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.

Consider a solution consisting of the following two buffer systems. {:(H_(2)CO_(3)hArrHCO_(3)^(-)+H^(+),pKa=6.4),(H_(2)PO_(4)^(-)hArrHPO_(4)^(2-)+H^(+),pKa=7.2):} At pH = 6.4, which one of the following is true for the concentration of acid and conjugate base present

Answer»

`[H_(2)CO_(3)]gt[HCO_(3)^(-)]&[H_(2)PO_(4)^(-)]gt[HPO_(4)^(2-)]`
`[H_(2)CO_(3)]=[HCO_(3)^(-)]&[HPO_(4)^(2-)]gt[H_(2)PO_(4)^(-)]`
`[HCO_(3)^(-)]gt [H_(2)CO_(3)[&[HPO_(4)^(2-)]gt[H_(2)PO_(4)^(-)]`
`[H_(2)CO_(3)]=[HCO_(3)^(-)]&[H_(2)PO_(4)^(-)]gt[HPO_(4)^(2-)]`

SOLUTION :`"At pH"=6.4`
As in CASE of
`pH=pKa` it will be only when
`"[weak acid]"="[conjugate base]"`
In case of `H_(2)PO_(4)^(-)|HPO_(4)^(2-)`
`pH=pKa+log.([HPO_(4)^(2-)])/([H_(2)PO_(4)^(-)])`
`6.4=7.2+log.([HPO_(4)^(2-)])/([H_(2)PO_(4)^(-)])`
`log.([HPO_(4)^(2-)])/([H_(2)PO_(4)^(-)])=-ve""([HPO_(4)^(2-)])/([H_(2)PO_(4)^(-)])LT1""rArr[HPO_(4)^(2-)]LT[H_(2)PO_(4)^(-)]`
Discussion
2.

Consider a setup of two urea solution of concentration C_(1) and C_(2)(C_(2)gtC_(1)), both at temperature T, separated by a semi permeable membrane. External pressure P_(1) and P_(2) respectively are applied on the two solutions. For what values of P_(1) and P_(2), osmosis does not occur through the semi permeable membrane? (R= universal gas constant)

Answer»

`P_(1)=C_(1)RT` and `P_(2)=C_(2)RT`
`P_(1)=` zero and `P_(2)=(C_(2)-C_(1))RT`
`P_(1)=C_(2)RT` and `P_(2)=C_(1)RT`
`P_(1)=2C_(1)RT` and `P_(1)=(C_(2)+C_(1))RT`

Solution :(A). Osmotic pressures of the TWO solutions are `pi_(1)=C_(1)RT` and `pi_(2)=C_(2)RT`. So, if we apply these external pressures on both solutions, they will not show osmosis with a pure solvent since they both will be in equilibrium with pure solvent (as far as osmatic flow is concerned). This implies that for these values of external pressures. both solutions will also be in equilibrium with each other and osmosis will not occur between them. If we add or subtract any common VALUE of pressure from their values in OPTION (A). the two solutions will still REMAIN in equilibrium. in option (B), we subtract `C_(1)RT` from each side's external pressures of option (A). in option (D), we add `C_(1)RT` on both sides (starting from option (A)), and the two solutions remain in equilibrium.
Discussion
3.

Consider a reversible reaction, which Statement about this reaction is correct? (P) The reaction will never reach equilibrium (Q) IF ((k_(1))/(k_(2)))=2"then at"t=oo,[A]=(2a)/(3) if initially only A was present and at t=0,[A]=a. (R ) At any instant ,If [A] =(a)/(5)"then" [B] =(2a)/(5), if(k_(1)/(k_(2)=2.

Answer»

Only P
P and Q
P,Q,R
Q and R

Answer :A
Discussion
4.

Consider a room that is 20 xx 10 m with 15 m eciling. If pollutant present is 2 ppm, how many grams of pollutant are present in this room. (Density of air = 1 gm//"litre"

Answer»


Solution :`V = L xx B xx H = 20 xx 10 xx 15`
`= 3000 m^(3) = 3000 xx 10^(3)` litre
`= 3 xx 10^(6)` litre
2 grams of pollutant in `10^(6)` grams of air
x grams of pollutant in `3 xx 10^(6)` GRAM of air`""` (because `d = 1 gm//"litre"` )
`x = (3XX 10^(6) xx 2)/(10^(6)) = 6`
Discussion
5.

consider a reaction that occurs by the following mechanism A + BC to AC+B AC + Dto A + CD the potential energy profile for the reaction is shown: a) Write the equation for the overall reaction. b) What are the different species present at reactions stagest 1 to 5? c) What is the rate determinig step?

Answer»

Solution :a) The equation for OVERALL reaction is:

b) State 1: BE (reactant), D (reactant), A (CATALYST)
stage 2: B………….C…………A (transition state), D (reactant)
stage 3: AC ( intermediate), B(products) , D(products)
stage 4: A..............C................D ( transition state), B (products)
Stage 5: CD (products) , B(products) ,A (catalyst)
c) The rate determining step is GIVEN by stage 3 since the potential energy is the minimum.It is: `A+BCto AC+B`
Discussion
6.

Consider a reaction aG + bH to products. When concentration of both the reactants G and H is doubled, the rate increases by eight times. However, when concentration of G is doubled keeping the concentration of H fixed, the rate is doubled. The overall order of the reaction is

Answer»

0
1
2
3

Answer :D
Discussion
7.

Consider a reaction aG + bH toProducts . When concentration of both the reactants G and H is doubled , the rate increase by eight times . However , when concentration of G is doubled keeping the concentration of H fixed , the rate is doubled . The overall order of the reaction is

Answer»

0
1
2
3

Solution :
Applying I'= k `[G]^(x) [H]^(y)`
x= 1 , y = 2 `THEREFORE ` overall order is 3 .
Discussion
8.

Consider a reaction aG + bH to products.When concentration of both the reactants G and H is doubled, the rate increases by eight times. However, when concectration of G is double keeping the concentrationof H fixed, the rate is doubled . The overall order of the reaction is

Answer»

0
1
2
3

Solution :`aG + bH to ` Product
Suppose order of REACTION = n
When concentration of both C and H is DOUBLED then the increase by eight time
rate `=" k (reactants)"^n`
`(8) = k(2)^n`
`n = 3`
When concentration of G is doubled keeping the concentration of H fixed, the rate is double.
rate `prop[G]^1`
Than rate `prop[G]^1[H]^2`
Discussion
9.

Consider a reaction aG+bH toproducts. When concentration of both the reactants G and H is doubled, the rate increase by eight times. However, when concentration of G is doubled keeping the concentration of H fixed, the rate is doubled. The overall order of the reaction is

Answer»

`0`
`1`
`2`
`3`

SOLUTION :Overall order=sum of ORDERS w.r.t each reactant. Let the order be x and y for G and H respectively

`because` For (1) and (3), the rate is doubled when CONC. Of G is dobled keeping that ofH constant i.e., rate`prop[G] therefore x=1`
From (2) and (3), y=2
`therefore` Overall order is 3.
Discussion
10.

Consider a reaction a G + bH rarr Products . When concentration of both the reactants G and H is doubled the rate increases by eight times . However when the concentration of G isdoubled keeping the concentration of H fixed the rate is doubled . The overall order of the reaction is :

Answer»

0
1
2
3

Solution :(D) If order of REACTION w.r.t G is x and w.r.t H is y, then on doubling [G] and [H] the rate becomes 8 TIMES i.e
`2^(x) 2^(y) =8 "or " 2^((x+y))= 2^(3)`
`:. ` x+y=3
`:.` Order of reaction =3
Discussion
11.

Consider a reaction A+B rarr C , in which boht reactants are in the same phase, may be

Answer»

UNIMOLECULAR ELEMENTARY reaction
Exothermic
Heterogeneous
Photochemical

Answer :B::C::D
Discussion
12.

Consider a prototypicalfullerene, C_(60) Let, a = Number of 5-membered rings:b = Number of 6-membered rings c = Number of pi bonds in C_(60) Find the value of (3a-2b+c)

Answer»


SOLUTION :N//A
Discussion
13.

Consider a general reaction aA+bBto cC+dDThe rate expression for the reaction is Rate =k[A]^x[B]^y.Establish the significance of (a+b)and (x+y)term in terms of order and molecularity.

Answer»

SOLUTION :(a+b) REPRESENT MOLECULARITY,(x+y)represent ORDER
Discussion
14.

Consider a large number of hydrogen atoms with electron randomly distributed in the n=1,2, 3 and 4 orbits. How many different wavelengths of light are emitted by these atoms as the electrons fall into lower energy orbits

Answer»

1
3
6
9

Answer :C
Discussion
15.

Consider a general, single-step reaction of the type A + BhArrC. Show that the equilibrium constant is equal to the ratio of the rate constant for the forward and reverse reaction, K_(c)=(k_(f))/(k_(r))

Answer»

Solution :`k_(F)[A][B]=k_(R)[C],(k_(f))/(k_(r))=([C])/([A][B])=k_(c)`
Discussion
16.

Consider a general chemical change 2A+3B to products. The rate with respect to A is r_(1) and that with respect to B is r_(2). The rates r_(1) and r_(2) are related as

Answer»

`3r_(1)=2r_(2)`
`r_(1)=r_(2)`
`2r_(1)=3r_(2)`
`r_(1)^(2)=2r_(2)^(2)`

Solution :For the CHANGE `2A+3B to` PRODUCTS
`-(1)/(2)(d[A])/(dt)=-(1)/(3)(d[B])/(dt), (1)/(2)r_(1)=(1)/(3)r_(2), 3r_(1)=2r_(2)`
Discussion
17.

Consider a gaseous reaction, the rate of which is given by k[A][B], the volume of the reaction vessel containing these gases is suddenly reduced to 1//4^(th) of the initial volume. The rate of reaction relative to the original rate would be

Answer»

`16//1`
`1//16`
`8//1`
`1//8`

Solution :By REDUCING VOLUME to `(1)/(4)th` the CONCENTRATION will BECOME `4` times hence rate `16` times .
Discussion
18.

A fuel cell is supplied 1 mole of H_(2) gas and 10 moles O_(2)gas. If the fuel cell is operated at 96.5 mA currecnt, how long will it deliver power?

Answer»

`1xx10^(6) s`
`0.5xx10^(6)s`
`2xx10^(6) s`
`4XX10^(6) s`

Solution :(c ) It is the CORRECT answer.For solution, consult EXAMPLE 71.
Discussion
19.

Consider a first order gas phase decomposition reaction given below: A_((g))toB_((g))+C_((g)) The initial pressure of the system before decomposition of A was p_(i) .After lapse of time .t. total pressure of the system increased by x units and became .p_(t).The rate constant k for the reaction is given as.......

Answer»

`K=(2.303)/(t)` log `(p_(i))/(p_(i)-x)`
`k=(2.303)/(t)` log `(p_(1))/(2p_(i)-p_(t))`
`k=(2.303)/(t)` log `(p_(i))/(2p_(i)+p_(t))`
`k=(2.303)/(t)` log `(p_(i))/(p_(i)+x)`

Solution :Given REACTION :`A_((g))toB_((g))+C_((g))`
INITIAL pressure :`p_(i)` 0.00.0
Pressure at t time: `(p_(i)-x)` xx
Total pressure =`(p_(i)-x)+(x)+(x)=(p_(i)+x)`
But total pressure =`p_(i)` is given
`therefore p_(t)=(p_(i)+x)` and x=`(p_(t)-p_(i))`
Pressure at t time:
`p_(A)p_(i)-(p_(t)-p_(i))=(2p_(i)-p_(t))=p_(t)=[R]_(t)`
`p_(B)=p_(t)-p_(i)` Initial `p_(A)=p_(i)=[R]_(0)`
`p_(C )=p_(t)-p_(i)`
k of first ORDER reaction =`(2.303)/(t)` log `([R]_(0))/([R]_(t))`
`=(2.303)/(t)` log `((p_(i))/(2p_(x)-p_(t)))`
Thus OPTION (B) is correct
Discussion
20.

Consider a first order gas phase decomposition reaction given below : A(g)toB(g)+C(g) The initial pressure of the system before decomposition of A was P_(i). After lapse of time 't', total pressure of the system increased by x units and became 'P_(t)'. The rate constant k for the reaction is given as ..................... .

Answer»

`k=(2.303)/(t)LOG""(P_(i))/(P_(i)-x)`
`k=(2.303)/(t)log""(P_(i))/(2P_(i)-P_(t))`
`k=(2.303)/(t)log""(P_(i))/(2P_(i)+P_(t))`
`k=(2.303)/(t)log""(P_(i))/(P_(i)+x)`

ANSWER :B
Discussion
21.

Consider a first order gas phase decomposition reaction given below : A(g) rarr B(g) + C(g) The initial pressure of the system before decomposition of A was P_(i). After lapse of time 't', total pressure of the system increased by x units and became 'p_(t)' The rate constant k for the reaction is given as :

Answer»

<P>`K=(2.303)/(t)"LOG"(p_(i))/(2p_(i)-x)`
`k=(2.303)/(t)"log"(p_(i))/(2p_(i)-p_(t))`
`k=(2.303)/(t)"log"(p_(i))/(2p_(i)+p_(t))`
`k=(2.303)/(t)"log"(p_(i))/(p_(i)+x)`

ANSWER :B
Discussion
22.

Consider a class room of dimensions 5 xx 10 xx 3 m^3at temperature 20^@C and pressure 1 atm. There are 50 peoples in the room, each losing energy at the average of 150 Joule/sec. Assuming that the walls ceiling, floor and furniture perfectly insulated and none of them absorbing heat, how much time will be needed for rising the temperature of air in the room to body temperature, i.e., 37^@C. For air C_p = 7/2 R . Loss of air to the outside as the temperature rise may be neglected.

Answer»

502.3 SEC
411.3 sec
602.2 sec
702.4 sec

Solution :Given P= 1ATM, T = 273 + 20 = 293 K
Volume of air in the room ` = 5 xx 10 xx 3 m^3`
` = 150m^3 = 150 xx 10^6cm^3`
As ` PV = nRT`
` therefore = n = (PV)/(RT) ` where n = moles of air
`= (1 xx 150 xx 10^6)/(10^3 xx 0.0821 xx 293) = 6.236 xx10^3 , Delta H = n.C_P . Detla T`
` = 6.236 xx 10^3 xx 7/2 xx 8.314 xx (310 - 293) = 3.085 xx 10^6 J`
Therefore , HEAT needed to heat the room to `37^@C`
`= 3.085 xx 10^6 J`
Also heat released by 50 people
` = 150 xx 50 J//sec = 7500 J//sec`
`because` 7500 J heat is produced in 1 sec
` therefore 3.085 xx 106 J` heat will be produced in `(1 xx 3.085 xx 10^6)/(7500)`
` = 411.3 sec `
Discussion
23.

Consider a chemical reaction 2X + Y rarr X_2Y. The reactant X will decrease at

Answer»

TWICE the RATE at which Y will DECREASE
the same rate at which Y will decrease
twice the rate at which `X_2Y` will form
half the rate at which Y will decrease.

Solution :(a,c) : For a chemical reaction `2X+Y=X_2Y`
Therefore `-(1)/(2)(d[X])/(dt)=-(d[Y])/(dt)=(d[X_2Y])/(dt)`
Discussion
24.

Consider a certain reaction Ato" Products with "k=2.0xx10^(-2)s^(-1) Calculate the concentration of A remainig after 100 s if the initial concentration of A is 1.0" mol L"^(-1).

Answer»

SOLUTION :The units of k SHOW that the reaction is of first order. HENCE, `k=(2.303)/(t)log""([A]_(0))/([A])`
`:.2.0xx10^(-2)s^(-1)=(2.303)/(100s)log""(1.0" MOL L"^(-1))/([A])" or "log[A]=-0.8684`
`:.[A]=" Antilog"(-0.8684)=" Antilog "(1.1316)=.1354" mol L"^(-1).`
Discussion
25.

Consider a certain reaction A to products with k=2.0xx10^(-2)s^(-1).Calculate the concentration of A remining after 100 s if the initial concentration of A is 1.0 mol L^(-1)

Answer»

Solution :Reaction A `to` product
K=`2.0xx10^(-2)s^(-1)`
t=100 s
Initial concentration =`[A]_(0)=1.0 mol L^(-1)`
After 100S, `[R]_(t)`=(?)
The unit of K is `SEC^(-1)`
`therefore` The reaction will be first order
`K=(2.303)/(t)`LOG `([R]_(0))/([R]_(t))`
`therefore 2.0xx10^(-2)=(2.303)/(100s)` log `((1.0))/(log[R]_(t))`
`therefore (2.0xx10^(-2)xx100)/(2.303)`=log 1=log `[R]_(t)`
`therefore0.8684=-log[R]_(t)`
`therefore log [R]_(t)=-0.8684`
`therefore[R]_(t)`=Antilog (1-0.8684)=0.1354 mol `L^(-1)`
Discussion
26.

Consider a certain reaction A to Products with k=2.0xx10^(-2)s^(-1). Calculate the concentration of A remaining after 100 s if the initial concentration of A is 1.0 mol L"^(-1).

Answer»

Solution :The units of K indicate that the REACTION is of FIRST order.
Hence, `k=(2.303)/(t)"log"([A]_(0))/([A])`
Substituting the values, we GET
`2.0xx10^(-2)s^(-1)=(2.303)/(100s)"log"(1.0"mol L"^(-1))/([A])`
or `log[A]= -0.8684`
or ` [A]="Antilog" (-0.8684)="Antilog"(bar1.1316)=0.1354" mol L"^(-1)`.
Discussion
27.

Consider a certain reaction A rarr Products with k=2.0xx10^(-2)s^(-1). Calculate the concentration of A remaining after 100s if the initial concentration of A is 1.0 mol L^(-1).

Answer»

SOLUTION : [A] = 0.135 M
Discussion
28.

Consider a cell given below: Cu | Cu^(2+) || Cl^(-) | Cl_(2),Pt Write the reactions that occur at anode and cathode.

Answer»

SOLUTION :`{:("Anode", Cu to Cu^(2+) + 2E^(-)),("Cathode", Cl_(2) + 2e^(-) to 2Cl^(-)):}`
Cu acts as anode as it is GETTING oxidised. `Cl_(2)` , acts as cathode as it is getting REDUCED.
Discussion
29.

Consider a cell given below Cu|Cu^(2+)||Cl^(-)|Cl_(2),Pt, write the reactions that occur at anode and cathode.

Answer»

Solution :* OXIDATION of Cu and reduction of `Cl_(2)`
Oxidation reaction on ANODE:
`Cu_((S)) to Cu_((AQ))^(2+)+2e^(-)`
Reduction reaction on cathode :
`Cl_(2)+2e^(-) to 2Cl_((aq))^(-)`
* Cu of anode dissolve and `Cl_(2)` on cathode liberated as `CL^(-)` due to reduction.
Discussion
30.

Consider a binary mixture of volatile liquids . If at X_(A)=0.4 the Vapour pressure of solution is 580 torr than the mixture could be (P_(A)^(0)=300 torr, P_(B)^(0)=800 torr)

Answer»

`CHCl_(3) & CH_(3)COCH_(3)`
`C_(6)H_(5)CL & C_(6)H_(5)Br`
`C_(6)H_(6) & C_(6)H_(5)CH_(3)`
`nC_(6)H_(14) & nC_(7) H_(16)`

Answer :A
Discussion
31.

Consider ""^(228)Th overset(-alpha)rarr ""^(224)Ra overset(-alpha)rarr ""^(220)Rn where t_((1)/(2)) (""^(228)Th)=1.913 yr, t_((1)/(2)) (""^(224)Ra)=3.64d. Determine the N (Th)/N (Ra) ratio once transient equilibrium in which t_((1)/(2)) (parent) gt t_((1)/(2)) (daughter) or lamda (parent) lt lamda (daughter) has been established

Answer»

Solution :We have, `(N_(1))/(N_(2))= (lamda_(2)-lamda_(1))/(lamda_(1))`
Now, `lamda_(1)= (0.6932)/((t_((1)/(2)))_(1))= (0.6932)/(1.913 xx 365)=9.927 xx 10^(-4) d^(-1)`
`lamda_(2)= (0.6932)/((t_((1)/(2)))_(2))= (0.6932)/(3.64)= 0.1900 d^(-1)`
Thus, `(N(Th))/(N(RA))= (0.1900-9.927 xx 10^(-4))/(9.927 xx 10^(-4))= 190`
Discussion
32.

Consider ""^(224)Ra overset(-alpha)rarr ""^(220)Rn overset(-alpha) ""^(216)Po where t_((1)/(2)) (Ra)= 3.64 yr, t_((1)/(2)) (Rn)= 55s. Determine the N (Ra)/N (Rn) ratio once secular equilibrium in which t_((1)/(2)) (parent) gt gt t_((1)/(2)) (daughter) or lamda (parent) lt lt lamda (daughter) has been established.

Answer»

Solution :We have, `(N_(1))/(N_(2))= (lamda_(2))/(lamda_(1))`
Now, `lamda_(1)= (0.6932)/((t_((1)/(2)))_(1))= (0.6932)/(3.64 xx 60 xx 60)= 2.20 xx 10^(-6) s^(-1)`
`lamda_(2)= (0.6932)/((t_((1)/(2)))_(2))= (0.6932)/(55)= 1.26 xx 10^(-2) s^(-1)`
THUS, `(N(Ra))/(N(RN))=(1.26 xx 10^(-2))/(2.20 xx 10^(-6))=5727`
Discussion
33.

Consider a 20 W light source that emits monochromatic light of wavelength 600 nm. The number of photons ejected per second in terms of Avogadro's constant (N_(A)) is approximately

Answer»

`N_(A)`
`10^(-2) N_(A)`
`10^(-4)N_(A)`
`10^(-5) N_(A)`

ANSWER :C
Discussion
34.

Connsider thefollowingreaction , CH_(3) - underset(D) underset(|) CH-underset(CH_(3))underset(|) CH_(3)+overset(*)(Br) toX +HBr identify thestruturefo majorproductX:

Answer»

`CH_(3)-underset(D) underset(|)(CH)-underset(CH_(3))underset(|) (CH)-CH_(3)`
`CH_(3) -underset(D) underset(|)(CH)-underset(CH_(3))underset(|)overset(*)( C) -CH_(3) `
`CH_(3)-underset(D)underset(|)overset(*)-underset(CH_(3))underset(|)(CH)-CH_(3)`
`CH_(3)-overset(*) (CH)-underset(CH_(3))underset(|)(CH)-CH_(3)`

ANSWER :B
Discussion
35.

Conjugation of electron withdrawing groups, e.e., -CHO, -overset(O)overset(||)C-R,-overset(O)overset(||)CRgt-overset(O)overset(||)C-OR,-C-=N,-NO_(2) activates nudeophilic attack in halobenzene. The order of reactivity of these groups is :

Answer»

`-NO_(2)gt-C-=gt-OVERSET(O)overset(||)C-Hgt-overset(O)overset(||)C-Rgt-overset(O)overset(||)C-OR`
`-overset(O)overset(||)C-Hgt-overset(O)overset(||)C-Rgt-overset(O)overset(||)C-ORgt-C-=Ngt-NO_(2)`
`-C-=Ngt-NO_(2)-overset(O)overset(||)C-Hgt-overset(O)overset(||)C-Rgt-overset(O)overset(||)C-OR`
`-overset(O)overset(||)-Hgt-NO_(2)gt-C-=Ngt-overset(O)overset(||)-ORgt-overset(O)overset(||)C-R`

SOLUTION :N//A
Discussion
36.

Conjugation of electron withdrawing groups, e.g., -CHO, -overset(O)overset("|")C-R","-overset(O)overset(||)C-OR","-C-=N","-NO_(2) activates nucleophilic addition. The order of reactivity of these groups is:

Answer»

`-NO_(2)gt-C-=Ngt-OVERSET(O)overset(||)C-ORgt-overset(O)overset(||)C-Rgt-overset(O)overset(||)C-H`
`-overset(O)overset(||)C-Hgt-overset(O)overset(||)C-Rgt-overset(O)overset(||)C-ORgt-C-=Ngt-NO_(2)`
`-C-=Ngt-NO_(2)gt-overset(O)overset(||)C-Hgt-overset(O)overset(||)C-Rgt-overset(O)overset(||)C-OR`
`-overset(O)overset(||)C-Hgt-NO_(2)gt-C-=Ngt-overset(O)overset(||)C-ORgt-overset(O)overset(||)C-R`

ANSWER :B
Discussion
37.

Conjugated double bonds are present in :

Answer»

Propylene
isobutane
butane
butadiene

Answer :D
Discussion
38.

Conjugated double bond is present in

Answer»

1, 2-butadiene
1, 3-butadiene
1, 3-pentadiene
`BETA`-butylene

Answer :B
Discussion
39.

Conjugated double bond is present in:

Answer»

Propylene
BUTADIENE
Isobutylene
Butylene

Solution :Butadiene `CH_2=CH-CH-CH_2`
Double BONDS SEPARATED by a single BOND is known as CONJUGATED double bond.
Discussion
40.

what is conjugate base Of HSO_4^- ?

Answer»


ANSWER :`SO_4^(2-)
Discussion
41.

Conjugate base of which of the following acid is weak ?

Answer»

`CH_(3)CH_(2)CH(BR)COOH`
`CH_(3)CH_(2)CH(F)COOH`
`CH_(3)CH_(2)CH(I)COOH`
`CH_(3)CH_(2)CH(CL)COOH`

Answer :B
Discussion
42.

Conjugate base of SO_(4)^(-) is

Answer»

`SO_(4)^(2-)`
`H_(2)SO_(4)`
`H_(3)SO_(4)^(+)`
None of these

Solution :`underset("CONJUGATE acid")(HSO_(4)^(-)) RARR H^(+) + underset("Conjugate BASE")(SO_(4)^(2-))`
Discussion
43.

Conjugate base of hydrazoic acid :

Answer»

`HN_3^-`
`N_3^-`
`N^(3-)`
`N_2^-`

ANSWER :B
Discussion
44.

conjugate base of HCO_3^- ion is :

Answer»

`CO_2`
`CO_3^2-`
`H_2CO_3`
`HCO_3^-`

ANSWER :C
Discussion
45.

Conjugate base of NH_(3) is

Answer»

`NH_(4)^(o+)`
`NH_(2)^(o+)`
`NH_(2)^(o+)`
`N_(2)`

SOLUTION :`NH_(3) HARR NH_(2)^(-) + H^(+)`
Discussion
46.

Conjugate base of HBr is

Answer»

`H_(2)Br^(+)`
`H^(+)`
`Br^(-)`
`Br^(-)`

Solution :`HBR + H_(2)O hArr H_(3)O^(+) + Br^(-)`
Discussion
47.

The conjugate base of H_(2)PO_(4)^(-) is :

Answer»

`HPO_(4)^(-)`
`HPO_(4)^(2-)`
`H_(3)PO_(4)`
`PO_(4)^(3-)`

Answer :B
Discussion
48.

Conjugate base of H_(2)PO_(4)^(-) is

Answer»

`HPO_(4)^(-)`
`HPO_(4)^(2-)`
`H_(3)PO_(4)`
`PO_(4)^(3-)`

Solution :`H_(2)PO_(4)^(-) underset(-H^(+))RARR HPO_(4)^(2-)`
Discussion
49.

Conjugate base for bronsted acids H_2O and HF are………

Answer»

`OH^- and H_2FH^+` RESPECTIVELY
`H_3O^+ and F^-` respectively
`OH^- and F^-` ,respectively
`H_3O^+ and H_2F^+`, respectively

Solution :`H_2O_(ACID 1)+H_2O_(BASE 1)=H_3O_(acid2)^(+)+OH_(base2)^-`
`HF_(acid 1)+H_2O_(base 1) leftrightarrowH_3O_(acid2)^(+)+F_(base 2)^(-)`
`therefore` CONJUGATE base are `OH^- and F^-` respectively
Discussion
50.

Conjugate base of Bronsted acids H_2 O and HF are

Answer»

`OH^(-)andH_(2)FH^(+)`, respectively
`H_(3)O^(+)ANDF^(-)`, respectively
`OH^(-)andF^(-)`, respectively
`H_(3)O^(+)andH_(2)F^(+)`, respectively

Solution :`OH^(-)andF^(-)`, respectively
`UNDERSET("acid "1)(H_(2)O)+underset("base "1)(H_(2)O)hArrunderset("acid "2)(H_(3)O^(+))+underset("base "2)(OH^(-))`
`underset("acid "1)(HF)+underset("base "1)(H_(2)O)hArrunderset("acid "2)(H_(3)O^(+))+underset("base "2)(F^(-))`
`therefore` Conjugate bases are `OH^(-)andF^(-)` respectively
Discussion