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92701.

2 g of benzoic acid (C_(6)H_(5)COOH)dissolved in 25 g of benzene shows a depression in freezing point equal to 1.62 K. Molar depression constant for benzene is 4.9 K kg mol^(-1). What is the percentage association of acid if it forms dimer in solution ?

Answer»

Solution :The given quantities are :
`w_(2)=2g , "" K_(f)=4.9 "K kg mol"^(-1)`,
`w_(1)=25 g, "" Delta T_(f)=1.62 K`
Substituting these values in equation,
`M_(2)=(K_(f)xx w_(2)xx1000)/(Delta T_(f)xx w_(1))`
`= (4.9"K kg mol"^(-1)xx 2g xx 1000 g kg^(-1))/(25 g xx 1.62 K)`
`= 241.98 g mol^(-1)`
Thus, experimental molar mass of benzoic acid in benzene is `= 241.98 g mol^(-1)`
Now consider the following equilibrium for the acid : `2C_(6)H_(5)COOH HARR (C_(6)H_(6)COOH)_(2)`
If x represents the DEGREE of association of the solute then we would have (1 - x) mol of benzoic acid left in unassociated form and correspondingly `(x)/(2)` as asociated moles of benzoic acid at equilibrium. Therefore, total number of moles of particles at equilibrium is :
`1-x+(x)/(2)=1-(x)/(2)`
Thus, total number of moles of particles ast equilibrium equals vant.s Hoff factor (i).
But `i=("Normal molar mass")/("ABNORMAL molar mass")`
`=(122 g mol^(-1))/(241.98 g mol^(-1))`
or `(x)/(2)=1-(122)/(241.98)=1-0.504 = 0.496`
or `x=2xx0.496=0.992`
Therefore, degree of association of benzoic acid in benzene is 99.2 %.
Discussion
92702.

2 g of benzoic acid (C_(6) H_(5) COOH) dissolved in 25 g of benzene shows a depressionc in freezing point equal to 1.62 K. Molal depression constant for benzene is "4.9 K kg mol"^(-1). What is the percentage association of acid if it forms dinner in solution? How many mL of "0.1 M HCl" are required to react completely with 1 g mixture of Na_(2) CO_(3) and NaHCO_(3) containing equimolar amounts of both?

Answer»

SOLUTION :`"LET MASS of "Na_(2)CO_(3)=xg`
`therefore"mass of "NaHCO_(3)=(1-x)g`
`therefore"moles of "Na_(2)CO_(3)="moles of "NaHCO_(3)`
`(x)/(106)=(1-x)/(84)`
`x=0.558g`
`"Number of moles of "Na_(2)CO_(3)=(0.558)/(106)="0.00526 mol"`
`"Number of moles of "NaHCO_(3)="0.00526 mol"`
`Na_(2)CO_(3)+2HCl rarr 2NaCl +H_(2)O +CO_(2)`
`NaHCO_(3)+HCl rarr NaCl+H_(2)O+CO_(2)`
Total number of moles of HCl required
`=(2xx0.00526)+0.00526`
`="0.01578 mol"`
`therefore"VOLUME 0.1 M HCl"=("Number of moles")/("molarity")`
`=(0.01578)/(0.1)=0.1578L`
`=157.8mL`
Discussion
92703.

2 g of a radioactive sample having half life of 15 days was synthesised on 1^(st) Jan 2009. The amount of the sample left behind on 1^(st) March, 2009 (including both the days)

Answer»

0.125 g
1 g
0.5 g
0g

Solution :No. of days from `1^(st)` Jan 09 to `1^(st)` March '09 = 60
GIVEN, `t_(1//2) = 15` days
Total time, `T = 60` days
`:.` No. of half-lives, n = ?
Also given, initial AMOUNT of sample, `N_(0) = 2g`
`:.` Sample PRESENT after 4 half lives, `N= ?`
`N = ((1)/(2))^(n) N_(0)`
`rArr N = ((1)/(2))^(4) xx 2 = (2)/(16) = 0.125g`
Discussion
92704.

2 g of a radioactive sample having half life of 15 days was synthesised on 1st Jan 2009. The amount of the smaple left behind on 1st March, 2009 (including btoh the days)

Answer»

0.125 g
1 g
0.5 g
0 g

Solution :No of days from 1st jan 09 to 1st March 09
`=60`
GIVEN `t_(1//2)=15` days
total time `T=60` days
`:.`no of HALF lives N=?
`T=nxxt_(1//2)implies60=nxx10impliesn=4`
Also given initial amount of sample `N_(0)=2g `
`:.` sample PRESENT after 4 half lives N=?
`N=(1/2)^(n)N_(0)`
`impliesN=(1/2)^()xx2=2/16=0.125g`
Discussion
92705.

2 g of a base whose eq.wt is 40 reacts with 3 g an aid .The eq.wt of the acid is

Answer»

40
60
10
80

Answer :B
Discussion
92706.

2 g each of the solutes A and B ( mol mass of A gt B) are dissolved separately in 20 g each of the same solvent C. Which will show gr eater lowering of vapour pressure and why ?

Answer»

SOLUTION :`"for dilute solutions, "(p^(@)-p_(s))/(p^(@))=(w_(2)M_(1))/(w_(1)M_(2)) or (DELTAP)/(p^(@))=(w_(2)M_(1))/(w_(1)M_(2))`
`"For same SOLVENT, "p^(@)=" constant, "M_(1)="constant. Also, given that "w_(2)=w_(A)=w_(B)=2g, w_(1)=w_(C)=20 g`
Hence, `(Deltap_(A))/(Deltap_(B))=(M_(B))/(M_(A))." As "M_(A) gt M_(B)`, therefore, `Deltap_(B) gt Deltap_(A)`, i.e., B will show greater lowering of vapour pressure.
Discussion
92707.

2 F of electricity will liberate 1 gram atom of the metal from the solution of

Answer»

NaCl
KCl
`BaCl_(2)`
`AlCl_(3)`

SOLUTION :`BA^(2+)+2e^(-) to Ba`
`therefore 2-=1mol.` of Ba`-=` 1 G of atom of Ba.
Discussion
92708.

2-ethoxy propane is formed from ethyl bromide and what ?

Answer»

SODIUM ETHOXIDE
Sodium iso-propoxide
iso-butyraldehyde
iso-propyl ALCOHOL

ANSWER :B
Discussion
92709.

2-ethoxy propane is effectively prepared from

Answer»

ISOBUTYL IODINE
ISOPROPYL iodine
ethyl iodine
n-propyl iodine

Answer :C
Discussion
92710.

Write the relavant chemical equations : 2-chloropropane to l-propanol

Answer»

Solution :`CH_(3)- overset(overset(Cl)(|))(CH)-CH_(3) overset("ALC KOH")rarr CH_(3)-CH=CH_(2) underset(2. H_(2)O_(2)//OH^(-))overset(1. BH_(3))rarr CH_(3)-CH_(2)-overset(overset(OH)(|))(CH_(2))`
Discussion
92711.

2-Chlorobutane was treated with alc. KOH and the product formed was reacted with dil. KMnO_(4) to give the product B. The structure of B is :

Answer»

`CH_(3) - underset(OH)underset(|)(CH) - CH_(2) - CH_(3)`
`CH_(3) - CH = CH - CH_(2)OH`
`CH_(3) - underset(OH)underset(|)(CH) - underset(OH)underset(|)(CH)CH_(3)`
`underset(OH)underset(|)(CH_(2)) - underset(OH)underset(|)(CH)CH_(2)CH_(3)`

Solution :`CH_(3) - CH_(2) underset(CL)(CH)CH_(3) OVERSET(alk.KOH)rarr CH_(3)CH = CHCH_(3) overset(alk.KMnO_(4))rarr CH_(3) - underset(OH)underset(|)(CH)-underset(OH)underset(|)(CH) - CH_(3)`
Discussion
92712.

2-chlorobutane is heated with alcoholic NaOH, the product formed in larger amount is

Answer»

1-butene
1-butyne
2-butene
All of these

Solution :`CH_3-undersetundersetH|CH-undersetunderset"CL"|CH-undersetundersetH|CH_2-underset"-HCL"OVERSET"ALC. NAOH"to CH_3CH=CH-CH_3`
Discussion
92713.

2-Chloro-2-methylpentane on reaction with sodium methoxide in methanol yields (I). C_(2)H_(5)CH_(2)underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-OCH_(3). (II). C_(2)H_(5)CH_(2)underset(CH_(3))underset(|)(C)=CH_(2) (III). C_(2)H_(5)CH=underset(CH_(3))underset(|)(C)-CH_(3).

Answer»

all these
I and III
III only
I and II

Solution :The REACTION can PROCEED both through `S_(N^(1))` and `E_(2)` MECHANISM. THEREFORE, all the three products can be formed.
`S_(N^(1))`
.
Discussion
92714.

2-Chloro-2-methylpentane on reaction with sodium methoxide in methanol yields (I) C_(2)H_(5)CH_(2)underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-OCH_(3) (II) C_(2)H_(5)CH_(2)underset(CH_(3))underset(|)(C)=CH_(2) (III) C_(2)H_(5)CH=underset(CH_(3))underset(|)(C)-CH_(3)

Answer»

all of these
I and III
III only
I and II

Solution :2-Chloro-2-methylpantane being a `3^(@)` alkyl halide can undergo either `S_(N)1` or `E_(1) ` REACTION. Further, the intermediate carbocation (I) can undergo elimination in two DIFFERENT ways, to give 2-methyl-2-pentene (major) and 2-methyl-1-pentene (minor)
`underset("2-Chloro-2-methylpentene")(CH_(3)CH_(2)CH_(2)-underset(Cl)underset(|)overset(CH_(3))overset(|)(C)-CH_(3))underset(-Cl^(-))overset("Ionization")to underset(3^(@)" carbocation (I)")(CH_(3)CH_(2)CH_(2)-underset(+)overset(CH_(3))overset(|)(C)-CH_(3))`
`(I) underset(S_(N)1)overset(CH_(3)ON a)to underset("2-Methoxy-2-methylpentane (Substitution PRODUCT)")(CH_(3)CH_(2)-underset(OCH_(3))underset(|)overset(CH_(3))overset(|)(C)-CH_(3))`
`(I)underset(-H^(+))overset(E_(1))to underset("2-Methyl-2-pentene (major product)")(CH_(3)CH_(2)CH=overset(CH_(3))overset(|)(C)=CH_(2))+underset("2-Methyl-1-pentene (minor product)")(CH_(3)CH_(2)CH_(2)-overset(CH_(3))overset(|)(C)=CH_(2))`
Discussion
92715.

2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields (A) C_(2)H_(5)CH_(2)underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-OCH_(3) (B) C_(2)H_(5)CH_(2)underset(CH_(3))underset(|)(C)=CH_(2) (C) C_(2)H_(5)CH=underset(CH_(3))underset(|)(C)-CH_(3)

Answer»

(A) and (C)
(C) only
(A) and (B)
All of these

Solution :`CH_(3)-underset(CH_(3))underset(|)OVERSET(|)overset(Cl)overset(|)(C)-CH_(2)-CH_(2)-CH_(2)-CH_(3) underset(MeOH)overset(MeONa)toCH_(3)-underset(CH_(3))underset(|)(C)=CH-CH_(2)-CH_(3)+CH_(2)=underset(CH_(3))underset(|)(C)-CH_(2)-CH_(2)-CH_(3)+CH_(3)-underset(CH_(3))underset(|)overset(OCH_(3))overset(|)(C)-CH_(2)-CH_(2)-CH_(3)`
Elimination dominate over substitution in the given REACTION but all the PRODUCTS are possible.
Discussion
92716.

2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields- I.CH_2H_5CH_2-overset(CH_3)overset(|)underset(CH_3)underset(|)C-OCH_3 II.C_2H_5CH_2-underset(CH_3)underset(|)C=CH_2 III. C_2H_5CH=underset(CH_3)underset(|)C-CH_3

Answer»

both I and III
only II
both I and II
all of these

Solution :
Discussion
92717.

2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields: (a) C_(2)H_(5)CH_(2)overset(CH_(3))overset(|)underset(CH_(3))underset(|)(C)-OCH_(3) (b) C_(2)H_(5)CH_(2)underset(CH_(3))underset(|)(C)=CH_(2) (c) C_(2)H_(5)CH_(2)=underset(CH_(3))underset(|)(C)-CH_(3)

Answer»

all of these
a and C
c only
a and B

ANSWER :A
Discussion
92718.

2-Chloro -2-methylpentane on reaction with sodium methoxide in methanal yields :

Answer»

(a) and (B)
all of these
(a) and ( C)
(c) only

Solution :
Discussion
92719.

2-cetoxybenzoic acid is an

Answer»

antipyratic, analgesic
antibiotic
antioxidant
antiseptic

Answer :A
Discussion
92720.

2-Butyne undergoes following reactions in steps as indicaed. Identify A to H.

Answer»


SOLUTION :N//A
Discussion
92721.

2-Butyne on oxidation with alkaline KMnO_(4) at 298-303 gives :

Answer»

Oxalic ACID
2-Butanone
2, 3-Butanedione
2-Keto propanoic acid

Solution :`underset("2-Butyne")(CH_(3) - C -= C - CH_(3)) OVERSET(Alk.KMnO_(4))rarr underset("2, 3-Butanedione")(CH_(3) - underset(O)underset(||)(C) - underset(O)underset(||)(C) - CH_(3))`
Discussion
92722.

2-butyne is reduced to trans-but-2-ene using

Answer»

`H_(2)//Ni`
NA in liq. `NH_(3)`
`H_(2)`/PD – C
Zn in dil. HCI.

Solution :(b): Birch reduction of non-terminal ALKYNES with Na or Li in liquid `NH_(3)` at 196-200 K gives trans-alkenes.
Discussion
92723.

2-Butyne can be prepared from which of following compounds?

Answer»

`CH_3-UNDERSET(CL)underset(|)OVERSET(Cl)overset(|)C-underset(Cl)underset(|)overset(Cl)overset(|)C-CH_3`<BR>`CH_3-CH_2-underset(Br)underset(|)overset(Br)overset(|)C-CH_3`
`underset(Cl)underset(|)(CH_2)-CH_2-CH_2-underset(Cl)underset(|)(CH_2)`
Both A and B

Answer :D
Discussion
92724.

2-butanone can be obtainedheating a mixture of calcium salt of

Answer»

FORMIC ACID and BUTYRIC acid
propionic acid and formic acid
propionic acid and acetic acid
acetic acid and formic acid

ANSWER :C
Discussion
92725.

2-butanol overset("conc. " H_2SO_4)to undersetA"1-butene"+undersetB"2-butene" Which are correct statements ?

Answer»

A is Saytzelf PRODUCT B is not
B is Saytzelf product A in not
Either' a' and 'b'
Neither

Answer :B
Discussion
92726.

2-butanol on dehydration mainly gives,

Answer»

`CH_3CH_2CH=CH_2`
`CH_3CH=CHCH_3`
`(CH_3)_2C=CH_2`
`CH_3CH=CH_2`

ANSWER :B
Discussion
92727.

2-Bromopentane is heated with potassium ethoxide in ethanol. The major product obtained is

Answer»

pent-1-ene
cis pent-2-ene
trans-pent-2-ene
2-ethoxypentane

Answer :C
Discussion
92728.

2-bromopentane is heated with pottasium ethoxide in ethanol the major product is :

Answer»

TRANS PENT 2 ENE
2-ethoxy pentane
pent-1-ene
cis-pent-2 -ene

Answer :A
Discussion
92729.

2-bromopentane is treated with alcoholic KOH solution the reaction respectively are :

Answer»


Solution :`CH_(3)-underset(2-"Bromopentane")OVERSET(Br)overset(|)CH-CH_(2) - CH_(3) overset(KOH(AQ)) to underset(+"mixture two products")(CH_(3)-underset(OH)underset(|)overset(H)overset(|)C-CH_(2)-CH_(2)-CH_(3)`
`underset("2-Bromopentane")(CH_(3)-overset(Br)overset(|)CH-CH_(2)-CH_(2)-CH_(3) overset(KOH (alc))to(CH_(2) = CH -CH_(2)-CH_(2) - CH_(3) + CH_(3) - CH= CH-CH_(2)CH_(3)` (E & Z)
HENCE, 5 products will formed.
Discussion
92730.

2-Bromobutane (A) is treated with Nal in the presence of dry acetone to give compound 'B'. The compound 'B' is boiled with moist silver oxide to give compound 'C'. Identify compounds 'B' and 'C' . How many optical isomers are possible for compound 'C' /

Answer»

2-iodobutane, Butan-2-ol, 2 isomers
3-Iodopentane, pentan-3-ol , 4 isomers
2-Iodobutane , Butan-2-ol , 4 isomers
3-Iodopentane-Pentan-3-ol, 2 isomers

Answer :A
Discussion
92731.

2-Bromobutane reacts with OH^(-) in H_(2)O to give 2-butanol. The reaction involves

Answer»

retention in configuration
inversionin configuration
racemization
mutarotation

SOLUTION :2-Bromobutane, `CH_(3)-underset(Br)underset(|)(CH)CH_(2)CH_(3)` is `2^(@)` ALKYL HALIDE. It REACTS with `OH^(-)` in PROTIC solvent `(H_(2)O)` through `S_(N)1` mechanism. Therefore, it involves racemisation.
Discussion
92732.

2-Bromo-2, 3-dimethylbutane is treated with alcoholic potash. Write the major product

Answer»

Solution :APPLYING ZAITSEV rule, bromine must be lost from C-2 and hydrogen from C-1 or C-3, during DEHYDROBROMINATION REACTION
The major PRODUCT is 2, 3-dimethyl-2-butane
Discussion
92733.

2-Acetoxy benzoic acid is used as an

Answer»

antimalarial
antidepressant
antiseptic
ANTIPYRETIC.

Solution :2-acetoxy BENZOIC ACID or ASPIRIN is USED as antipyretic.
Discussion
92734.

2-Acetoxy benzoic acid is used as

Answer»

Antimalaria
Antidepressant 
An analgesic
Antipyretic 

ANSWER :D
Discussion
92735.

2-Acetoxy benzoic acid is

Answer»

ANTISEPTIC
ANTIPYRETIC
ANTIBIOTIC
MORDANT dye

Answer :B
Discussion
92736.

2-Acetoxy benzoic acid can be used as

Answer»

ANTISEPTIC
ANTIPYRETIC
Antibiotic
Mordant DYE

ANSWER :B
Discussion
92737.

2 . 76 g of silver carbonate on beingstrongly heated yields a residue weighing .

Answer»

2. 16 G
2 . 48 g
2 . 32 g
2 . 64 g

Answer :A
Discussion
92738.

2.5 of a mixture of BaO and CaO when heated with H_(2)SO_(4) , produced 4.713 g of the mixed sulphates . Find the percentage of BaO present in the mixture.

Answer»


ANSWER :`60%`
Discussion
92739.

2, 4-Dinitrophenylhydrazine test is used for :

Answer»

carboxylic ACID GROUP
phenolic group
aldehyde group
ESTER group.

Answer :C
Discussion
92740.

2 : 4 Dinitrophenyl hydrazine is an example for

Answer»

TRIDENTATE LIGAND
MONODENTATE ligand
Polydentate ligand
Didentate ligand

Solution :
is a monodentate ligand.
Discussion
92741.

2, 4-Dinitroaniline underset((ii)" Anisole")overset((i) NaNO_(2)//HCl (273 K))(rarr) Z. Compound Z is

Answer»




SOLUTION :
Discussion
92742.

2, 4-dichlorophenoxy acetic acid is the important weed killer known as 2,4 - D. Outline its synthesis.

Answer»

SOLUTION :
Discussion
92743.

2, 4-dichloro phenoxyacetic acid is used as a

Answer»

Insecticide
Fungicide
HERBICIDE
Moth REPELLENT

SOLUTION :It is herbicide (KILL HERBS).
Discussion
92744.

2,4- dichlorophenoxyacetic acid is used as a:

Answer»

Fungicide
Insecticide
Herbicide
Moth REPELLENT

ANSWER :C
Discussion
92745.

2, 4, 6 -trinitrochorobenzene is warmed with water gives

Answer»

piciric acid
2, 4, 6-trinitrobenezoic acid
phenol
p-nitrophenol

Answer :A
Discussion
92746.

2-3% solution of iodine in alcohol-water is called _______

Answer»


ANSWER :TINCTURE of IODINE
Discussion
92747.

2, 3-dimethyl-2-butene, on reductive ozonolysis gives

Answer»

Acetone
Acetaldehyde
Butanone
Fomaldehyde

Answer :A
Discussion
92748.

2, 2-dichloro butane on boiling with aq. Potash gives,

Answer»

BUTANAL
2-butanone
2-butanol
BUTANOIC acid

Answer :B
Discussion
92749.

1^(st) law of thermodynamics introduces the concept of _______ .

Answer»

temperature
pressure
change in DIRECTION of reaction
conservation of energy

Answer :D
Discussion
92750.

1st law of thermodynamic is represented by the equation.

Answer»

`/_\E=Q+W`
`/_\E=Q-W`
`W=Q+/_\E`
NONE of these

Answer :B
Discussion