Explore topic-wise InterviewSolutions in Current Affairs.

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1.

State and explain Kohlrausch's law ofindependent migration of ions.

Answer»

Solution :Statement of Kohlrausch's LAW: This statesthat of INFINITE dilutionof the solution,each ionof an electrolytemigratesindependentlyof its co-ionsand contributesindependentlyto thetotalmolarconductivityof the electrolyte, irrespectiveof the natureof otherions presentin te solution .
Explanation :Both theions, cationand anionof the electrolytemake a definitecontributionto the molar conductivity ofthe electrolyte at infinite dilution or ZERO concentration`(wedge_(0))`.
Of `lambda_(+)^(0)` and `lambda_(-)^(0)`are the molarconductivities of cation and nation respectively at infinite dilution, then
`wedge_(0) = lambda_(+)^(0) + lambda_(-)^(0)` .
This is known as Kohlrausch's law of independentmigrationof ions.
`wedge_(0KCl) = lambda_(K^(+))^(0) + lambda_(Cl^(-))^(0)` and `wedge_(0NaCl) = lambda_(Na^(+))^(0) + lambda_(Cl^(-))^(0)`
`:. wedge_(0KCl) - wedge_(0NaCl)`
`= (lambda_(K^(+))^(0) + lambda_(Cl^(-))^(0)) - (lambda_(Na^(+))^(0) + lambda_(Cl^(-))^(0)) = lambda_(K^(+))^(0) - lambda_(Na^(+))^(0) + lambda_(I^(-))^(0)`
`:. wedge_(0KI) - wedge_(0NAl) = (lambda_(K^(+))^(0) + lambda_(I^(-))^(0)) - (lambda_(Na^(+))^(0) + lambda_(I^(-))^(0))`
`= lambda_(K^(+))^(0) - lambda_(Na^(+))^(0) = 150.3- 126.9 = 23.4`
Thus the difference in `wedge_(0)` valuesfor `K^(+)` and `Na^(+)` saltsis constant and is independentof the nature of other ANIONS.
This provesthe validity of Kohlrausch's law.
Discussion
2.

State and explain Freundlich adsorption isotherm.

Answer»
Discussion
3.

State and explainFaraday's secondlaw of electrolysis .

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Solution :Explanation :
(1) Faraday's second law of ELECTROLYSIS : It states thatwhen the same quantity of electricityis passedthroughdifferentelectrolyticcells containingdifferentelectrolytes,arrangedin series,then the amounts of the substancesproducedby oxidationor reductionat therespectiveelectrodesare directly proportional to theirrespectivechemical equivalent masses.
(2) Mathematical EXPRESSION : If `W_(1)` and `W_(2)` are themasses of two substances A and B havingmolar masses`M_(1)` and `M_(2)` respectively.
Number of moles of `A = W_(1)//M_(1)`
Number of moles of `B = W_(2)//M_(2)`
Now from half cell reactions at two ELECTRODES,
Moles of A produced = Moles of electrons passed `xx` mole RATIO of A in half reaction
Moles of B produced = Moles of electrons passed `xx` molesratio of B in half reaction
Hence Faraday's second law can be mathematicallyrepresentedas.
`("moles of A produced")/("moles of B produced") = ("moles ratio of A halfreaction")/("moles ratio of B half reaction")`
Discussion
4.

State Faraday's first law of electrolysis.

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ANSWER :
Discussion
5.

State and derive Ostwaid dilution law for a weak electrolyte.

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SOLUTION :At constant temperature the degree of dissociation of weak electrolyte PROCESS with dilution in such a way that the value of the `(alpah^2 c)/(1- alpha)` remains constant.
LET US CONSIDER weak acid
`HA leftrightarrow H^(+)+ A^-`
c 0 0
`c-a c ca ca `
Initial concentration
Equilibrium concentration
Applying law of mass action
`K_a=([H^+][A^-])/([HA])`
`K_a=(ca times ca)/(c (1-a))=(c a^2)/(1-a)`
Discussion
6.

State an example and function of the following: wide spectrum antibiotics.

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Solution :Chloramphenicol is a WIDE spectrum antibiotic. It is used to treat acute fever, typhoid, dysentry, urinary INFECTIONS, MENINGITIS and pneumonia. VANCOMYCIN and OFLOXACIN are other broad spectrum antibiotics.
Discussion
7.

State an example and function of the following : wide spectrum antibiotics.

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Solution :Chloramphenicol is a wide spectrum antibiotic. It is used for the treatment of TYPHOID, dysentery, meningitis and acute FEVER. It is effective against a NUMBER of MICROORGANISMS.
Discussion
8.

Why the use of Aspartame is limited to cold foods and soft drinks?

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SOLUTION :It is UNSTABLE at HIGH TEMPRATURE.
Discussion
9.

State a consequence of lanthanoid contraction shown by transition elements.

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SOLUTION :The PROPERTIES of 4D and 5d-transition ELEMENTS rasemble CLOSELY.
Discussion
10.

Starting with toluene, outline a synthesis of (a) 1-bromo- 2-trichloromethylbenzene,(b) 1-bromo-3-trichloromethylbenzene, and (c) 1-bromo-4-trichloromethylbenzene.

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Solution : Compounds (a) and (c) can be obtained by ring BROMINATION of toluene followed by benzylic radical CHLORINATION of the side CHAIN USING three molar equivalents of CHLORINE:
Discussion
11.

State a condition under which a bimolecular reaction is kinetically first order reaction.

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SOLUTION :A bimolecular reaction BECOMES first order reaction when one of the reactants is in EXCESS.
Discussion
12.

Starting with three different amino acid molecules, how many different tripeptide molecules are formed

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12
9
8
6

Solution :SIX TYPES of TRIPEPTIDE MOLECULES are FORMED.
Discussion
13.

Starting with one mole of a compound X it is found that the reaction is 3/4 complete in 1 min . What is the rate constant if the reaction follows first order ?

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`0.213 SEC^(-1)`
`2.31 sec^(-1)`
`0.0231 sec^(-1)`
`2.31 xx10^(-3) sec^(-1)`

Solution :(C ) `t_(3//4) = (2.303)/(K)` LOG 4
`:. K = (2.303)/(60 sec)"log" 4 = (2.303xx0.602)/(60)`
`=0.0231 sec^(-1)`
Discussion
14.

Starting with alcohol, outline a synthesis of cyclohexyl chloride

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SOLUTION :
Discussion
15.

Starting with alcohol, outline a synthesis ofbutyl bromide

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SOLUTION :
Discussion
16.

Starting with alcohol, outline a synthesis of benzyl bromide

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SOLUTION :
Discussion
17.

Starting materials for the preperation of inorganic benzene is

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`B_2 H_6 + HN_3`
`BH_3 + NH_2 NH_2`
`B_2 H_6 + NH_3`
`BH_3 + NH_2 OH`

ANSWER :C
Discussion
18.

Starting with 10 g of a radioactive substance 0.1 gis left after 10 days. The rate constant for the decay is:

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`0.9212"DAY"^(-1)`
`0.4606"day"^(-1)`
`100"day"^(-1)`
`0.001"day"^(-1)`

ANSWER :B
Discussion
19.

Starting from the following reagents, how will your prepare ethanenitrile. CH_3CONH_2

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SOLUTION :By DEHYDRATION of primary amides and ALDOXIMES with `P_2O_5`
`{:(CH_3-CONH_2underset(-H_2O)OVERSET(P_2O_5)(to)CH_3-CN),("AcetamideEthanenitrile"):}`.
Discussion
20.

Starting materials for nylon -66 are

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hexmethylene diamine and ethylene GLYCOL
hexamethylene diamine and ADIPIC acid
hexamethylene diamine and DMT
hexamethylene diamine and GLYCEROL

Answer :B
Discussion
21.

Starting material for Nylon - 6 is

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`ALPHA` - CAPROLACTUM
`BETA`- caprolactum
`in`- caprolactum
`GAMMA`- caprolactum

SOLUTION :
Discussion
22.

Starting from the following reagents, how will your prepare ethanenitrile. CH_3CH=NOH

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SOLUTION :`{:(CH_3-CH=NOHunderset(-H_2O)OVERSET(P_2O_5)(to)CH_3-CN),("AcetaldoximesEthanenitrile"):}`
Discussion
23.

Starting from SiCl_(4), prepare the following in steps not exceeding the number given in parentheses. Silicon

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SOLUTION :`3SiCl_(4)+4AL to 4AlCl_(3)+3Si`
Discussion
24.

Starting from the following reagents, how will your prepare ethanenitrile. CH_3Br

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Solution :Aryl cyanide cannot be prepared in this METHOD because of their less reactivity TOWARDS NUCLEOPHILIC substitution. Aryl cyanides are prepared using Sandmeyers reactions.
`{:(KCN+CH_3-Br to CH_3-CN+KBr),("methyl bromideethanenitrile"):}`.
Discussion
25.

Starting from SiCl_(4), prepare the following in steps not exceeding the number given in parentheses. Na_(2)SiO_(3).

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Solution :`SiCl_(4)+4H_(2)O to underset("orthosilicic acid")(H_(4)SiO_(4)+4HCL)`
`H_(4)SiO_(4)OVERSET(Delta)to SiO_(2)+2H_(2)O`
`SiO_(2)+Na_(2)CO_(3)to Na_(2)SiO_(3)+H_(2)O`.
Discussion
26.

Starting from SiCl_(4), prepare the following in steps not exceeding the number given in parentheses. Linear silicon containing methyl groups only

Answer»

Solution :`n(CH_(3))_(2)SI(OH)_(2)OVERSET(Delta)to-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(Si)-O-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(Si)-O-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(Si)-O-`
Discussion
27.

Starting from radium, the radioactive disintegration process terminates when the following is obtained

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Lead
Radon
Radium A
Radium B

Solution :RA belongs to `(4n +2)` SERIES. END PRODUCT will also belong to the same series.
Discussion
28.

Starting from propanoic acid, the following reactions were carried out what is the compound Z

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`CH_(3)-CH_(2)-BR`
`CH_(3)-CH_(2)-NH_(2)`

`CH_(3)-CH_(2)-CH_(2)-NH_(2)`

Solution :`CH_(3)CH_(2)COOH overset(SOCl_(2))toCH_(3)CH_(2)COCl overset(NH_(3))toCH_(3)CH_(2)CONH_(2) overset(Br_(2)+KOH)to CH_(3)CH_(2)NH_(2)`
Discussion
29.

Starting from radium, the radioactive disintegration process terminates when the following is obtained:

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LEAD
RADON
RADIUM A
Radium B

Answer :A
Discussion
30.

Starting from phenol, how would you prepare the following compounds. (i)Benzene (ii) Aniline (iii) Anisole (iv) 1,4, benzoquinone (v) Cyclohexanol

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SOLUTION :
Discussion
31.

Starting from methyl magnesium iodide, how would you prepare acetic acid?

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SOLUTION :
Discussion
32.

Starting from diethyl malonate, what type of acids can prepared from it?

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Succinic ACID
Barbituric acid
ADIPIC acid
Monoalkyl acetic acid

Answer :A::B::C::D
Discussion
33.

Starting from ethyl ethanoate, how would you prepare ethanol?

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SOLUTION :`UNDERSET("ETHYL ethanoate")(CH_3COOCH_2CH_3)underset(H_2O)OVERSET(LiAlH_4)to underset("ETHANOL")(2CH_3-CH_2OH)`
Discussion
34.

Starting from methyl magnesium bromide, how would you obtain ethane nitrile?

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Solution :`underset("METHYL magnesium bromide")(CH_(3)MGBR)+underset("Cyanogaen chloride")(Cl-CN)tounderset("Ethane nitrile")(CH_(3)-CN)+ `
Discussion
35.

Starting from butyl magnesium bromide, how would you obtain 2-methyl hexan-2-o1?

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SOLUTION :
Discussion
36.

Starting from benzene diazonium chloride how would you get bright orange azo dye?

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SOLUTION :
Discussion
37.

Starting frombenzene diazonium chloride, how would you prepare (i) Biphenyl (ii) Phenyl hydrazine (iii) p-hydroxy azo benzene (iv) p-amino azo benzene (v) Chloro benzene

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SOLUTION :(i) `underset("Benzene DIAZONIUM chlorid")(C_(6)H_(5)N_(2)^(+)Cl^(-))+underset("Benzene")(C_(6)H_(6))overset(NaOH)(to)underset("BIPHENYL")(C_(6)H_(5))-C_(6)H_(5)+N_(2)uarr+HCl`
(ii) `C_(6)H_(5)N_(2)^(+)Cl^(-)underset(("or")Na_(2)SO_(3))overset(Sncl_(2)+HCl)(to)underset("Phenol hydrazine")(C_(6)H_(5)NH-NH_(2))`

(V) `C_(6)H_(5)N_(2)^(+)Cl^(-)overset(Cu//HCl)(to) underset("Chloro benzene")(C_(6)H_(5)Cl)+N_(2)uarr`
Discussion
38.

Starting from Benzene diazonium chloride, how would you prepare (i) Benzen (ii) Phenol (iii) Nitro benzene (iv)( Benzoic acid (v) Fluorobenzene

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SOLUTION :(i) `underset("Benzene diazonium chloride")(C_(6)H_(5)N_(2)^(+)Cl^(-))+H_(3)PO_(2)+H_(2)Otounderset("Benzene")(C_(6)H_(6))+H_(3)PO_(3)+HCl+N_(2)uarr`
(ii) `C_(6)H_(5)N_(2)^(+)Cl^(-)+H_(2)O overset(DELTA//283K)(to)underset("Phenol")(C_(6)H_(5)OH)+N_(2)uarr+HCl`
(iii) `C_(6)H_(5)N_(2)^(+)Cl^(-)+HBF_(4)toC_(6)H_(5)N_(2)^(+)BF_(4)^(-)underset(Delta)overset(NaNO_(2)//Cu)(to)underset("Nitro benzene")(C_(6)H_(5)NO_(2))+N_(2)uarr +NaBF_(4)`
(iv) `C_(6)H_(5)N_(2)^(+)Cl^(-)+HBF_(4)toC_(6)H_(5)N_(2)^(+)BF_(4)^(-)+CH_(3)COOHtounderset(Benzoic acid")(C_(6)H_(5)COOH)+BF_(3)+CH_(3)F`
(v) `C_(6)H_(5)N_(2)^(+)Cl^(-)+HBF_(4)toC_(6)H_(5)N_(2)^(+)BF_(4)^(-) overset(Delta)(to) underset("FLUORO benzene")(C_(6)H_(5)F)+BF_(3)+N_(2)uarr`
Discussion
39.

Starting from aniline how can the following be prepared ? (a) Chloro benzene (b) p-hydroxy azobenzene (c) Benzonitrile (d) (d) p-amino azo benzene

Answer»

SOLUTION :(a) `C_(6)H_(5)NH_(2) OVERSET(HNO_(2))underset(HCl)rarr C_(6)H_(5)N_(2)CL overset(HCl)underset(Cu_(2)Cl_(2))rarr C_(6)H_(5)Cl`
(b) `C_(6)H_(5)NH_(2) overset(HNO_(2))underset(HCl)rarr C_(5)H_(5)N_(2)Cl overset("Phenol")rarr C_(6)H_(5)-N=N-C_(6)H_(4)-OH`
(c ) `C_(6)H_(5)NH_(2) overset (HNO_(2))underset(HCl)rarr C_(6)H_(5)N_(2)Cl overset(KCN) underset(Cu_(2)(CN)_(2))rarr C_(6)H_(5)CN`
(d) `C_(6)H_(5)NH_(2) overset (HNO_(2)) underset(HCl)rarr C_(6)H_(5)N_(2)Cl overset("Aniline")rarr C_(6)H_(5)N=NC_(6)H_(4)NH_(2)`
Discussion
40.

Starting from aniline how can the following be prepared ? (a) Chloro benzene (b) p-hydroxy azobenzene (c) Benzonitrile (d) p-amino azo benzene

Answer»

Solution :(a) `C_(6)H_(5)NH_(2) OVERSET(HNO_(2))underset(HCl)rarr C_(6)H_(5)N_(2)CL overset(HCl)underset(Cu_(2)Cl_(2))rarr C_(6)H_(5)Cl`
(b) `C_(6)H_(5)NH_(2) overset(HNO_(2))underset(HCl)rarr C_(5)H_(5)N_(2)Cl overset("PHENOL")rarr C_(6)H_(5)-N=N-C_(6)H_(4)-OH`
(c ) `C_(6)H_(5)NH_(2) overset (HNO_(2))underset(HCl)rarr C_(6)H_(5)N_(2)Cl overset(KCN) underset(Cu_(2)(CN)_(2))rarr C_(6)H_(5)CN`
(d) `C_(6)H_(5)NH_(2) overset (HNO_(2)) underset(HCl)rarr C_(6)H_(5)N_(2)Cl overset("ANILINE")rarr C_(6)H_(5)N=NC_(6)H_(4)NH_(2)`
Discussion
41.

Starting form Benzene diazonium chloride, how will you get Nitrobenzene?

Answer»

Solution :When benzene diazonium CHLORIDE is treated with fluororic acid we get diazonium fluorobroate which on treated with sodium nitrite solution in the PRESENCE of copper, nirobenzene is obtained.
`underset("Benzene diazonium chloride")(C_(6)H_(5)N_(2)""^(+)Cl^(-))+underset("Hydro fluoro boric acid")(HBF_(4))to underset("Benzenediazonium fluoroborate")(C_(6)H_(5)-N_(2)^(+)BF_(4)^(-)BF_(4)^(-)underset(DELTA)overset(NaNO_(2)//Cu)(to) underset("Nitrobenzene")(C_(6)H_(5)NO_(2))+N_(2)uarr+NaBF_(4)`
Discussion
42.

Starch when heated with enzyme diastase yields

Answer»

GLUCOSE
SUCROSE
MALTOSE
GLYCOGEN

SOLUTION :maltose
Discussion
43.

Starch reacts with iodine to give :

Answer»

blueblack COLOUR
BLACK colour
RED colour
yellow colour.

ANSWER :A
Discussion
44.

Starch paper is used to test for the presence of

Answer»

Iodine
Oxidising agent
Iodine ion
Reducing agent

Solution :STARCH paper are USED for iodine test
as: `I^(-)+" oxidant "rarr I_(2)`
`I_(2)+" starch "rarr "BLUE COLOUR"`
Discussion
45.

Starch paper moistened with KI solution turns blue in ozone because of:

Answer»

IODINE liberation
Oxygen liberation
Alkali formation
Ozone REACTS with LITMUS paper

Answer :A
Discussion
46.

Starch undergoes hydrolysis in presence of mineral acids to give fructose.True/false

Answer»


ANSWER :GLUCOSE
Discussion
47.

starch on hydrolysis produces

Answer»

`ALPHA`-D-glucose
`BETA`-D-glucose
`alpha`-D-fructose
`beta`-D-fructose

Answer :A
Discussion
48.

Starch on hydrolysis by a dilute inorganic mineral acid gives :

Answer»

SUCROSE
GLUCOSE
FRUCTOSE
MALTOSE

ANSWER :D
Discussion
49.

Starch is used to test even a small of

Answer»

Urea in BLOOD
Iodine in AQUEOUS SOLUTION
PROTEIN in blood
Glucose in aqueous solution

Solution :Iodine in aqueous solution
Discussion
50.

Starch is used for …………………………in plants.

Answer»

SOLUTION :energystorage
Discussion