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92201.

4 mL of a gas at 1 atm pressure and 300 K is dissolved in 1 L of solution. Calculate the volume of the gas dissolved at 4 atm pressure and 300 K in (1)/(2)L of solution in mL.

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ANSWER :8
Discussion
92202.

Write chemical reactions of affect the 4-Methylacetophenone to benzene -1,4 -dicarboxylic acid transformations.

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SOLUTION :
Discussion
92203.

4-methoxy acetophenone is obtained from

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ANSWER :C
Discussion
92204.

4 L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water. The molality of the resultant solution is ……

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`0.004`
`0.008`
`0.012`
`0.016`

SOLUTION :EXPLANATION :
APPLY the reaction : `M_(1)V_(1)=M_(2)V_(2)`
GIVEN : `M_(1)=0.02 M, V_(1)=4 L, M_(2)=? V_(2)=5L`
Therefore, `0.02xx4L=M_(2)XX 5L`
`M_(2)=0.08//5=0.016 M`
Discussion
92205.

4-Methoxyacetophenone can be prepared from anisole by

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Reimer-Tiemann reaction
Kolbe's reaction
Friedel-Crafts reaction
Wurtz reaction

Solution :
Discussion
92206.

4-Hydroxy-4-methylpentanal on heating with excess of methanol in the presence of an acidcatalyst followed by dehydration of the products gives

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ANSWER :B
Discussion
92207.

4 gm of the chloroplatinate salt of triacidic base 'B' on ignition gave 1 gm of Platinum then moar mass of base 'B' in gm is -

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585
555
1110
1170

Solution :`M = (N)/(2) [(w xx 195)/(x) - 410] g mol^(-1)`
`M = (3)/(2) [(4 xx 195)/(1) - 410]`
`M = (3)/(2) [780 - 410] rArr M = 555 gm//mol`
Discussion
92208.

4 gm of NaoH are added in 1 litre. The PH value of the solution will be :

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1
0
7
13

Answer :D
Discussion
92209.

4 g of copper was dissolved in concentrated nitric acid. The copper nitrate solutio on strong heating gave 5g of its oxide. The equivalent weight of copper is

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23
32
12
20

Solution :In 5 gm `CuO,4gm` Cu and 1 gm O be PRESENT.

EMPIRICAL formula `=CuO` of oxide
In this oxide, oxidation no. of Cu=+2
EQUIVALENT weight=`("Molecular weight")/("Oxidation no.")=(63.5)/(2)~~31.75`
But equivalent weight should be an INTEGRAL no.=32.
=9.65A
Discussion
92210.

4 g of copper was dissolved in concentrated nitric acid. The copper nitrate on strong heating gave 5 g of its oxide. The equivalent weight of copper is

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23
32
12
20

Solution :Cu = 4gCuO = 5G` THEREFORE ` Oxygen = 1 g
1G of oxygen combines with Cu = 4g
8g of oxygen will combine with `Cu = 4 xx 8 = 32g`
` therefore` Eq. wt. of Cu = 32
Discussion
92211.

4 g of caustic soda (molar mas = 40 g mol^(-1)) are dissolved in water and the solution is made to 200 mL. Calculate the molarity of the solution.

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Solution :`"Molarity of solution (M)"=("Mass of SOLUTE/Molar mass of solute")/("VOLUME of solution in litres")`
`=(4g//(40" G mol"^(-1)))/((200//1000)L)=0.5" mol L"^(-1)(0.5 M)`
Discussion
92212.

4 g carbon were heated with 8 g of sulphur. How much carbon disulphide (CS_(2)) will be formed when that reaction is complete ? What will be its percentage purity?

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Solution :`underset(12g)(C)+underset(2xx32=64g)(2S)rarrunderset(12+2xx32=76g)(CS_(2))`
Obviously, sulphur will be the limiting reactant.
`"8 g sulphur will produce CS"_(2)=(76)/(64)xx8=9.5g`
`"CARBON reacted "=(12)/(64)xx8=1.5g`
`"Carbon left "=4-1.5=2.5g`
`"Total mass of products"=9.5+2.5=12g`
`therefore %" purity of "CS_(2)" in the product "=(9.5)/(12)xx100=79.2%.`
Discussion
92213.

4-chloro-3, 5-dimethyl phenol is called

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Chloramphenicol
Paracetamol
Barbital
Dettol

Answer :D
Discussion
92214.

4-chloro-3,5-dimethyl phenol is called

Answer»

chloramphenicol
paracetamol
barbital
DETTOL

Solution :
4-chloro-3,5-dimethyl PHENOL is CALLED dettol.
Discussion
92215.

4 A neutral, resolvable organic compound A has molecular formula (C_(8)H_(16)O_(2)). A on treatment with LiAlH_(4) gives isomeric B and C(C_(4)H_(10)O) of which only B is optically active. B on treatment with acidified dichromate solution gives D (C_(4)H_(8)O), which on refluxing with dilute NaOH followed by acidification of product gave E (C_(8)H_(14)O). E on heating with N_(2)H_(4) in alkaline medium affords F(C_(8)H_(16)) which on treatment with B.H/H,O,/HO produced a resolvable G (C_(8)H_(18)O).G on treatment with acidified dichromate solution produced H (C_(8)H_(16)O), which on treatment with MCPBA produces A. Compound (H) is

Answer»




ANSWER :B
Discussion
92216.

4-bromo-9-methyltricyclo[5.3.0.0^(2.6)]

Answer»

SOLUTION :
Discussion
92217.

4-bromophenol is mainly formed, when phenol is reacted with,

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`Br_2`/WATER
`Br_2`/INERT solvent
HBr/water
HBr/inert solvent

Answer :B
Discussion
92218.

4-alkoxy alkyl benzene is obtained from

Answer»

FRIEDEL - CRAFT reaction
Ulmann reaction
Wurtz - FITTIG reaction
Fittig reaction

Answer :A
Discussion
92219.

4 A neutral, resolvable organic compound A has molecular formula (C_(8)H_(16)O_(2)). A on treatment with LiAlH_(4) gives isomeric B and C(C_(4)H_(10)O) of which only B is optically active. B on treatment with acidified dichromate solution gives D (C_(4)H_(8)O), which on refluxing with dilute NaOH followed by acidification of product gave E (C_(8)H_(14)O). E on heating with N_(2)H_(4) in alkaline medium affords F(C_(8)H_(16)) which on treatment with B.H/H,O,/HO produced a resolvable G (C_(8)H_(18)O).G on treatment with acidified dichromate solution produced H (C_(8)H_(16)O), which on treatment with MCPBA produces A. Compound (A) is

Answer»




ANSWER :B
Discussion
92220.

4 A neutral, resolvable organic compound A has molecular formula (C_(8)H_(16)O_(2)). A on treatment with LiAlH_(4) gives isomeric B and C(C_(4)H_(10)O) of which only B is optically active. B on treatment with acidified dichromate solution gives D (C_(4)H_(8)O), which on refluxing with dilute NaOH followed by acidification of product gave E (C_(8)H_(14)O). E on heating with N_(2)H_(4) in alkaline medium affords F(C_(8)H_(16)) which on treatment with B.H/H,O,/HO produced a resolvable G (C_(8)H_(18)O).G on treatment with acidified dichromate solution produced H (C_(8)H_(16)O), which on treatment with MCPBA produces A. Compound (B) is

Answer»




ANSWER :B
Discussion
92221.

4 A neutral, resolvable organic compound A has molecular formula (C_(8)H_(16)O_(2)). A on treatment with LiAlH_(4) gives isomeric B and C(C_(4)H_(10)O) of which only B is optically active. B on treatment with acidified dichromate solution gives D (C_(4)H_(8)O), which on refluxing with dilute NaOH followed by acidification of product gave E (C_(8)H_(14)O). E on heating with N_(2)H_(4) in alkaline medium affords F(C_(8)H_(16)) which on treatment with B.H/H,O,/HO produced a resolvable G (C_(8)H_(18)O).G on treatment with acidified dichromate solution produced H (C_(8)H_(16)O), which on treatment with MCPBA produces A. Compound (E)

Answer»




NONE of these

Answer :B
Discussion
92222.

4 : 4 coordination is noticed in:

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ZnS
CuCI
AgI
All

Answer :D
Discussion
92223.

3xx10^(-3) kg acetic acid is added into 500 cm^(3) water. If dissociation of acetic acid is 23% then find out depression in freezing ? K_(f) of water = 1.86 K kg / mol and density = 0.997 gm cm^(-3).

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ANSWER :`0.229`
Discussion
92224.

3HCHO+CH_(3)CHOoverset(NaOH)rarrA. A found can

Answer»

REDUCE Tolen's reagent
Give Cannizaro reaction
React wit Na
Give green colour with `Cr_(2)O_(7)^(2-)//H^(+)`

Solution :`2CH_(3)-UNDERSET(O)underset(||)C-CH_(3)underset(H^(+))overset(Mg//Hg)RARR` Product, product in the reaction is :
Discussion
92225.

3g of metal ions were discharged at cathode using a current of 3 amperes for 2 hours from aqueous cupric sulphate solution. Calculate the current efficiency. At wt . of Cu is 63.5 .

Answer»


ANSWER :0.422
Discussion
92226.

3g of solute is dissolved in 22g of water. The depression if freezing point shown by the solution is 1.45K. What is the molar mass of solute ? K_f for water is = 1.86 "K kgmol"^(-1)

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ANSWER :174.9
Discussion
92227.

3g of hydrocarbon on combustion with 11.2 g of oxygen produce 8.8 g of CO_(2) and 5.4 g of H_(2)O. The data illustrate the law of:

Answer»

conservation of MASS
multiple proportions
constant proportions
RECIPROCAL proportions

Solution :`sum` masses of REACTANTS =`sum` Masses of products
`(3+11.2)G (8.8+5.4)g`
Hence, law of conservation of mass is verified.
Discussion
92228.

3g of an oxide of a metal is converted to chloride completely and it yields 5g chloride. The equivalent weight of metal is:

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33.25
3.325
12
20

Answer :A
Discussion
92229.

3g of an oxide of a metal is converted to chloride completely and it yielded 5g of chloride. The equivalent weight of metal is

Answer»

12
20
33.25
2.325

Solution :3 GRAM of an oxide of metal is converted to chloride COMPLETELY.
It YIELDED 5 gram of chloride.
Equivalent weight of metal E
`therefore (E+35.5)/(5)=(E+8)/(3)`
`implies5E+40=3E+106.5impliesE=33.25`.
Discussion
92230.

3g of an oxide of a metal is converted completely to 5 g chloride. Equivalent mass of metal is:

Answer»

33.25
3.325
12
20

Answer :A
Discussion
92231.

3d-series elements exhibit vaiable oxidation states. Why ?

Answer»

Solution :(i) DUE to PRESENCE of vacant d-orbitals
(ii) The energy GAP 4S and 4d is LESS.
Discussion
92232.

Derive an integrated rate equation for the rate constant of a first-order reaction.

Answer»

Solution :Consider a general first order reaction.
`R to ` Products.
The differential rate equation is
rate`=(d[R])/(dt)=d[R]`
Where K, is the rate CONSTANT of the first order reaction.
On REARRANGING`(d[R])/([R])= -k dt`
Integrating on both the sides, we get.
`ln[R]= -Kt +I ""...(i)`
Where I is the constant of Integration.
When `t = 0, [R] = [R]_(0)`and hence using equation (i)
`I= ln[R]_(0)""...(ii)`
Rearranging this equation.
`ln[R]_(0)-ln[R] = Kt`
or `K=(1)/(t)"ln" ([R]_(0))/([R])""...(iii)`
or `K= (2.303)/(t)"log"([R]_(0))/([R])""...(iv)`
Discussion
92233.

3ClO^(-)toClO_(3)^(-)+2Cl The rate of reaction of above reaction is given by…….

Answer»

`K_(1)[CLO^(-)]`
`K_(1)[Cl^(-)]^(2)`
`K_(1)[ClO_(2)^(-)]`
`K_(1)[ClO^(-)]^(2)`

ANSWER :D
Discussion
92234.

3CH_(3)CH_(2)underset(CH_(3))underset(|)CHCH_(2)OH underset(-H_(3)PO_(3))overset(PBr_(3))to 3X+ H_(3)PO_(3) What is X?

Answer»

1- bromo - 2 - METHYL butane<BR>3- methyl - 4 - bromo butane
1- bromo - 3- methyl butane
3- bromo -1-methyl butane

Solution :`3CH_(3)CH_(2)underset(CH_(3))underset(|)CHCH_(2)OH underset(-H_(3)PO_(3))overset(PBr_(3))to 3CH_(3)CH_(2)underset(CH_(3))underset(|)CH_(2)Br`
Discussion
92235.

3ClO^-(aq) rarr ClO_3^(-) + 2CI^- is an example of

Answer»

OXIDATION reaction
reduction reaction
disproportionation reaction
decomposition reaction.

Solution :
Discussion
92236.

3(CH_(3)(CH= CH_(2))) underset(" Hydroboration")overset((BH_(3))_(2))to X Identify X.

Answer»

`(CH_(3)CH_(2)CH_(2))_(3).B`
`(CH_(3)CH_(2)CH_(2))_(2)B`
`CH_(3)CH_(2)CH_(2)B`
`CH_(3)-CH(OH)CH_(3).B`

SOLUTION :`3(CH_(3)CH= CH_(2)) underset("HYDROBORATION")OVERSET(BH_(3))to (CH_(3)CH_(2)CH_(2))B`
Discussion
92237.

3C_2H_5OH+PCl_3 rarr 3C_2H_5Cl + X,where 'X' is

Answer»

`H_3PO_2`
`H_3PO_4`
`H_3PO_3`
`H_4 P_2 P_7`

ANSWER :C
Discussion
92238.

3A to 2B , rate of reaction +(d[B])/(dt) is equal to

Answer»

`-(3)/(2)(d[A])/(DT)`
`-(2)/(3)(d[A])/(dt)`
`+2(d[A])/(dt)`
`-(1)/(3)(d[A])/(dt)`

Answer :B
Discussion
92239.

3AtoB + Citwouldbe a zeroorderreaction , when

Answer»

the rate ofreactionisproportinaltosquareofconcentrationof A
the rateof reactionremains sameat anyconcentration of A
the rateremainsuncharged at any concentration of B and C
the rateor reaction DOUBLES ifconcentration of B isincreased todouble

Solution :forreaction
`3A to B+C`
if it iszeroreaction then the rateremains same at anyconcentrationof `AOR (dx)/(dt)=K[A^(0)][A^(0)=1]`
itmeamns thatfor zeroorderreactionrate is independent of concentration of reactants
Discussion
92240.

3A t0 2B ,rate of reaction + (d[B])/(dt)is equal to

Answer»

`-(3)/(2) (d[A])/(dt)`
`-(2)/(3)(d[B])/(dt)`
`+2(d[A])/(dt)`
`-(1)/(3)(d[A])/(dt)`

Solution : `:-(1)/(3)(d[A])/(dt)=+(d[B])/(dt)`
`-(2)/(3)(d[A])/(dt)=+(d[B])/(dt)`
Discussion
92241.

3.92 g. of Mohr salt [mol. wt = 392] is present in 100 ml of an aqueous solution. The sulphate ion concentration of the resulting solution is

Answer»

1M
2M
0.2M
0.02 M

ANSWER :C
Discussion
92242.

3.92g//Lofa sampleofferrousammoniumsulphatereactscompletelywith50 mL(N )/(10 )KMnO_4solutionthe percentagepurityof the sampleis

Answer»

50
`78.4`
`80.0`
`39.2`

SOLUTION :`N_1xxV_1= N_2 xxV_2`
` [FeSO_4(NH_4)_2 SO_4 .^H_2O] [KMnO_4]`
` N_1xx 100 =1/100=1/10xx 50orN_1 =(1)/( 200)`
eq WT of ` FeSO_4.(NH_4)_2 SO_4 .^H_2O= Molwt= 392`
` THEREFORE ` Strengthofpuresalt`= 392 xx (1)/(200 ) = 1.96gL^(-1)`
` therefore` % purity = `( 1.96)/ ( 3.92 ) xx 100 =50 % `
Discussion
92243.

3.9 g of benzoic acid dissolved in 49 g of benzne shows a depreesionin freezingpointof 1.62 K. Calculate the van't Haff factorand predict the natureof solutei(associated or dissociated ). [Given:Molarmass of benzoic acid= 122g mol^(-1), K_(f) for benzene = 4.9 K kgmol^(-1)].

Answer»

Solution :`Delta T_(F) = ik_(f) XX m`
` = (ixxk_(f) xx w_(B) xx 100)/(M_(B) xx w_(A))`
` 1.62= (ixx4.9 xx 3.9 xx 1000)/(122xx49)`
` or "" i= (1.62 xx122xx49)/(4.9 xx 3.9 xx 1000) = 0.506`
Since .i.is lessthanone the solute is associated .
Discussion
92244.

3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of solute (associated for dissociated). ("Given : Molar mass of benzoic acid = 122 g mol"^(-1),K_(f)" for benzene = 4.9 K kg mol"^(-1))

Answer»


Solution :`DeltaT_(f)"(calculated)"=(1000K_(f)w_(2))/(w_(1)M_(2))=("1000 g kg"^(-1)xx"4.9 K kg mol"^(-1)xx3.9 g)/(49g xx"122 g mol"^(-1))=3.197K`
`""("Mol. mass of BENZOIC acid "C_(6)H_(5)COOH="122 g mol"^(-1))`
`i=(Delta_(f)"(observed)")/(DeltaT_(f)"(calculated)")=(1.62)/(3.197)=0.506`
Alternatively, MOLALITY `(m)=(3.9)/(122)xx(1)/(49)xx1000=0.652, DeltaT_(f)=iK_(f)m`
`"or"i=(DeltaT_(f))/(K_(f)m)=(1.62)/(4.9xx0.652)=0.507.`
As i lt 1, solute is associated.
Discussion
92245.

3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van't Hoff factor and predict the nature of solute (associated or dissociated). [Given : Molar mass of benzoic acid = 122 g "mol"^(-1), K_f for benzene = 4.9 K kg "mol"^(-1) ]

Answer»

Solution :Apply the relation, `M_2 = (1000 xx w_2 xx K_f)/(DeltaT_f xx w_1)`
SUBSTITUTING the values, we have
`M_2 = (1000 G KG^(-1) xx 3.9 g xx 4.9 K kg "MOL"^(-1))/(1.62 k xx 49 g) = 240 g "mol"^(-1)`
van.t Hoff factor = Normal molar mass / ABNORMAL molar mass
` = 122/240 = 0.508`
Benzoic acid gets associated (dimerised).
Discussion
92246.

3.9 g of benzoic acid dissoved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate Van't Hoff factor and predict the nature of solute (associated or dissociated) . (Given. Molar mass of benzoic acid = 122 g mol^(-1) , K_(f) for benzene = 4.9 K kg mol^(-1)).

Answer»


Solution :`i=(DeltaT_(f))/(K_(f)xxm)DeltaT_(f)=1.62 K, K_(f)=4.9" K kg mol"^(-1), M=((3.9g)//(122" g mol"^(-1)))/((0.049 kg))`
`i=(1.62Kxx(0.049" kg")XX(122" g mol"^(-1)))/((4.9"K kg mol"^(-1))xx(3.9g))=0.506`
SCINCE Van't Hoff factor is LESS than 1, SOLUTE has UNDERGONE has undergone associated in benzene solvent.
Discussion
92247.

3.8 gm of a tribasic carboxylic acid derivative of a saturated hydrocarbon required 100 ml of 0.6M NaOH solution to reach equivalence point. Calculate molar mass of tribasic carboxylic acid derivative of saturated hydrocarbon.

Answer»


ANSWER :`0190`
Discussion
92248.

380 mL of a gas at 27^@C, 800 mm of Hg weight 0.455 g. The mol.wt. of gas is :

Answer»

27
28
29
30

Answer :B
Discussion
92249.

375 mg of an alcohol reacts with required amount of methyl magnesium bromide and releases 140 mL of methane gas at STP. The alcohol is

Answer»

ethanol
n-butanol
methanol
n-propanol

Solution :`underset("1 MOL")(ROH)+CH_(3)MgBrrarr underset("= 22400 mL at STP")underset("1 mol")(CH_(4)+Mg(OR)Br)`
140 mL of `CH_(4)` is PRODUCED from alchohol
= 375 mg = 0.375 g
`therefore"22400 mL of CH"_(4)" will be produced from"`
`"alcohol"=(0.375)/(140)xx22400=60g`
`therefore" MOLAR mass of alcohol = 60 g mol"^(-1)`
`M(C_(2)H_(5)OH)=46, M(C_(4)H_(9)OH)=74,`
`M(CH_(3)OH)=32, M(CH_(3)CH_(2)CH_(2)OH)=60`
Hence, the alcohol is n-propanol.
Discussion
92250.

375 mg of an alcohol reacts with required amount of methylmagnesium bromide and releases 140 mL of methane gas at STP. The alcohol is

Answer»

ethanol
n-butanol
methanol
n-propanol

Solution :140 mL of `CH_(4)` is produced from ALCOHOL=375mg
`therefore22400`mL of `CH_(4)` is produced from alcohol`=(375)/(1000)xx(22400)/(140)=60` G `mol^(-1)`
Thus, alcohol with molecular mass=60g `mol^(-1)`
is `CH_(3)CH_(2)CH_(2)OH` (n-propanol).
Discussion