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.

pH of 0.5 M aqueous NaCN solution is (pK_a of HCN = 9.3, log 5 = 0.7)

Answer»

10.3
9.5
10.6
11.5

Answer :D
Discussion
2.

pH of 0.1 N NaOH(aq) solution

Answer»

2.7
4.7
7
13

Answer :1
Discussion
3.

pH of 0.1 M NH_(3) aqueous solution is(K_(b) = 1.8 xx 10^(-5))

Answer»

11.13
12.5
13.42
11.55

Solution :`NH_(4)OH HARR NH_(4)^(+) + OH^(-)`
`K_(b) = C alpha^(2) , (1.8 xx 10^(-5))/(.1) = alpha^(2), alpha = 1.34 xx 10^(-3)`
`[OH^(-)] = alpha. C = 1.34 xx 10^(-3) xx .1`
`POH = LOG 10(1)/(1.34 xx 10^(-4)) , pOH = 2.87`
`pH = pOH = 14 , pH + 2.87 = 14`
`pH = 14 - 2.87, pH = 11.13`.
Discussion
4.

What is the PH OF 0.1 M HCl.

Answer»


ANSWER :[1(ONE)]
Discussion
5.

pH of 0.01 M (NH_4)_2 SO_4 and 0.02 M NH_4OH buffer (PK_a of NH_(4)^(+)= 9.26) is

Answer»

`4.74 + log2`
`4.74 - log 2`
`4.74 + LOG1`
`9.26+ log 1`

Solution :Itsisan example of basicbufferfor that
` PH =pK_a+log(["Base"])/(["salt"])`
`=9.26 + log(0.02 )/( 2 xx 0.001 )= 9.26+Log1`
Discussion
6.

PH of 0.01 M HCl solution is______.

Answer»


ANSWER :2
Discussion
7.

ph -NO_2overset((I ))toPh NH_2underset( NaNO_2 //HCI ) overset( 0 ^@C )IIunderset( H_2 O)overset(Delta )III I ,II ,III are resoectively ,

Answer»

`Sn//HCI , PhN_(2) ^(+)Cl^(-), C_6H_5OH`
`C_(6)H_5OH, PhN_2 ^(+)Cl^(-), Sn //HCI`
`PhN_2^(+) CL^(-), Sn//HCI, C_6H_5 OH`
None ofabove

Answer :A
Discussion
8.

pH of 0.0002 M formic acid [K_(a) = 2xx 10^(-4)] approximatly is

Answer»

1.35
0.5
3.7
1.85

Answer :C
Discussion
9.

Ph-NH_(2)underset(0^(@)C)overset(HNO_(2))to A underset(BF_(3))overset(HF)toB overset(Delta)to C, C is

Answer»

`Ph-overset(+)(N)-=NBF_(4)^(-)`


Ph-F

Solution :`Ph-NH_(2)underset(0^(@)C)overset(HNO_(2))to underset((A))(Ph-overset(+)(N)-=N)underset(BF_(3))overset(HF)to underset((B))(Ph-overset(+)(N)-=NBF_(4)^(-))overset(DELTA)to underset((C))(Ph-F)+BF_(3)+N_(2)`.
Discussion
10.

Ph - NH_(2) overset(NaNO_(2)//HCl) underset(0-5^(@) C) to overset(Ph-NH_(2))underset(dil.HCl) to (X) Find the sum of number of nitrogen atoms present in (X) and the total number of stereoisomers of (X) formed.

Answer»


Solution :`pH -NH_(2) overset(N NO_(2)//HCL)UNDERSET(0-5^(@)C) to Ph - N_(2)^(+)Cl^(-)`

No. of N-atoms in X is 3:
No. of stereo isomers = 2 (syn/anti).
Discussion
11.

Ph-N=C=Ounderset((ii)H_2O)overset((i)CH_3MgBr)toPoverset(LiAH_4)toQ

Answer»

<P>P is `Ph-NH-UNDERSET(O)underset(||)C-CH_3` and Q is `Ph-NH-CH_2-CH_3`
P is `Ph-NH-CH_3` and Q is `NH_3`
P is `PhN=oversetoverset(CH_3)(|)C-OH` and Q is `Ph-NH-CH_3`
P is `Ph-NH-undersetunderset(OH)(|)CH-CH_3` and Q is `Ph-NH-undersetunderset(CH_3)(|)CH-CH_3`

SOLUTION :
Discussion
12.

pH in stomacth is approximately

Answer»

7
2
6.5
10

Solution :In STOMACH MEDIUM is STRONGLY ACIDIC . HENCE, pH=2
Discussion
13.

pH for the solution of salt undergoing anionic hydrolysis (say CH_3COONa) is given by:

Answer»

PH= `1/2 [PK_w + PK_a + LOGC]`
pH= `1/2 [PK_w + PK_a - logC]`
pH= `1/2 [PK_w + PK_b + logC]`
NONE of these

Answer :A
Discussion
14.

Ph-CH=CH-underset(O)underset(C)-H overset((i) CH_(3)CH_(2)MgBr)to overset((ii) H_(3)O^(o+))to(X) overset(Cu,Delta)to(Y)

Answer»

X is 1,4-ADDITION product , Y is `Ph-UNDERSET(O)underset(||)(C)-CH=CH-CH_(2)-CH_(3)`
X is 1,2-addition product , Y is `Ph-CH=CH -underset(O)underset(||)(C)-CH_(2)-CH_(3)`
X is 1,4-addition product , Y is `Ph-CH=CH-underset(O)underset(||)(C)-CH_(2)-CH_(3)`
X is 1,2 addition product , Y is `Ph-underset(O)underset(||)(C)-CH=CH-CH_(2)-CH_(3)`

ANSWER :B
Discussion
15.

Ph-CH=CH-CH_(2)-COOHunderset((2)Br_(2)"CC"l_(4).Delta)overset((1)AgOH)to Product can be :

Answer»

`Ph-CH=CH-CH_(2)-BR`
`Ph-underset(Br)underset(|)CH-CH=CH_(2)`
`Ph-CH=CH-CH_(3)`
`Ph-CH=CH-Br`

ANSWER :A::B
Discussion
16.

Ph-CH_2-oversetoverset(CH_3)(|)CH-CH_3underset(("monochlorination"))overset(Cl//hv)to Which statements is/are correct about photochemical chlorination of the above compound ?

Answer»

The major product will be chiral carbon atom having OPTICALLY inactive compound
The intermediate free radical of the major product is resonance STABILISED
The intermediate free radical is tertiary
The intermediate free radical is planar , and stabilised by only hyperconjugation

Solution :Free radical MECHANISM `Ph-CH_2-undersetunderset(CH_3)(|)CH-CH_3` give THREE type of alkyl radicals
Discussion
17.

Ph-CH_(2)-overset(CH_(2))overset(|)underset((+))CH-OHoverset(TaCl)rarrXoverset(KSH)rarrY:

Answer»

`X " is "Ph-underset((+))CH_(2)-overset(CH_(3))overset(|)CH-OTs`
`X " is "Ph--CH_(2)-underset((-))overset(CH_(3))overset(|)CH-OTs`
`Y " is "Ph-underset((-))CH_(2)-overset(CH_(3))overset(|)CH-SH`
`Y" is "Ph-underset((+))CH_(2)-overset(CH_(3))overset(|)CH-SH`

ANSWER :A::C
Discussion
18.

Ph-CH_(2)-O-overset(O)overset("||")C-CH_(3)+overset(Θ18)(OH)rarrX+Y:

Answer»

`X " is"Ph-CH_(2)-overset(18)(OH)`
`X " is "Ph-CH_(2)-OH`
`Y " is "CH_(3)-underset(O)underset("||")C-O^(Θ)`
`Y " is "CH_(3)-underset(O)underset("||")C-overset(18)(O^(Θ))`

ANSWER :B::D
Discussion
19.

Ph-CH_(2)-Cloverset(Mg)underset(Ether)rarroverset(CO_(2))underset(H^(oplus)//HOH)rarroverset(SOCl_(2))rarroverset(NH_(3))rarroverset(LiAlH_(4))rarr

Answer»


ANSWER :B::C
Discussion
20.

Ph-CH_2-C-=CHunderset(Y)overset(X)hArrPh-C-=C-CH_3. The reagents X and Y respectively are

Answer»

Lindlar CATALYST , `NaNH_2`
`NaNH_2` and ALC . KOH
PT catalyst , Wilkison's catalyst
Alc. KOH and `NaNH_2`

ANSWER :D
Discussion
21.

pH calculation upon dilute of strong acid solution is generally done by equating n_(H) in original solution & diluted solution.However . If strong acid solution is very dilute then H^(+) from water are also to be considered take log3.7=0.568 and answer the following questions. A 1 litres solution of pH=4(solution ofa strong acid ) is added to the 7//3 litres of water.What is the pH of resulting solution?

Answer»

`4.52`
`4.365`
`4.4`
`4.432`

Solution :INITIAL `pH=4`
`[H^(+)]=10^(-4)`
`N_(1)V_(1)=N_(2)V_(2)rArr 10^(-4)=N_(2)xx[1+(7)/(3)]`
`10^(-4)=N_(2)xx(10)/(3)rArr N_(2)=3xx10^(-5)"" gt10^(-6)`
so `[H^(+)]` of WATER is not consider
`[H^(+)]=3xx10^(-5)`"" so""`pH=5-log(3)`""`=5-0.48=4.52`
Discussion
22.

Ph-C-=C-Ph overset(Na//NH_3)toAoverset(HC Cl//H^+)toB

Answer»




B and Cboth

Solution :
Discussion
23.

Ph-C-=C-CH_(3) overset(Hg^(2+)//H^(+))toA A is

Answer»




SOLUTION :
Discussion
24.

Ph-C-=C-CH_3overset(Hg^(2+)//H^+)to A . A is

Answer»




SOLUTION :`C_6H_5-C-=C-CH_3underset(H_2SO_4)OVERSET(HgSO_4)to C_6H_5-oversetoverset(O)(||)C-CH_2-CH_3`
Discussion
25.

Ph-C -= C-CH_(3) overset(Hg^(2+)//H^(+))rarr A,A is

Answer»




SOLUTION :
Discussion
26.

PF_(5) is known but NF_(5) is not. Explain.

Answer»

SOLUTION :Nitrogen cannot extend its valency from 3 to 5 DUE to absence of d-orbitals while PHOSPHORUS SHOWS penta covalency as.d-orbitals are present in it.
Discussion
27.

PF_(3) reacts with XeF_(4) to give PF_(5) underset((g))(2PF_(3))tounderset((s))(XeF_(4))tounderset((g))(2PF_(5))+underset((g))(Xe) If 100.0gm of PF_(3) and 50.0 gm XeF_(4) react, then which of the following statement is true?

Answer»

`XeF_(4)` is the limiting reagent
`PF_(3)` is the limiting reagent
1.127 MOL of `PF_(5)` are produced
0.382 mol of `PF_(5)` are produced

Answer :A::D
Discussion
28.

PF_3 molecule is:

Answer»

SQUARE PLANAR
TRIGONAL bipyramid
Tetradedral
Trigonal pyramidal

Answer :D
Discussion
29.

Purest from of iron is ______ .

Answer»

SOLUTION :WROUGHT IRON
Discussion
30.

Petroleum refining involves:

Answer»

VACUUM DISTILLATION
Steam disillation
FRACTIONAL distillation
Passing over activated charcoal

Answer :C
Discussion
31.

Petroleum is mainly a mixture of

Answer»

Alkanes
Cyclohexane
Benzenoid hydrocarbons
Alkenes

Answer :A
Discussion
32.

Petroleumisobtainedformwatergasname of thereactioninvolved is :

Answer»

FISCHER -Tropsch
Begius
DOW's
Kjeldahl's

Answer :A
Discussion
33.

Petroleum refining and naturalgas plant are the main sources of

Answer»

`H_(2) S`
`NH_(3)` pollutants
`SO_(2) `
`CI_(2)`

ANSWER :A
Discussion
34.

Petroleumis amixtureof :

Answer»

ALKANES
cycloalkanes
aromatichyrocarbons
all of these

Answer :D
Discussion
35.

Pertrolis amixturehydrocabonsfromC_(6)"to "C_(5) thequalityofpetrolis determindedin termsofoctanenumber,Thehigher the octanenumber betteris thequalityof fuel . Thecorrectorderof octane numberis :

Answer»

cycloalknes LT ALKANES ltalkanes lt AROMATIC hydrocarbons
alkanes lt aromatic hydrocarbons lt cycloalkanes
alkanes ltaromatic hydrocarbons lt cycloalkanes ltalkanes
alkanes ltalkanes lt cycloalkanes lt aromatic hydrocarbons

ANSWER :D
Discussion
36.

Petroleum ether can be used as

Answer»

Solvent for fat, OIL, VARNISH and RUBBER
As a fuel
Both (a) and (b)
NONE of these

Solution :Solvent for fat, oil , varnish and rubber
Discussion
37.

Petroleum consists mainly of

Answer»

ALIPHATIC hydrocarbons
Aliphatic ALCOHOLS
Aromatic hydrocarbons
NONE of these

Answer :A
Discussion
38.

Peroxy links are present in H_(2)SO_(5) and H_(2)S_(2)O_(8). How manynumber of peroxy bonds are presentin each acid?

Answer»


Solution :`H_(2)SO_(5) rArr HO-underset(O)underset(||)overset(O) overset(||)S-underset("Peroxy LINKAGE")(ubrace(O-O))-H,H_(2)S_(2)O_(8) rArr HO-underset(O)underset(||)overset(O)overset(||)S-underset("linkage")underset("Peroxy")(ubrace(O-O)) -underset(O)underset(||)overset(O)overset(||)S-O-H`
Discussion
39.

Persons working in cement plants and lime stone quarries are prone to disease like :

Answer»

Cancer
Asthma
Silicosis
Pneumoconiosis

Answer :C
Discussion
40.

Peroxy linkage is present in:

Answer»

Caro.s acid
Pyrophosphoric acid
Sulphurous acid
Dithonic acid

Answer :A
Discussion
41.

Perptization is a process of

Answer»

reducing the IMPURITIES of the electrolytes
purification of colloids
dispersing precipiatate into COLLOIDAL sols
MOVEMENT of colloidal PARTICLES in the electrical field.

Solution :dispersing precipiatate into colloidal sols
Discussion
42.

Peroxydisulphuric acid has the following bond

Answer»

`O LARR O =O`
`larr O=O to`
`GT O to O lt`
`-O-O-`

SOLUTION :The structure of peroxydisulphuric acid `(H_(2)S_(2)O_(8))` is

Thus, peroxydisulphuric acid contains PEROXIDE (-O-O-) bond.
Discussion
43.

Peroxy linkage is present in

Answer»

`H_(2)S_(2)O_(2)`
`H_(2)S_(2)O_(2)`
`H_(2)S_(2)O_(6)`
`H_(2)S_(2) O _(8)`

ANSWER :D
Discussion
44.

Peroxoacids of sulphur are:

Answer»

`H_(2)SO_(4)`
`H_(2)SO_(3)`
`H_(2)SO_(5)`
`H_(2)S_(2)O_(8)`

ANSWER :C::D
Discussion
45.

Peroxodisulphuric acid has the following bond

Answer»

`O LARR O = O`
`larr O = O RARR`
`GT O rarr O LT`
`-O-O-`

ANSWER :D
Discussion
46.

Peroxide plays a vital role in producing

Answer»

carbocation
carbonation
free radical
carbene

Answer :C
Discussion
47.

Peroxide linkage is present in

Answer»

`MnO_(2)`
`CrO_(5)`
`H_(2)SO_(5)`
`BaO_(2)`

ANSWER :B::C::D
Discussion
48.

Peroxide ion....... (i) Has five completely filled antibonding molecular orbitals (ii) Is diamagnetic (iii) Has bond order one (iv) Is isoelectric with neon Which one of these is correct

Answer»

(iv) and (iii)
(i),(ii) and (iv)
(i),(ii) and (iii)
NONE of these

Solution :(None of the given options is correct)
Peroxide ion is `O_(2)^(2-)`
TOTAL electrons=18
E.c.=`sigma1s^(2)sigma^(**)1S^(2)sigma2s^(2)sigma^(**)2S^(2)sigma(2p_(X))^(2)pi (2p_(y))^(2)`
`pi^(**)(2p_(x))^(2)pi^(**)(2p_(y))^2`
`B.O. =(10-8)/2=1`
options (ii) and (iii) are correct.
Discussion
49.

Peroxide ion (i) has five completely filled antibonding molecular orbitals (ii) is diamagnetic (iii) has bond order one (iv) is isoelectronic with neon. Which one of these is correct ?

Answer»

(iv) and (iii)
(i),(II) and (iv)
(i), (ii) and (iii)
(i) and (iv)

Solution :Peroxide ion is `O_(2)^(2-)`
Total electrons `=18`
`E.C.=sigma1s^(2)SIGMA^(**)1s^(2)SIGMA2S^(2)sigma^(**)2s^(2)sigma(p_(z))^(2)pi(2pi_(X))^(2)pi(2p_(y))^(2)pi^(**)(2p_(x))^(2)pi^(**)(2p_(y))^(2)`
`B.O.=(10-8)/(2)=1`
Options (ii) and (iii) are correct.
Discussion
50.

Peroxide effeis observed

Answer»

only with HBr
with both HCl and HBr
only with HI
with both HCI and HF

Solution :HCl and HF do not give peroxid effect (oranti Markownikoff'srule) because they have high bond energies. They are not brokenby alkoxy free radicals obtained from peroxide. Hl also do not given peroxide effect, although it is broken by the alkoxy free radical. It is because iodine atoms so formed radily combine with each other to yiedl`I_(2)`rather than attacking the DOUBLE bond.
Discussion