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91151.

(A) C_(7)H_(14) an organic compound decolourises Br_(2) solution to produce (B) which combines with strong base NaNH_(2) to produce (C) which does not give white ppt. with AgNO_(3)//NH_(4)OH. (A) again when treated with Hg(OA c)_(2)//H_(2)O+NaBH_(4) toproduce (D) and when treated with HBr/ROOR then OH^(-) to produce (E). Here (D) and (E) are isomeric alcohol, one is optically acetic and other is optically inactive. Q. What is (A)

Answer»

`C-C-C-UNDERSET(C)underset(|)(C)=C-C`
`C-C-C-underset(C)underset(||)(C)-C-C`
`C-C-underset(C)underset(|)underset(C)underset(||)(C)-C-C`
`C-C-underset(C)underset(C-Br)underset(||)(C)-C-C`

ANSWER :C
Discussion
91152.

A C_(7)H_(12)O_(2) optically active alcohol is oxidised by jones reagent to an optically inactive ketone. The molecule is.

Answer»




ANSWER :C
Discussion
91153.

A C_(7)H_(12)O_(2) compound gives a positive Tollen's silver mirror test as well as positive iodoform test.Which of the following would not satisfy these facts?

Answer»

2-hydroxy-3,3-dimethylcyclopentanone
2,5-heptanedione
2,2-dimethyl-3-oxopentanal
2,2-dimethyl-4-oxopentanol

Solution :N//A
Discussion
91154.

A C_(6)H_(12)O compound does not react with Br_(2) in C Cl_(4), produces a flammable gas on treatment with LiAlH_(4), and reacts with H_(2)CrO_(4) changing the color from orange to green. Which of the following compounds best agrees with these facts?

Answer»

1-methylcyclopentanol
methoxycyclopentane
2-cyclopropyl-2-propanol
2-cyclobutylethanol

Answer :D
Discussion
91155.

A C_(5)H_(13)N compound gives a base soluble derivative on reaction with Hinsberg test. Which of the following fits these facts best?

Answer»

1,1-Dimethylpropylamine
Isopropyldimethylamine
2,2-Dimethylpropylamine
N-Methyl-2-methylpropylamine

Solution :`CH_(3)-OVERSET(CH_(3))overset(|)underset(CH_(3))underset(|)C-CH_(2)-NH_(2)+ClSO_(2)H_(5)C_(6)toCH_(3)-overset(CH_(3))overset(|)underset(CH_(3))underset(|)C-CH_(2)-NH-overset(C_(6)H_(5))overset(|)SO_(2)`
Discussion
91156.

A (C_(5)H_(12)O) produces, on reaction PCl_(5) from alkyl chloride B and C. B and C both on reaction with aqueous KOH form alcohol D and E . Both D and E give iodoform test. Identify the correct answers.

Answer»



C is `CH_(3)CH_(2)CH_(2)CI`

ANSWER :A::B::D
Discussion
91157.

(A) C_(2)H_(5)OH (B) CH_(3)CHO (C) CH_(3)COCH_(3) (D) C_(6)HCHO (E) C_(6)H_(5)CH_(2)CHO. Number of compounds which not only give yellow ppt with NaOH+I_(2) but also give red ppt with Fehling's reagent are _________

Answer»

4
3
2
1

Answer :D
Discussion
91158.

(A): C CI_4 can be used as a fire extinguisher. (R): C CI_4 is insoluble in water.

Answer»

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

Answer :B
Discussion
91159.

(A) C Cl_(3)-CHO on treatment with an alkali to form chloroform and formate ion but not undergo Cannizzaro's reaction even through no alpha-hydrogen. (R) H^(-) is not transfer does not take place because C Cl_(3), is good leaving group.

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) is false

Answer :A
Discussion
91160.

(a). C-C-overset(O)overset(||)(C)-Cunderset(Delta)overset([O])to ltb rgt (b). Me_(2)CH-overset(O)overset(||)(C)-Meunderset(Delta)overset([O])to (c). Me_(3)C-overset(O)overset(||)(C)-Meunderset(Delta)overset([O])to (d). .

Answer»

SOLUTION :
Discussion
91161.

A bydrogenation reaction is carried out at 500 K. If the same reaction is carried out in the presence of a catalyst at the same rate, the temperature required is 400 K. Calculate the activation energy of the reaction if the catalyst lowers the activation energy by 20 kJ mol^(-1).

Answer»

Solution :Suppose activation energy in the absence of catalyst is `E_(a)` and that in the presence of catalyst, it is `E_(C)`. Then`K=Ae^(-E_(a)//400"R")("as k is same in both the cases").`
Hence, `(E_(a))/(500" R")=(E_(c))/(400" R")" or "E_(c)=(4)/(5)E_(a)`
But `E_(a)-E_(c)=20" kJ"" or "E_(a)-(4)/(5)E_(a)=20" or "(1)/(5)E_(a)=20" or "E_(a)=100" kJ MOL"^(-1).`
Discussion
91162.

A buttonn cell used in watchesfunction as followingZn(s) +Ag_(2)O(s) +H_(2)O(l) rarr 2Ag(s)+Zn^(2+)(aq)+2OG^(-)(aq), E^(@)=0.34 V the cellpotential will be

Answer»

1.10 V
0.42 V
0.84 V
1.34 v

Solution :Anodeis always sthe SITE of oxidation THUS anode half cell us
`Zn^(2+) (aq)+2e^(-) RARR Zn(s) , E^(@)= -0.76 V`
cathodehalfcell is
`Ag_(2) O(s)+H_(2)O(I)+2e^(-) rarr 2Ag(s)+2OH^(-) (aq)`
`E^(@)=0.34 V`
`E_(cell)^(@)=E_(CATHODE)^(@) - E_(anode)^(@) =0.34 -(-0.76) =+1.10 V`
Discussion
91163.

A button cell used in watches functions as follows: Zn(s)+Ag_(2)O(s)+H_(2)O(l)hArr2Ag(s)+Zn^(2+)(aq)+2OH^(-)(aq) The half-cell potential are Zn^(2+)(aq)+2e^(-)toZn(s),E^(@)=-0.76V Ag_(2)O(s)+H_(2)O(l)+2e^(-)to2Ag(s)+2OH^(-)(aq),E^(@)=0.34V The cell potential will be

Answer»

1.10 V
0.42 V
0.84 V
1.34V

Solution :`E_(CELL)^(@)=E_("CATHODE")^(@)-E_("ANODE")^(@)=0.34-(-0.76)V=1.10V`
Discussion
91164.

A button cell used in watches function as following Zn(s)+Ag_(2)O(s)+H_(2)O(l)hArr2Ag(s)+Zn^(2+)(aq)+2OH^(-)(aq) if half cell potential are Zn^(2+)(aq)+2e^(-)toZn(s), Ag_(2)O(s)+H_(2)O(l)+2e^(-)to2Ag(s)+2OH^(-)(aq),E^(@)=0.34V The cell potential will be

Answer»

1.34V
1.10V
0.42V
0.84V

Solution :]`ZN^(2+)(AQ)+2e^(-)toZn(s),E^(o)=-0.76V`
`Ag_(2)O(s)+H_(2)O(l)+2e^(-)to2Ag(s)+2OH^(-)(aq),E^(o)=0.34V`
`Zn(s)+Ag_(2)O(s)+H_(2)O(l)+2e^(-)to2Ag(s)+2OH^(-)(aq)`
`+2OH^(-)(aq),E_(CELL)=?`
`E_(cell)^(o)=(E_(R.P.)^(o))_("cathode")-(E_(R.P.)^(o))_("anode")`
`E_(cell)^(o)=0.34(-0.76)=1.10V`
`E_(cell)=E_(cell)^(o)=1.10V`
Discussion
91165.

A button cell used in watches functions as following : Zn_((s)) + Ag_(2)O_((s)) + H_2O_((l))

Answer»

1.10 V
0.42 V
0.84 V
1.34 V

Answer :A
Discussion
91166.

A : But-2-ene is more stable than propene. R : Heat of hydrogenation of but-2-ene is lesser than that of propene.

Answer»

If both Assertion & Reason are TRUE and the reason is the correct explanation of the assertion, then MARK (1).
If both Assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark (2)
if Assertion is true STATEMENT but Reason is false, then mark (3)
If both Assertion and Reason are false statements, then mark (4)

Answer :1
Discussion
91167.

A bulb of unknown volume contained an ideal gas at 650 mm pressure. A certain amount of gas was withdraw and found to occupy 1.52 cc at 1 atm pressure. The pressure of the gas remaining in the bulb was 600 mm. If all measurements were made at a constant temperature, find the volume of the bulb.

Answer»

SOLUTION :When a certain amount of the gas is WITHDRAWN, the pressure dropped from 650 mm to 600 mm. So the pressure difference (650 - 600) ot 50 mm will be the pressure of the gas withdrawn which occupied the bub of volume `V_(1)` cc (suppose). SINCE the same amount of withdrawn gas occupied 1.52 cc at 1 atm (760 mm) pressure at the same TEMPERATURE, we can apply Boyle.s law :
`p_(1)V_(1) = p_(2)V_(2)`
`50 xx V_(1) = 760 xx 1.52`
`V_(1) = 23.1` cc.
Discussion
91168.

A bulb of unknown volume V contains an ideal gas at 1 atm pressure. This bulb was connected to another evacuated bulb of volume 0.5 litre through a stopcock. When the stopcock was opened the pressure at each became 530 mm while the temperature remained constant. Find V in litres.

Answer»

Solution :Since on openingthe stopcock, the same NUMBER of moles of the gas occupy a VOLUME of (V + 0.5) litre at the same temperature,
We apply Boyle.s LAW :
`p_(1)V_(1) = p_(2)V_(2)`
`1 xx V = (530)/(760) xx (V + 0.5) ""(p_(2) = (530)/(760)"atm")`
`V = 1.152` litres.
Discussion
91169.

A bulb emits light of wavelength 4500Å. The bulb is rated as 150 watt and 8% ofthe energy is emitted as light. How many photons are emitted per sec?

Answer»

SOLUTION :`2.72xx10^(19)`
Discussion
91170.

A bulb contains 2 moles of H_2at a pressure of 0.8 atm and temperature T K. 0.6 mole of O_2is added to this bulb and the temperature of the bulb is lowered by 15 K to keep the same pressure. Calculate the volume of the bulb and its temperature T. Also, calculate the partial pressure of each gas.

Answer»

<P>

SOLUTION :`13.34 dm^3, 65 K, p_(H_2) = 0.615 ATM, p_(O_2) = 0.185 atm`
Discussion
91171.

A buffer solution with pH 9 is to be prepared by mixing NH_(4)Cl and NH_(4)OH. Calculate the number of moles of NH_(4)Cl that should be added to one litre of 1.0 M NH_(4)OH. [ K_(b) = 1.8 xx 10^(-5)]

Answer»

3.4
2.6
1.5
1.8

Solution : `pH = -LOG K_(B) + log. (["Salt"])/(["ACID"])`
`pH = -log[1.8 xx 10^(-5)] + log.(["Salt"])/(1.0)`
`9 = 4.7 + log.(["Salt"])/(1.0)`
`log.(["Salt"])/(1.0) = 9 - 4.7 = 4.3`
`(["Salt"])/(1.0)= "Antilog" (1)/(4.3), ["Salt"] = 1.8`.
Discussion
91172.

A buffer solution is prepared in which the concentration of NH_(3) is 0.30 M and the concentration of NH_(4)^(+) is 0.20 M. If the equilibrium constant, K_(b) for NH_(3) equals 1.8 xx 10^(-5), what is the pH of this solution (log 2.7 = 043)

Answer»

8.73
9.08
9.43
11.72

Solution :`pOH = pK_(b)LOG.(["SALT"])/(["Base"])`
`= 4.74 + log.(0.20)/(0.30) = 4.74 + (0.301 - 0.477)`
`= 4.74 - 0.176 = 4.56:. PH = 14- 4.56 = 9.44`.
Discussion
91173.

A buffer solution is prepared in which the concentration of NH_(3) is 0.30 M and concentration of NH_(4)^(+) is 0.20 M. If equilibrium constant K_(b) for NH_(3) equals 1.8xx10^(-5), what is the pH of the solution ?

Answer»

`9.08`
`9.43`
`11.72`
`8.72`

Solution :`pOH=pK_(a)+(["SALT"])/(["Base"])`
`pK_(b)=-LOG K_(b)=-log(1.8xx10^(-5))=4.74`
`pOH=4.74+log((0.20)/(0.30))=4.56`
`pH=14-4.56=9.44`
Discussion
91174.

A buffer solution is prepared by mixing equal concentration of acid (ionisation constant K_(a)) and a salt. The pH of buffer is

Answer»

`pK_(a) + 7`
`14 - pK_(a)`
`pK_(a)`
`pK_(a) + 1`

Solution :PH of buffer is
`pH = pK_(a) + LOG.(["SALT"])/(["acid"])`
Since, [salt] = [acid], `pH = pK_(a)`.
Discussion
91175.

A buffer solution is prepared by mixing 'a' moles of CH_(3)COONa and 'b' moles of CH_(3)COOH such that (a+b) = 1, into water to male 1L buffer solution. If the buffer capacity of this buffer solution is plotted against moles of salt CH_(3)COONa 'a' then the plot obtained will be (to the scale) approximately. (As shown in fig. in options)

Answer»




ANSWER :B
Discussion
91176.

A buffer solution helps in maintaining the :

Answer»

ALKANITY of solution
Acidic NATURE of solution
pH of medium
None of these

Answer :C
Discussion
91177.

A buffer solution has equal volumes of 0.2 M NH_(4)OH and 0.02 M NH_(4)Cl. The pK_(b) of the base is 5. The pH is

Answer»

10
9
4
7

Solution :`pOH = pK_(b) + LOG.(["salt"])/(["base"])`
`= 5 + log.(0.02)/(0.2) = 5 + log.(1)/(10) = 5 + (-1) = 4`
`pH = 14 - pOH = 14 - 4 = 10`.
Discussion
91178.

A buffer solution has equal volumes of 0.2 M NH_(4)OH and 0.02 M NH_(4) CI The pk_(b) of the base is 5. The pH is

Answer»

A) 10
B) 9
C) 4
D) 7

Solution :`pOh=pK-(b)+log[sal t]/([base])`
`=5+log(0.02)/(0.2)=5+log(1)/(10)=5+(-1)=4`
`pH=14-pOH=14-4=10`
Discussion
91179.

A buffer solution contains 0.1 mole of sodium acetate in 1000 cm^(3) of 0.1 M acetic acid. To the above buffer solutions, 0.1 mole of sodium acetate is further added and dissolved. The pH of the resulting buffer is equal to............

Answer»

`pK_(a) - log_(2)`
`pK_(a)`
`pK_(a) + 2`
`pK_(a) + log 2`

SOLUTION :`PH = pK_(a) + log.(["Salt"])/(["Acid"])`
`["Salt"] = (0.1 + 0.1)/(1000 ml) = (0.2)/(1) = 0.2`
`["Acid"] = (0.1)/(1000 ml) = (0.1)/(1) = 0.1`
`pH = pK_(a) + log 2`
`:. pH = 14 - 2 RARR pH = 12`.
Discussion
91180.

A buffer solution contains 0.1 mole of sodium acetate in1000 cm^(3) of 0.1 M acetic acid. To the above buffer solution, 0.1 mole of sodium acetate is further added and dissolved. The pH of the resulting buffer is equal to

Answer»

`pK_(a) + LOG 2`
`pK_(a)`
`pK_(a) + 2`
`pK_(a) + log 2`

ANSWER :D
Discussion
91181.

A buffer solution contains 0.1 mole of sodium acetate dissolved in 1000 cm^(3) of 0.1 M acetic acid. To the above buffer solution, 0.1 mole of sodium acetate is further added and dissolved.The pH of the resulting buffer is

Answer»

`pK_(a)`
`pK_(a)+2`
`pK_(a)-log 2`
`pK_(a)+log2`

SOLUTION :TOTAL MOLES of sodium acetate
`0.1+0.1=0.2`
`pH=pK_(a)+"log"([CH_(3)COONa])/([CH_(3)COOH])`
`pH=pK_(a)+"log"(0.2)/(0.1)=pK_(a)+log2`.
Discussion
91182.

A buffer solutioncontains 0.1 M of acetic acid and 0.1 M of sodium acetate. What will be its pH, it pK_(a) of acetic acid is 4.75

Answer»

`4.00`
4.75
`5.00`
5.25

Solution :`pH = pK_(a) + log.(["Salt"])/(["ACID"]), pH = 4.75 + log.(0.1)/(0.1)`
`pH = 4.75 + log 1, pH = 4.75`.
Discussion
91183.

A buffer solution contains 0.1 M each of acetic acid and sodium acetate. The pH of the buffer solution is (pK_(a)(CH_(3)COOH)=4.75) :

Answer»

`4.75`
`5.75`
`6.75`
`9.25`

SOLUTION :`pH=pK_(a)+"LOG"(["SALT"])/(["ACID"])`
`=4.75+log 1=4.75`
Discussion
91184.

A buffer solution containing 0.1 mole of ammonium hydroxide and 0.15 mole of ammonium chloride per litre of the solution. Calculate the pH of the buffer solution. K_b for ammonium hydroxide is 1.8 xx10^(-5).

Answer»

SOLUTION :This is a buffer mixture containing a weak base and its salt. Hence the equation to be USED is
`pOH=pK_b+log.(["salt"])/(["base"])`
`pK_b=log K_b`
`=-log 1.8-log 10^(-5)=4.7447`
`therefore pOH=4.7447+log.(0.15)/(0.10)`
`pOH=4.7447+log 1.5`
`=4.7447+0.1761=4.9208`
`pH+POH=14` `pH+4.9208=14`
`pH=14-4.9208=9.0792`
`PH=9.08`
Discussion
91185.

A buffer mixture of acetic acid and potassium acetate has pH= 5.24. The ratio of [CH_3COO^-]/[CH_3COOH] in this buffer is, (pK_a = 4.740):

Answer»

3:1
1:3
1:1
1:2

Answer :A
Discussion
91186.

A bufer solution was prepared by dissolving 0.05 mol formic acid & 0.06 mol sodium formate in enough water to make 1.0 L of solution. Ka for formic acid is 1.80 xx 10^(-4) If the solution in (b) were diluted to 10 times its volume, what would be the pH ?

Answer»


ANSWER :PH =3.99
Discussion
91187.

A bubble of air is underwater at temperature 15^(@)C and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25^(@)C and the pressure is1.0 bar, what will happen to the volume of the bubble?

Answer»

Volume will become GREATER by a factor of 1.6
Volume will become greater by a factor of 1.1
Volume will become smaller by a factor of 0.70
Volume will become greater by a factor of 2.5

Solution :WE know that from IDEAL EQUATION
`Vprop T/P`
Given `T_(1)=15+273=287, P_(1)=1.5` BAR
`T_(2)=25+273=298, P_(2)=1` bar
`V_(1)prop 288/1.5` i.e. `V_(1)prop 192`
`V_(2)prop 298/1implies(V_(2))/(V_(1))=298/192=1.55~~1.6`
Discussion
91188.

A bubble of 8 moles of helium is submerged at certain depth n water. The temperature of water increases by 30^(@)C. How much heat is added approximately to helium (in kJ) during expansion?

Answer»


ANSWER :3
Discussion
91189.

A+Bto products, the kinetic data of the reaction is {:(A("m. lit"^(-1)),B("m. lit"^(-1)),"Rate"("m. lit"^(-1)"Sec"^(-1))),((1)0.5,0.5,2.xx10^(-4)),((2)0.5,1.0,1.99xx10^(-4)),((3)1.0,0.5,2.01xx10^(-4)):} The order of the reaction is

Answer»

ONE
zero
two
fractional

Answer :B
Discussion
91190.

A+Bto products is an elementary reaction. When excess of A is taken in this reaction, then the molecularity and other are respectively

Answer»

2 & 2
2 & 1
1 & 2
1 & 1

Answer :B
Discussion
91191.

A brown ring is formed in the ring test for NOg ion. It is due to the formation of

Answer»

`[Fe(H_(2)O)_(5) (NO)]^(2+)`
`FeSO_(4).NO_(2)`
`[Fe(H_(2)O)_(4)(NO)_(2)]^(2+)`
`FeSO_(4).HNO_(3)`

SOLUTION :`[Fe(H_(2)O)_(6)]^(2+) + NO tounderset("Brown ring")([Fe(H_(2)O)_(5)(NO)]^(2+) + H_(2)O)`
Discussion
91192.

A brown ring is formed in the ring test for NO_3^(-) ion. It is due to the formation of

Answer»

`[FE(H_2O)_5(NO)]^(2+)`
`FeSO_(4).NO_(2)`
`[Fe(H_2O)_4(NO)_2]^(2+)`
`FeSO_(4).HNO_(3)`

ANSWER :A
Discussion
91193.

A brown coloured mixture of two gases is obtained by the reduction of 6N nitric acid with metallic copper. The mixture on cooling condenses to a blue liquid which on freezing (-30^@ C) gives a blue solid.The correct choice for blue liquid or solid is :

Answer»

It is REFERRED to as the ANHYDRIDE of nitrous acid
It is an acidic oxide and hence dissolves in alkalies producing nitrites
It can also be prepared by the action of 50% `HNO_3` on arsenious oxide and then cooling to 250 K
All of these

Solution :`2Cu+6HNO_3 to 2Cu(NO_3)_2 + NO +NO_2 +3H_2O`
`NO+NO_2 OVERSET(-20^@C)HARR N_2O_3 ("blue"), (A) 2HNO_2 to N_2O_3 +H_2O`
(B)`N_2O_2+2KOH to 2KNO_2+ H_2O`
(C )`2NHO_3 + As_2O_3 + 2H_2O to`

All options are correct.
Discussion
91194.

A brown ring appears in the test for

Answer»

Nitrate
Nitrite
Bromide
Iron

Solution :At the junction of salt solution and `FeSO_(4)` solution with conc. `H_(2)SO_(4)` a BROWN RING is OBTAINED `UNDERSET("Brown ring")((FeSO_(4).NO))`
Discussion
91195.

A brown coloured mixture of two gases is obtained by the reduction of 6N nitric acid with metallic copper. Condeness to a blue liquid which on freezing at -30^(@)C gives a blue solid. The correct choice for blue liquid or solid is .

Answer»

It is referred to as the anhydride of nitrous acid
It is an acidic OXIDE and hence dissolves in alaklies producing nitries.
It can ALSO be PREPARED by the action of `50% HNO_(3)` on arsenious oxide and then cooling to 250 K.
All of these

Answer :D
Discussion
91196.

(A): Bromine does not displace chlorine from its salt solution (R): Chlorine is displaced from its oxysalt by bromine

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 :B
Discussion
91197.

A brown colloidal solution of platinum was prepared by ........... in 1898.

Answer»

SOLUTION :GEORGE BREDIG
Discussion
91198.

A broad spectrum antibiotic is

Answer»

PARACETAMOL
Penicillin-G
ASPIRIN
CHLORAMPHENICOL

SOLUTION :chloramphenicol is broad spectrum
Discussion
91199.

What are broad spectrum antibiotics.

Answer»

Paracetamol
Penicillin
Aspirin
Chloramphenicol.

Answer :D
Discussion
91200.

A bright cone of light is called .......... in Tyndall effect.

Answer»

TYNDALL CONE
Aggregate cone
Light cone
None of above 

SOLUTION :Tyndall cone
Discussion