Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

In the case of immiscible liquids, the addition of one liquid to another does not after the properties of either liquid. Hence liquid exerts its own vapour pressure independently of the pressure of the other. Since boiling point of any system is the temperature at which its total vapour pressure becomes equal to the prevailing pressure, so by knowing the miscibility of two liquids, we can find out boiling temperature.If the vapour pressure of two liquids X and Y in their pure state at 300 K are 85 mm of Hg and 120 mm of Hg respectively and if their mole fraction in liquid solution are 0.45 and .55 respectivel. Then find out the total vapour pressure above the mixture containing two immiscible liquids at 300 K.

Answer»

<P>`104.25 mm` of HG
`205 mm` of Hg
`27.75 mm` of hg
35 mm of Hg

Solution :`p_("Total")=p_(A)^(@)+p_(B)^(@)`
Discussion
2.

In the following reaction: RMgBr + HC(O E t)_(3) overset("Ether")rarr overset(H_(3)O^(+))rarr P The product 'P' is

Answer»

RCHO
`R_(2)CHOEt`
`R_(3)CH`
`RCH(O E t)_(2)`

Solution :
Discussion
3.

In the case of immiscible liquids, the addition of one liquid to another does not after the properties of either liquid. Hence liquid exerts its own vapour pressure independently of the pressure of the other. Since boiling point of any system is the temperature at which its total vapour pressure becomes equal to the prevailing pressure, so by knowing the miscibility of two liquids, we can find out boiling temperature. If we mix two immiscible liquids A and B then boiling point of the mixture will be (Given p_(A)^(@) gt p_(B)^(@))

Answer»

greater than `T_(B)` and lower than `T_(A)`
greater than EITHER of `T_(A)` and `T_(B)`
lower than `T_(B)` and higher than `T_(A)`
lower than either of `T_(A)` and `T_(B)`

SOLUTION : LIQUID will boil when
`p_(A)^(@)+p_(B)^(@)=` Atmospheric PRESSURE.
Discussion
4.

In the following reaction :R-OH overset(HCl) rarr R-Cl+H_(2)Othe, ZnCl_(2) is not required if R is …….

Answer»

`CH_(3)-`
`CH_(3)-overset(CH_(3))overset("|")underset(CH_(3))underset("|")("C ")-`
`CH_(3)-underset(C_(2)H_(5))underset("|")("C")H-`
`CH_(3)-CH_(2)-`

SOLUTION :`3^(@)`- alcohols can be converted EASILY to `3^(@)`- alkyl halides in ABSENCE of `ZnCl_(2)` catalyst.
Discussion
5.

In the case of immiscible liquids, the addition of one liquid to another does not after the properties of either liquid. Hence liquid exerts its own vapour pressure independently of the pressure of the other. Since boiling point of any system is the temperature at which its total vapour pressure becomes equal to the prevailing pressure, so by knowing the miscibility of two liquids, we can find out boiling temperature. In question 6, if we decrease the amount of liquid X in the mixture then what will bethe effect on the boiling point of the mixture at constant temperature?

Answer»

It will remain constant
Decrease
Increases
Can't predict

Solution :There will be no effect on changing the CONCENTRATION of any COMPONENT as they are IMMISCIBLE.
Discussion
6.

In the following reaction RCH_(2)COOH overset(Br_(2)//P)rarr X overset("excess "NH_(3))rarr Y The major compounds X and Y are

Answer»

`RCH(BR)CONH_(2), RCH(NH_(2))COOH`
`RCH(Br)COOH,RCH(NH_(2))COOH`
`RCH_(2)COBr,RCH_(2)COONH_(4)`
`RCH(Br)COOH,RCH_(2)CONH_(2)`

Solution :`X=R-UNDERSET(Br)underset(|)CH-COOH, Y=R-underset(NH_(2))underset("|")"CH"COOH`
Discussion
7.

In the following reaction, RCH_2CH=CH_2+I Cl to [A] Markownikoff's product [A] is

Answer»

`RCH_2undersetunderset(Cl)(|)CH-CH_2I`
`RCH_2-undersetunderset(I)|CH-CH_2Cl`
`RCH-undersetundersetI|CH=CH_2`
`RCH=CH-CH_2I`

Solution :Markownikoff's addition : The NEGATIVE part of the unsymmetrical REAGENT adds to a less hydrogenated (more substituted ) carbon atom of the double bond.
In IC L , Cl is more electronegative. So it will take negative CHARGE . i.e., `I^(+)Cl^(-)`
So, the PRODUCT is
`RCH_2CH=CH_2+IC l to R-CH_2-undersetunderset(Cl)(|)CH-CH_2I`
Discussion
8.

In the case of d-block elements :

Answer»

OUTERMOST and PENULTIMATE SHELLS are incomplete
Both penultimate and prepenultimate shells are incomplete
Outermost SHELL is incomplete
Innermost shell is incomplete

Answer :A
Discussion
9.

In the following reaction : RCH_(2)CH = CH_(2) + Icl rarr [A] Markovinkoff's product [A] is :

Answer»

`RCH_(2)underset(Cl)underset(|)(CH) - CH_(2)I`
`RCH_(2) - underset(I)underset(|)(CH) - CH_(2)Cl`
`RCH - underset(I)(CH) = CH_(2)`
`RCH = CH - CH_(2)I`

Solution :`RCH_(2)CH = CH_(2) + ICL rarr RCH_(2) - underset(Cl)underset(|)(CH) - CH_(2)I`
In ACCORDANCE with Markovnikov.s rule, more ELECTRONEGATIVE chlorine atom goes to the carbon with LESSER number of hydrogens.
Discussion
10.

In the following reaction oxidation state of fluorine changes to 2F_(2)+H_(2)O to 4HF+O_(2)

Answer»

0 to -01
0 to +1
`-1 "to "0`
`+1" to "0`

Solution :In `F_2`, FLUORINE has O.S. zero. In HF, fluorine has O.S.-1.
Discussion
11.

In the case of a radioisotope, the value of …….and …..are indentical in magnitude. The value is

Answer»

1/0.693
`(0.693)^(2) `
`0.693`
`(0.693)^(1//2)`

SOLUTION :`t_(1//2) = (0.693)/(LAMBDA) i.e, t_(1//2) XX lambda = 0.693`.
or `X xx x = 0.693 i.e., x^2 = 0.693 " or " x = (0.693)^(1//2)`
Discussion
12.

in the following reaction, product 'P' is R-overset(O)overset(||)(C)-Cl underset(Pd-BaSO_(4))overset(H_(2))toP

Answer»

`RCH_(2)OH`
`RCOOH`
`RCHO`
`RCH_(3)`.

ANSWER :C
Discussion
13.

In the case of a radio isotope the value of T_(1//2) and lamda are identical in magnitude. The value is

Answer»

0.693
`(0.693)^(1//2)`
`1//0.693`
`(0.693)^(2)`

Solution :For a radioisotope,
half-life period, `T_(1//2)=0.696//LAMDA`
SINCE `T_(1//2)=lamda` (given), `T_(1//2)^(2)=0.693`
`impliesT_(1//2)=0.693^(1//2)`.
Then VALUE of `T_(1//2)` on `lamda` is `0.693^(1//2)`.
Discussion
14.

in thefollowingreaction, product P is

Answer»




ANSWER :B
Discussion
15.

In the following reaction, how is the rate of appearance of the underline product related to the rate of disappearance of the underlined reactant? BrO_(3)^(-)(aq)+ul(5Br^(-))(aq)+6H^(+)(aq)toul(3Br_(2))_((1))+3H_(2)O_((1))

Answer»

`(d[Br_(2)])/(dt)==5/3(d[Br^(-)])/(dt)`
`(d[Br_(2)])/(dt)=-(d[Br^(-)])/(dt)`
`(d[Br_(2)])/(dt)=-(d[Br^(-)])/(dt)`
`(d[Br_(2)])/(dt)=-3/5(d[Br^(-)])/(dt)`

SOLUTION :`(-d[BrO_(3)^(-)])/(dt)xx1/1=(-d[Br^(-)])/(dt)xx1/5=(d[Br_(2)])/(dt)xx1/3`
Discussion
16.

In the case of a first order reaction, the ratio of the time required for 99.9% completion of the reaction to its half life is nearly

Answer»

1
10
20
8

Answer :B
Discussion
17.

In the following reaction most stable intermediate is CH_3overset(CH_3)overset|CHCH=CH_2+H_2O overset(H^+)to

Answer»

`CH_3overset(CH_3)OVERSET|CHCH_2overset(o+)CH_2`
`CH_3oversetoverset(CH_3)|CH UNDERSET(o+)CHCH_3`
`CH_3underset(o+)overset(CH_3)overset(|)C CH_2CH_3`
`CH_3overset(CH_3)overset|CHCH_2overset(ѳ)CH_2`

ANSWER :C
Discussion
18.

In the case homologous series of alkanes, which one of the following statements is incorrect

Answer»

The members of the series are isomers of each other
The MEMBER of the series have similar chemical PROPERTIES
The members of the series have the general formula `C_n H_(2n+2)`, where n is an interger
The difference between any two successive members of the series correspondsto 14 units of relative ATOMIC mass

Solution :The difference between any two successive numbers of the homologous series `-CH_2`-i.e., the molecular weight of EVERY two ADJACENT members differ by 14.
`(CH_2=12+2=14)`
Discussion
19.

In the carbon cycle, from which hot stars obtain their energy, the C_6^14nucleusis

Answer»

COMPLETELY CONVERTED into energy
Regenerated at the end of the CYCLE
Combined with oxygen to FORM carbon monoxide
Completely converted into `D_1^2`

Answer :B
Discussion
20.

In the following reaction HC_(2)O_(4)^(-) + PO_(4)^(--) hArr HPO_(4)^(--) + C_(2)O_(4)^(--) Which are the two Bronsted bases

Answer»

`HC_(2)O_(4)^(-)` and `PO_(4)^(---)`
`HPO_(4)^(--)` and `C_(2)O_(4)^(--)`
`HC_(2)O_(4)^(-)` and `HPO_(4)^(---)`
`PO_(4)^(---)` and `C_(2)O_(4)^(--)`

Solution :Both possess the TENDENCY to accept PROTON.
Discussion
21.

In the carbon cycle from which stars hotter than the sun obtain their energy the ._(6)C^(12) isotope

Answer»

COMPLETELY CONVERTED into energy
Regenerated at the end of the cycle
Combined with oxygen to form carbon monoxide
Broken up into its constituent protons and neutrons

Answer :B
Discussion
22.

In the following reaction HCO_3^(-)+H_2OhArrCO_3^(2-)+H_3O^+ which two substances are Bronsted base ?

Answer»

`CO_3^(2-) and H_3O^+`
`HCO_3^- and H_3O^+`
`HCO_3^- and CO_3^(2-)`
`CO_3^(2-) and H_2O`

ANSWER :D
Discussion
23.

In the following reaction , HC -= CH underset(HgSO_4)overset(H_2SO_4)(to) X product 'X' will not give

Answer»

tollen's TEST
victor meyer test
iodoform test
fehling solution test

Solution :
(X) reduces TOLLENS REAGENT and fehling solution and it also answer iodoform test .
Discussion
24.

In the cannizzaro reaction given below, 2Ph-CHO overset(OH^(-))to Ph-CH_(2)OH+PhCO_(2)^(-) the slowest step is

Answer»

the attach of `.^(-)OH` at the carbonyl group
the transfer of hydride ion to the carbonyl group
the ABSTRACTION of a PROTON from the carboxylic acid
the deprotonation of `Ph-CH_(2)OH`.

Answer :B
Discussion
25.

In the Cannizzaro's reaction given below : 2Ph-CHOoverset (OH^-) rarr Ph-CH_2OH + PhCOO^- the slowest step is :

Answer»

The attack of OH^- at the carbonyl GROUP
The transfer of hydride to the carbonyl group
The ABSTRACTION of proton from the CARBOXYLIC acid
The DEPROTONATION of `Ph-CH_2OH`

Answer :A
Discussion
26.

In the following reaction H_3C-underset(CH_3)underset(|)oversetoverset(CH_3)|C-CH=CH_2overset(H_2O //H^+)to underset"Major Product"A+underset"Minor Product"B The major product is

Answer»

`H_3C-undersetunderset(OH)|oversetoverset(CH_3)|C-undersetunderset(CH_3)|CH-CH_3`
`undersetunderset(OH)|CH_2-undersetunderset(CH_3)|oversetoverset(CH_3)|C-CH_2-CH_3`
`H_3C-undersetunderset(CH_3)|oversetoverset(CH_3)|C-undersetunderset(OH)(|)CH-CH_3`
`H_3C-undersetunderset(CH_3)|oversetoverset(CH_3)|C-CH_2-undersetunderset(OH)|CH_2`

SOLUTION :
Discussion
27.

Inthe carbinol system , monohydric alcohols are named as derivatives of _________.

Answer»

METHYL ALCOHOL
ETHYL alcohol
n-Propyl alcohol
PHENOL

ANSWER :A
Discussion
28.

In the following reaction HC-=CH underset(Hg^(2+))overset(H_2SO_4)to 'P' Product 'P' will not give

Answer»

TOLLEN's reagent test
Brady's reagent test
VICTOR MEYER's test
Iodoform test

SOLUTION :`CH_3CHO` does not give Victor meyer test
Discussion
29.

In the Cannizzaro reaction given below: 2Ph-CHO overset(overset(Θ)OH)rarrPh-CH_(2)OH+PhCO_(2)^(-) the slowest step is:

Answer»

TGE attack of :` OH^(-)`at the carbonyl GROUP
the transfer of hydride to the carbonyl group
The abstraction of PROTON from the carboxylic group
The deprotonation of `PhCH_(2)OH`

ANSWER :B
Discussion
30.

In the following reaction, H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-CH=CH_(2) overset(H_(2)O //H^(+))to underset("Major product")(A)+underset("Minor product")(B) The major product is,

Answer»

`H_(3)C-UNDERSET(OH)underset(|)OVERSET(CH_(3))overset(|)(C)-underset(CH_(3))underset(|)(C)H-CH_(3)`
`underset(OH)underset(|)(C)H_(2)-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-CH_(2)-CH_(3)`
`H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-underset(OH)underset(|)(C)H-CH_(3)`
`H_(3)C-underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C)-CH_(2)-underset(OH)underset(|)(C)H_(2)`

Solution :
Discussion
31.

In the button cells widely used in watches and other devices, the following reaction takes place: Zn (s) + Ag_(2)O (s) + H_2O (l) to Zn^(2+) (aq) + 2Ag (s) + OH^(-) (aq) Determine E^(@) and Delta_(r)G^(@)for the reaction.

Answer»

Solution :Electrode reactions taking place in the button cell are as under :
`Zn(s) to Zn^(2+)(aq) + 2e^(-)`
`2Ag^(+)(aq) + 2e^(-) to 2Ag(s)`
---------------------------------------------------------------------
`Zn(s) + 2Ag^(+)(aq) to Zn^(2+)(aq) + 2Ag(s)`
THUS, n=2
`E^(@)` for the REACTION can be obtained as under:
`E_("cell")^(@)= (E_("cathode")^(@) - E_("anode")^(@))`
`=+0.80 V - (0.76 V) = + 1.56 V`
Use the following RELATION to calculate `Delta_(r)G^(@)`
`Delta_(r)G^(@) =-nE^(@) F`
`=-301.080 kJ mol^(-1)`
Discussion
32.

In the following reaction H_(3)C - underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C) - CH = CH_(2) overset(H_(2)O //H^(oplus))rarr underset("Major product")(A) + underset("Minor product")(B) The major product is

Answer»

`H_(3)C - underset(OH)underset(|)overset(CH_(3))overset(|)(C) - CH - CH_(3)`
`underset(OH)underset(|)(CH_(2)) - underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C) - CH_(2) - CH_(3)`
`H_(3)C - underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C) - CH - CH_(3)`
`H_(3)C - underset(CH_(3))underset(|)overset(CH_(3))overset(|)(C) - CH_(2) - underset(OH)underset(|)(CH_(2))`

SOLUTION :
Discussion
33.

In the button cells widely used in watches and other devices the following reaction takes place : Zn_((S))+Ag_(2)O_((S))+H_(2)O_((l)) to Zn_((aq))^(2+)+2Ag_((S))+2OH_((aq))^(-) Determine Delta_(r)G^(@) and E^(Theta) cell for the reaction.

Answer»

Solution :(a) CALCULATION of standard reduction potential `E_(cell)^(Theta)` :
The following reaction is carried out in button cell.
`Zn_((S))+Ag_(2)O_((S)) +H_(2)O_((l)) to Zn_((aq))^(2+) +2Ag_((S))+2OH_((aq))^(-)`
OXIDATION : `Zn_((S)) to Zn_((aq))^(2+)+2E^(-)`
`therefore E_(L)^(Theta)=E_(Zn^(2+)|Zn)^(Theta)=-0.76V`
Reduction : `Ag_(2)O_((S))+2e^(-) +H_(2)O_((l)) to 2Ag_((S))+2OH_((aq))^(-)`
`therefore E_(R)^(Theta)=E_(Ag_(2)O|Ag)^(Theta)=0.344V`
`therefore DeltaE_(cell)^(Theta)=E_(R)^(Theta)-E_(L)^(Theta)=E_(Ag_(2)O|Ag)^(Theta)-E_(Zn^(2+)|Zn)^(Theta)`
`=0.344-(-0.76)`
`+1.104V`
(b) Calculation of `Delta_(r)G^(@)`,`Delta_(r)G^(@)=-nFE_(cell)^(Theta)|" Where, "n=2" mol "`, `E_(cell)^(Theta)=1.104V`,`F=96500" C "mol^(-1)`
`therefore Delta_(r)G^(@)=-(2mol)xx(96500" C "mol^(-1))xx(1.104V)`
`=-213072CV`
`=-213072J`
`=-2.13xx10^(5)J`
`=-213.072kJ`
`=-213kJ`
Discussion
34.

In the button cells widely used in watches and other devices the following reaction takes plance : Zn(s)+Ag_(2)O(s)+H_(2)O(l) rarr Zn^(2+)(aq)+2Ag(s)+2OH^(-)(aq) Determine Delta_(r )G^(@) and E^(@) for the reaction.

Answer»

Solution :In the given cell, Zn is oxidised to `Zn^(2+)` and `Ag_(2)O` is REDUCED to Ag.
`therefore E_("cell")^(@)=E_("cathode")^(@)-E_("ANODE")^(0)=[0.344-(-0.76)V]=1.104 V`
`therefore Delta_(r )G^(@)=-nFE_("cell")^(@)=-2xx96500xx1.104=-2.13xx10^(5)J`
Discussion
35.

Inthefollowingreaction gamma- Hydroxy carboxylicacid underset((ii) ^(+)) overset((i) DIBALH) to A is

Answer»

`gamma ` - lactone
lactice
a hemicaetal
`alpha, BETA` - UNSATURATED acid

Answer :C
Discussion
36.

In the brown ring test, nitric oxide donates one electron to Fe^(2+) to form [Fe(H_(2)O)_(5)NO]^(2+), which contains ________ and ________ ions.

Answer»

SOLUTION :`NO^(+), FE^(+)`
Discussion
37.

In the following reaction equation, what is A and C ? (i) CH_(3)-overset(CH_(3))overset(|)CH-underset(OH)underset(|)CH-CH_(3) overset(H^(+)//"Heat")to underset("Main product") (A)+ underset("By product")(B) ii) A overset("HBr, dark")to underset("Main product") (C) +underset("By product")(D)

Answer»

`CH_(3)=OVERSET(CH_(3))overset(|)C-CH_(2)-CH_(3)` and `UNDERSET(Br)underset(|)CH_(2)-overset(CH_(3))overset(|)CH-CH_(2)-CH_(3)`
`CH_(3)=overset(CH_(3))overset(|)C-CH-CH_(3)` and `CH_(3)-overset(CH_(3))overset(|)underset(Br)underset(|)CH-CH_(2)-CH_(3)`
`CH_(3)=overset(CH_(3))overset(|)C-CH-CH_(3)` and `CH_(3)-overset(CH_(3))overset(|)CH-underset(Br)underset(|)CH-CH_(3)`
`CH_(3)=overset(CH_(3))overset(|)C-CH_(2)-CH_(3)` and `CH_(3)-underset(Br)underset(|)overset(CH_(3))overset(|)C-CH_(2)-CH_(3)`

Solution :`CH_(3)=overset(CH_(3))overset(|)C-CH-CH_(3)` and `CH_(3)-overset(CH_(3))overset(|)underset(Br)underset(|)CH-CH_(2)-CH_(3)`
Discussion
38.

In the brown ring test , the brown colour of the ring is due to

Answer»

a MIXTURE of NO and `NO_(2)`
nitrosoferrous SULPHATE
ferrous nitrate
ferric nitrate

Solution :No forms dark brown nitrosoferrous sulphate.
`underset("Impure GAS")(FeSO_(4)+NO)tounderset("dark brown")(FeSO_(4)*NO)`(Nitrosoferrous sulphate) When this solution is heated , pure nitric OXIDE is liberated .
`FeSO_(4)*NOoverset("heat")toFeSO_(4)+underset("pure gas")(NO)`
Discussion
39.

In the following reaction, final product is :

Answer»

`ClCH_(2)underset(OH)underset(|)CHoverset(14)CH_(2)OC_(2)H_(5)`
`ClCH_(2)underset(OC_(2)H_(5))underset(|)Coverset(14)H_(2)ON a`

Solution :N//A
Discussion
40.

In the brown ring complex [Fe(H_2 O)(NO)]SO_4, nitric oxide behaves as

Answer»

`NO^(+)`
NEUTRAL NO molecule
`NO^(-)`
none of these

Solution :No replaces a WATER ligand and by doing so , it is oxidised to a nitrosy ligand , `NO^(+)` and the COMPLEX formed is `[Fe(H_(2)O)_(5) NO]^(2+)`. In this complex iron is PRESENT in +1 oxidation state.
Discussion
41.

In the following reaction Compound X is :

Answer»




ANSWER :C::D
Discussion
42.

In the brown ring test, brown colour of the ring is due to ………………. .

Answer»

a MIXTURE of `NO and NO_(2)`
NITROSO FERROUS SULPHATE
Ferrous nitrate
Ferric nitrate.

Solution :Nitroso ferrous sulphate
Discussion
43.

In the following reaction CH_3-CH_2-CH_2-CH_3 underset"475 K"overset(H_2SO_4)to

Answer»

`CH_3CH=CHCH_3` predominates
`CH_2=CHCH_2CH_3` predominates
Both are FORMED in equal amounts
The amount of production depends on the nature of catalyst

Solution :`CH_3-CH_2-CH_2-CH_3underset"475 K"overset(H_2SO_4)to underset"More symmetrical (MAJOR PRODUCT )"(CH_3-CH=CH-CH_3)`
`CH_3-CH_2-CH_2-CH_3underset"475 K"overset(H_2SO_4)to underset"Less symmetrical for UNSYMMETRICAL (MINOR product)"(CH_2=CH-CH_2-CH_3)`
It is based onSaytzeff's rule . According to this more symmetrical or more alkylated alkene predominates.
Discussion
44.

In the Bohr's model , for unielectronic species following symbols are used r_(n,z)to Radius of n^"th" orbit with atomic number "z" U_(n,z)to Potential energy of electron in n^"th" orbit with atomic number "z" K_(n,z)toKinetic energy of electron in n^"th" orbit with atomic number "z" V_(n,z)to Velocity of electon in n^"th" orbit with atomic number "z" T_(n,z)to Time period of revolution of electon in n^"th" orbit with atomic number "z" Calculate z in all in cases. (i)U_(1,2):K_(1,z)=-8:9 (ii)r_(1,z):r_(2,1) =1:12 (iii)v_(1,z):v_(3,1)=15:1 (iv)T_(1,2):T_(2,z)=9:32 Report your answer as (2r-p-q-s) where p,q,r and s represents the value of "z" in parts (i),(ii),(iii),(iv).

Answer»


ANSWER :1
Discussion
45.

In the following reaction, CH_(3)NH_(2)+CHCl_(3)+KOH to Nitrogen containing compound +KCl+H_(2)O. The nitrogen containing compound is

Answer»

`CH_(3)-NH-CH_(3)`
`CH_(3)-C-=N`
`CH_(3)overset(-)N-=overset(+)C`
`CH_(3)overset(+)N-=overset(-)C`

SOLUTION :`H_(3)C-N^(+)-=C^(ODOT)`
Discussion
46.

In the Bohr's orbit, what is the ratio of total kinetic energy and the total energy of the electron?

Answer»

`-1`
`-2`
`+1`
`+2`

Solution :Kinetic energy `= 1/2mv^(2)`
POTENTIAL energy `= -(ZE^(2))/R`
But electrostatic force, `(Ze^(2))/(r^(2))=(MV^(2))/r` (centrifugal force)
`THEREFORE` Potential energy `=-mv^(2)`
Total energy `= 1/2mv^(2)-mv^(2)=-1/2mv^(2)`
`therefore` Kinetic energy/total energy `= -1`.
Discussion
47.

In the Borax bead test of Co^(2+), the blue colour of bead is due to the formation of :

Answer»

`B_(2)O_(3)`
`Co_(3)B_(2)`
`CO(BO_(2))_(2)`
`Co O`

ANSWER :C
Discussion
48.

In the following reaction, calculate the mass of Cl_(2) required to produce 1385 gof KCiO_(4) (K=39):- Cl_(2)+ 2KOH to KCl +KClO +H_(2)O 3KClO to 2 KCl +KClO_(3) 4KClOto 3KClO_(4)+ KCl

Answer»

40 G
2840 g
1420 g
710 g

Answer :B
Discussion
49.

In the Bohr’s model, for unielectronic species following symbols are used, r_(n,z)toRadius of n ^(th) orbit with atomic number "z". U _(n,z) to Potential energy of electron in n ^(th) orbit with atomic number "z". K _(n , z)to Kinetic energy of electron inn^(th) orbit with atomic number "z". v_(n,z) to Velocity of electron inn^(th) orbit with atomic number "z". T_(n,z)to Time period of revolution of electron in n^(th) orbit with atomic number "z". Calculate z in all in cases. (i) U _(1,2) : K _(1,z) =-8:1 (ii) r _(1,z) : r _(2,1) =1:8 (iii) v _(1,z):v _(3,1)=9:1 (iv)T_(1,2) : T_(2,z) =9:32 Represent your answer as abcd, where a,b,c and d represent number from 0 to 9. a,b,c and d represents the value of "z" in parts (i), (ii), (iii) & (iv). Suppose your answer is 1,2,3 & 4 then the same must be filled in OMR sheet as 1234.00.

Answer»

2233
1233
3233
4233

Answer :B
Discussion
50.

In the following reaction : CaC_(2) (s) + 2H_(2) O (l) rarr C_(2) H_(2) (g) + Ca (OH)_(2) (s) Suppose 16 gm CaC_(2) reacts with 9 gm water and acetylene formed is collected in 10 litre vessel at temperature 300 K then its pressure will be :

Answer»

`0.75` ATM
`0.5` atm
`0.6` atm
`0.25` atm

Solution :`{:(,CaC_(2) (s),+,2H_(2)O (L),rarr,C_(2)H_(2) (g),+,Ca(OH)_(2) (s),),("MOLE",(1)/(4) "mole",,(1)/(2)"mole",,-,,-,),(-,,,-,,(1)/(4) " mole",,(1)/(4) "mole",):}`
`PV = nRT`
`P = (0.25 xx 0.08 xx 300)/(10)`
`P = 0.6 atm`
Discussion