. On ozonolysis, it yields Propanedial. What is hydrocarbon ?

Answer»


SOLUTION :HYDROCARBON is CYCLOHEXA-`1,4`-DIENE.
Discussion
88983.

A liqud mixture of two volatile liquids has composition corresponding to point P in the temperature vs composition graph and is subjected to distillation from the composition P. Which of the following statements is incorrect?

Answer»

The composition of distillate will be different
the boiling point of the solution will keep on increasing.
The mixture will have highest vapour PRESSURE at the composition.
The composition of P will CHANGE if EXTERNAL pressure changes.

Answer :A::B
Discussion
88984.

A hydrocarbon of formula C_(10)H_(16) absorbs only one mole of H_(2) upon hydrogenation. On ozonolysis, it yields 1, 6-Cyclodecanedione. What is the structure of the hydrocarbon ?

Answer»


SOLUTION :
Discussion
88985.

A linear product of phenol and formaldehyde is used

Answer»

in PAINTS 
in EXPLOSIVE MIXTURE 
GEARS 
in utensils

Answer :C
Discussion
88986.

A hydrocarbon is composed of 75% carbon. The empirical formula of the compound is

Answer»

`CH_2`
`CH_3`
`C_2H_5`
`CH_4`

Solution :Carbon = 75%
Hydrogen = 100 - 75 = 25%

`THEREFORE ` EMPRICAL FORMULA ` = CH_4`
Discussion
88987.

A hydrocarbon K adds one mole of hydrogen in presence of platinum catalyst to form n-hexane. When K is oxidised vigorously with KMnO_4 a single carboxylic acid is isolated and no gas evolved.The correct stablement about 'K' is

Answer»

It is a chiral compound
It has two STEREOISOMERS
It has free ROTATION about all C-C BONDS
its ozononlysis product gives POSITIVE iodoform test

Solution :
Discussion
88988.

A linear molecule is:

Answer»

`Cl_2O`
`H_2 O`
`ClO_2`
`BeCl_2`

ANSWER :D
Discussion
88989.

A linear chain compound 'A' (C_(5)H_(10)O_(2)) gives brisk effervescence of CO_(2) when treated with saturated aqueous NaHCO_(3) solution. It reacts with chlorine gas in the presence of catalytic amount of red phosphorous to givecompound 'B' . Identify compound 'A' and 'B' . Identify the chiral isomer of compound 'A'/

Answer»

Pentanoic ACID , 2-chloropentanoic acid, 2-methylbutanoic acid
2-Methyl BUTANOIC acid, 2-chloro-2-methylbutanoic acid , pentnoic acid
Pentanoic acid, 2-chloropentanoic acid, 2,2-dimethylpropanoic acid
2,2-Dimethylpropanoic acid, 2-chloro-2-Methylpropanoic acid, pentanoic acid

Answer :A
Discussion
88990.

A hydrocarbon has the formula C_3H_4 To fine out whether it contains two double bond or triple bond, the following test is performed:

Answer»

PASSED through AMMONIACAL `AgNO_3`
TREATED with BAEYER's reagent
Treated with FEHLING's solution
Treated with `Br_2`water

Answer :A
Discussion
88991.

A hydrocarbon has the molar mass 86. The number of chain isomers possible for the compound is :

Answer»

Five
Six
Four
Ten

Answer :A
Discussion
88992.

A (Light pink colour complex) underset(Delta)overset(Pb_(2)O//dil." "HNO_(2)) to HMnO_(4) overset(H_(2)S //H^(+)) to A (Light pink colour complex). Calculate CFSE value in light pink colour complex.

Answer»


SOLUTION :`implies`Pink COLOUR complex`:[Mn(H_(20O)_(6)]^(2+)`
`Mn^(2+)(3d^(5))` in `[Mn(H_(2)O)_(6)]^(+ +),Delta_(0) LT P`

CFSE of pink colour complex
`=3xx(-(2)/(5)Delta_(0))+2xx[(3)/(5)Delta_(0)]=0`
Discussion
88993.

A hydrocarbon has density 1.25 gL^(-1)at STP . The hydrocarbonis

Answer»

`C_(2)H_(4)`
`C_(2)H_(2)`
`C_(2)H_(6)`
`C_(3)H_(6)`

SOLUTION :1L of hydrocarbon at STP weighs `= 1.25 g`
`22.4 L` of hydrocarbon at STP weigh
`= 1.25 xx 22.4 = 5/4 xx 22.4 = 28.0 g mol^(-1)`
Hydrocarbon will be `C_(2)H_(4) = 2xx 12 + 4 = 28 g mol^(-1)`
(B) , (c ) and (d) are not possible.
Discussion
88994.

A lighting flash through air may result in the formation of :

Answer»

nitrogen pentaoxide
AMMONIA
nitric acid
nitric oxide

Solution :`UNDERSET(("From air"))(N_(2)+O_(2))overset("Electric disch ARGE ")to 2NO`
Discussion
88995.

A hydrocarbon contains 91.3% carbon by mass. Find the empirical formula of hydrocarbon?

Answer»

`CH`
`C_2H_3`
`C_7H_8`
`C_3H_5`

ANSWER :C
Discussion
88996.

A light greenish coloured salt was soluble in water. On passing H_2 S into the solution, a black ppt. was obtained which dissolved readily in HCl. The metal ion present is

Answer»

`Co^(2+)`
`Fe^(2+)`
`Ni^(2+)`
`Mn^(2+)`

SOLUTION :`Fe^(2+)+H_(2)S to underset("Black PPT.")(FeS darr) +2H^(+)`
`FeS(s)+2HCl(aq) to FeCl_(2)(aq)+H_(2)S(g)`
Discussion
88997.

A hydrocarbon contains 10.5 g of carbon per gram of hydrogen. 1 L of vapour of the hydrocarbon at 127^(@)C and 1 atm pressure weighs 2.8 g. Find the molecular formula of the hydrocarbon.

Answer»

<P>

Solution :Volume at NTP`= (p_(1)V_(1))/(T_(1)) xx (T_(2))/(P_(2))`
`=(1 xx 1)/(400) xx (273)/(1) = (273)/(400)` litre.
`because (273)/(400)` litre weighs 2.8g at NTP
`therefore` 22.4 litres (1 mole) weights `=(2.8 xx 400)/(273) xx 22.4g`
=91.89g `~~92g`
`therefore` molecular weight of the HYDROCARBON =92
`therefore{(,"Weight","Mole",),(C,10.5g,10.5//12,=0.875),(H,1g,1//1,=1):}` or `C:H= 0.875:1`
`=0.875 xx 8: 1 xx`
`=7: 8`
Thus empirical formula is `C_(7)H_(8)`
Now empirical formula weight =92= molecular weight
Hence molecular formula is also `C_(7)H_(8)`
Discussion
88998.

A light greenish coloured salt was soluble in water. On passing H_(2)S into the solution a black precipiate was obatined which dissolve readily in HCl. The metal ion present is

Answer»

`CO^(+2)`
`Fe^(2+)`
`NI^(+2)`
`Mn^(+2)`

Solution :FeS is soluble in HCl. ALSO `Fe^(2+)` salts are green.
Discussion
88999.

A light green coloured salt was soluble in water. On passing H_(2)S into the solution a black precipitate was obtained which dissolved readily in HC. The metal ion present is

Answer»

`Co^(2+)`
`Fe^(2+)`
`NI^(2+)`
`MN^(2+)`

ANSWER :B
Discussion
89000.

A light green salt soluble in water gives blackprecipitate onpassing H_(2)S which dissolves readily in HCI .The metalion present is

Answer»

`Co^(2+)`
`Fe^(2+)`
`NI^(2+)`
`Ag^(o+)`

SOLUTION :`FES` issoluble in HCI, Also `Fe^(2+)` salts are GREEN
Discussion