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.

Phenol on exposure to air produces a red coloured product due to the formation of

Answer»

p-benzoquinone
o-benzoquinone
phenoquinone
o-and p-benzoquinones

Solution :This is DUE to the FORMATION of o-and p-benzoquinone
Discussion
2.

Phenol is treated with bromine water and shaken well to get white precipitate is:

Answer»

1-Bromophenol
2,4,6-Tribromophenol
2,4-Dibromophenol
Mixture of o-, p-bromophenol

Solution :BROMINATION of PHENOL in AQUEOUS MEDIUM GIVES 2,4,6-tribromophenol.
Discussion
3.

Phenol is more readily soluble in

Answer»

dil. HCl
Both NaOH and HCl
NaOH sol
Sodium bicarbonate solution

Solution :`C_(6)H_(5)OH+NaOH RARR C_(6)H_(5)ONa+H_(2)O`
Discussion
4.

Phenol is less acidic than

Answer»

Acetic acid
p-Methoxyphenol
p-Nitrophenol
Ethanol

Solution :CARBOXYLIC acids are stronger acids than PHENOL. Presence of electron withdrawing group `(-NO_(2))` in p-nitro phenol makes it more acidic than phenol. Alcohols are WEAKER acids tha phenol. Presence of electron donating group (METHOXY) in p-methoxyphenol makes it acidic than phenol.
Discussion
5.

Phenol is heated with C Cl_(4) and alkaline KOH when salicylic acid is produced. The reaction is known is

Answer»

Friedel-Craft REACTION
Riemer-Tiemann's reaction
Rosenmund's reaction
Sommelet reaction

Solution :Reimer-Tiemann reaction.
Discussion
6.

Phenol is heated with a solution of mixture KBr and KBrO_(3). The major product obtained the above reaction is :

Answer»

3-Bromophenol
4-Bromophenol
2,4,6-Tribromophenol
2-Bromophenol

Solution :
Discussion
7.

Phenol gives.......colour with neutral FeCl_(3) solution

Answer»

VIOLET
green
Red `rarr` green `rarr` white
blue

Solution :Violet COLOUR DUE to the FORMATION of `(C_(6)H_(5)O)_(3)Fe`.
Discussion
8.

Phenol gives two polymers on condensation with formaldehyde X and Y are:

Answer»


ANSWER :B
Discussion
9.

Phenol dimerises in benzene having Van't Hoff factor 0.54. What is the degree of association?

Answer»

SOLUTION :`I= 0.54 `
`ALPHA =((1-i)N)/( (n-1)) =( (1 -0.54) 2) /((2-1)) = 0.46xx 2`
` alpha= 0.92`
Discussion
10.

Phenol condenses with formaldehyde in the presence of acid or base to form

Answer»

bakelite
teflon
phenolphthalein
none of these

Answer :A
Discussion
11.

Phenol dismerises in benzene having Van,t Hoff factor 0.54 What is the degree of association?

Answer»

`0. 46`
`92`
46
`0.92`

SOLUTION :`I = 0.54`
` alpha = ( (1-i)N)/((n -1))= ((1- 0.54)2)/((2-1))= 0.46 xx 2 =0.92`
Discussion
12.

Phenol cannot be obtained by one of the following reactions

Answer»

`C_(6)H_(5)N_(2)^(+)Cl^(-)+H_(2)O overset(dil. H_(2)SO_(4))rarr`

`C_(6)H_(5)Cl+NaOH UNDERSET(underset((II) " Dil. HCl")(200 atm))overset((i) 633K) rarr`
`C_(6)H_(5)MgBr+H_(2)O overset("Dil.HCl)rarr`

Solution :In (D) benzene is FORMED.
Discussion
13.

Phenol can be distinguished from ethanol by the following reagents except

Answer»

SODIUM
`NaOH//I_(2)`
NEUTRAL `FeCl_(3)`
`Br_(2)//H_(2)O`

ANSWER :A
Discussion
14.

Phenol are more acidic than alcohols because

Answer»

phenoxide ion is stabilised by resonance
phenols are more SOLUBLE in polar solvents
phenols IONS do not exhibit resonance
ALCOHOLS donot lose H atoms at all.

Solution :Phenols are more acidic than alcohols because phenoxide ion is resonance stabilised.
Discussion
15.

Phenol and benzoic acid may be distinguished by their reaction with

Answer»

AQUEOUS `NAOH`
Aueous `NaHCO_(3)`
Neutral `FeCl_(3)`
Aqueous `NH_(3)`

Solution :(B) Benzoic acid gives brisk efferescence `(CO_(2))` with aqueous sod. Bicarbonate while phenol does not.
(C) Phenol gives characteristic colout (violet) with neutral aqueous `FeCl_(3)` while benzoic acid gives buff coloured precipitates
Discussion
16.

pH=7.40, K_(1) "of" H_(2)CO_(3)=4.5xx10^(-7). What will be the ratio of [HCO_(3)^(-)] to [H_(2)CO_(3)] ?

Answer»

SOLUTION :`H_(2)CO_(3) hArr H^(+) + HCO_(3)^(-)`
`K_(1) = ([H^(+)][HCO_(3)^(-)])/([H_(2)CO_(3)])`
`:. ([HCO_(3)^(-)])/([H_(2)CO_(3)])=(K_(1))/([H^(+)])`
pH = 7.40 means - LOG `[H^(+)] = 7.4`
or `log [H^(+)] = - 7.4 bar(8) . 6 or [H^(+)] = 3.981 xx 10^(-8)`
`:. ([HCO_(3)^(-)])/([H_(2)CO_(3)]) = (4.5xx10^(-7))/(3.981xx10^(-8))=11.3`
Discussion
17.

PH_(3) has lower boiling point than NH_(3). Why ?

Answer»

Solution :The electronegativity of N(3.0) is much higher than that of P(2.1). Therefore, `NH_(3)` undergoes extensive intermolecular H-bonding and hence it exsists as an ASSOCIATED molecule. To BREAK these H-bonds, a LARGE amount of energy is NEEDED. On the other hand, `PH_(3)` does not undergo H-bonding and thus exists as discrete molecules. Therefore, the boiling point of `PH_(3)` is much LOWER than that of `NH_(3)`.
Discussion
18.

PH_(3) forms bubbles when passed slowly in water but NH_(3) dissolves. Explain why ?

Answer»

Solution :DUE ot HIGH electronegativity (3.0) and SMALL size of `N,NH_(3)` forms H-bonds with water and H-bonds with water and hence it is water soluble. On the other hand, due to its LOWER electronegativity (2.1) of P and its bigger size than `N,PH_(3)` does not form H-bonds with `H_(2)O`. As a result, it does not dissolve in `H_(2)O` and hence escapes as bubbles.
Discussion
19.

PH_(3) and BF_(3) form an adduct readily because they form

Answer»

A COORDINATE bond
A COVALENT bond
An ionic bond
A hydrogen bond

ANSWER :A
Discussion
20.

pH value of which of the followingis NOT equal to one ?

Answer»

`0.1 M CH_(3) CO OH`
`0.1 M HNO_(3)`
`0.05 MH_(2) SO_(4)`
`50 cm^(3) 0.4 M HCl + 50 cm^(3) 0.2 M NAOH`

Solution :As `CH_(3)CO OH` does not DISSOCIATE completely, , its `10^(-1)M` solution does not have pH = 1
Discussion
21.

Ph-underset(Cl)underset(|)overset(Cl)overset(|)C-CH_3 overset(3NaNH_2)to (A) , What is product (A) ?

Answer»

`Ph-CH=CH_2 `
Ph-C = CH
`Ph-CH_2 - CH_3`
`Ph-C - C -= NA `

SOLUTION :
Discussion
22.

Ph-oversetoverset(CH_(3))(|)CH-undersetunderset(Br)(|)oversetoverset(CH_(3))(|)C-CH_(3) underset((S_(N)1))overset(EtoH)to(A) Major -product (A) is

Answer»

`PH-oversetoverset(CH_(3))(|)CH-undersetunderset(OEt)(|)oversetoverset(CH_(3))(|)C-CH_(3)`
`Ph-oversetoverset(CH_(3))(|)undersetunderset(OEt)(|)C-oversetoverset(CH_(3))(|)CH-CH_(3)`

`Ph - undersetunderset(H)(|)oversetoverset(OEt)(|)C-oversetoverset(CH_(3))(|)undersetunderset(CH_(3))(|)C-CH_(3)`

ANSWER :B
Discussion
23.

pH of the water coming out of cation exchange resin

Answer»

7
`GT 7`
`lt 7`
can not PREDICTED

ANSWER :C
Discussion
24.

pH of saturated aquoeus solution of Ba(OH)_(2) is 12. Identify the correct statement.

Answer»

The SOLUBILITY of `Ba(OH)_(2)` in a buffer SOLUTION of `pH = 13` is `5 xx 10^(-5)` moles/L
The solubility of `Ba(OH)_(2)` decreases in the presence of `NH_(4)Cl`
The solubility of `Ba(OH)_(2)` in a buffer solution of `pH = 1` is `5 xx 10^(-5)` moles/L
The solubility of `Ba(OH)_(2)` is `5 xx 10^(-7) M^(3)`

Solution :N//A
Discussion
25.

pH of rainy wates is around 5.6.

Answer»

SOLUTION :TRUE STATEMENT
Discussion
26.

pH of NH_4Cl solution is = 5.28 calculate degree of hydrolysis of its 0.02 M solution.

Answer»

SOLUTION :`h=2.624xx10^(-4)`
Discussion
27.

pH of heavy water at 25^(@)C

Answer»

7
`GT7`
`lt7`
0

Answer :B
Discussion
28.

pH of black coffee is 5.0 "at " 25^(@) C. Is black coffee acidic or basic ?

Answer»


ANSWER :ACIDIC
Discussion
29.

pH of Ba(OH)_(2) Ssolution is 12. Its solubility product is :

Answer»

`10^(-6) M^(3) `
` 4 XX 10^(-6) M^(3)`
` 0.5 xx 10 ^(-7) M^(3)`
` 5xx 10 ^(-7) M^(3)`

Solution :`pH =12 RARR [OH^(-) ] =10 ^(_2)=2S rArr S =5 xx 10 ^(-3)`
` Ksp = 4S^(3) =5 xx 10 ^(-7) `
Discussion
30.

pH of aqueous solution of 0.1 M , NH_4Cl is found to be 5. The equilibrium constant for the neutralization of NH_4 OHby HCl is 10 ^(y) . The value of .y. is

Answer»


SOLUTION :` [H^(+) ]= SQRT(CKa)=10 ^(-5) ,K_a =(10^(-10))/( 10^(-1)) =10 ^(-9)`
` K_ N =(1)/(K_n)=(K_b)/(K_w)=(1)/(K_a)=(1)/(10^(-9) ) =10 ^(9) `
Discussion
31.

pH of a weakacid in 0.1 M solution is 4.3. Which statements are correct ?

Answer»

its 0.01 M Solution has `PH gt 4.3`
its 0.01 M solution has `pH lt 4.3 `
its 0.01 M solution degree of DISSOCIATION is LESS than in 0.1 M solution
its 0.01M solution degree of dissociation in less than in 0.1 M solution

Solution :` pH =[(1)/(2)[P^(ka)- log] C]`
` 4.3 =(1)/(2)[P^(ka) -log 10 ^(-1)] rArrP^(ka) =7.6`
` pH =(1)/(2)[7.6-log 10 ^(-2)]rArr pH=4.8`
As DIL `uparrow ,alpha " ALSO" uparrow `
Discussion
32.

pH of a solution of CuSO_(4) is.......... Than7and that of solution of Na_(2)CO_(3) is ..........than 7 .

Answer»


ANSWER :LESS, GREATER
Discussion
33.

pH of a solution of a strong acid is 5.0. What would be the pH of the solution obtained after diluting the given solution a 100 times ?

Answer»

Solution :Given that, pH = 5
`[H^+]=10^(-5) "mol L"^(-1)`
On diluting the solution 100 TIMES
`[H^+]=10^(-5)/100= 10^(-7) "mol L"^(-1)`
On calculating the pH using the equation `pH = -log [H^+]`, value of pH comes out to be 7. It is not POSSIBLE. This indicates that solution is very dilute.
Hence, Total `H^+` ion concentration = `H^+` ions from ACID + `H^+` ion from water
`[H^+]=10^(-7)+10^(-7)=2XX10^(-7)` M
pH=-log `[2xx10^(-7)]`
pH = 7 - 0.3010 = 6.699
Discussion
34.

P^(H)of a solution of the mixture of 0.1 HCl and 0.1N CH_3 COOHis (K_a =2xx10 ^(-5))

Answer»

`1`
`2`
`1.7`
`0.7`

Solution :Contribhution of `H^(+)" from " CH_3COOH ` is negligeble ` therefore [H^(+) ] = 0.1 RARR PH = 1 `
Discussion
35.

pH of a solution of a strong acid is 5.0 . What will be the pH of the solution obtained after diluting the given solution 100 times ?

Answer»

Solution :PH = 5 means `[H^(+)]=10^(-5)M`. On diluting 100 times , `[H^(+)]=(10^(-5))/(100) = 10^(-7) M`.
Thisshould give pH = 7 but it cannot be so because solution is acidic and pH should be less than 7 Thereason is that `[H^(+)]` from `H_(2)O` cannot be neglected. THUS, total `[H^(+)]=10^(-7)M` (from HCl) + `10^(-7)` M (from `H_(2)O ` ) `= 2xx10^(-7)M`
`:. pH = - log (2xx10^(-7))=7-0.3010 = 6.699`.
Discussion
36.

P^(H)of a solution is independent of

Answer»

Temperature
Nature of the solution
DEGREE of dissociation of ACID (or)base
Volume of the solution

Solution :pH DEPENDS on nature `ALPHA ,and T `
Discussion
37.

Ph of a saturated solution of Ba(OH)_(2) " is "12. The value of solubility product is :

Answer»

`3.3 xx 10^(-7)`
`5.0 xx 10^(-7)`
`4.0 xx 10^(-6)`
`5.0 xx 10^(-6)`

Solution :for ANSWER CONSULT Example 7.73.
Discussion
38.

pH of a mixture of HA and A^(-)buffer is 5. K_b of A^(-)=10 ^(-10 ) .Hence [HA]/ [A^(-) ] will be

Answer»

1
10
`0.1`
`100`

SOLUTION :`pH =5 = pKa +log ""([S])/([A])`
`pKa +PKB =14 rArr pKa =1 4 -10 =4`
` 5=4 +log ""([S])/([A]) rArr ([A^(-)])/([H_A]) =10`
Discussion
39.

pH of a buffer solution decreases by 0.02 units when 0.12 g of acetic acid is added to 250 mL of a buffer solution of acetic acid and potassium acetate at 27^(circ)C. The buffer capacity of the solution is

Answer»

0.1
10
1
0.4

Answer :D
Discussion
40.

pH of a 1.0xx10^(-8) M solution of HCl is

Answer»

7.02
6.958
7.4
6.8

Answer :B
Discussion
41.

pH of 0.08 mol "dm"^(-3) HOCl solution is 2.85. Calculate its ionisation constant.

Answer»

Solution :pH of HOCl = 2.85
But, -pH = log `[H^+]`
`THEREFORE -2.85=log [H^+]`
`rArr 315 = log [H^+]`
`rArr [H^+]= 1.413xx10^(-3)`
For weak MONOBASIC ACID `[H^+]= SQRT(K_axxC)`
`K_a=[H^+]^2/C=(1.413xx10^(-3))^2/0.08`
`=24.957xx10^(-5) = 2.4957xx10^(-5)`
Discussion
42.

pH of 0.08 mol dm^(-3) HOCl solution is 2.85. Calculate its ionisation constant.

Answer»

Solution :pH of Hocl = 2.85, i.e, - LOG `[H^(+)]= 2.85 ` or log `[H^(+)]=-2.85= bar(3).15`
or `[H^(+)]`= antilog ` bar(3) .15 = 1.413 xx 10^(-3)M`
For weak monobasic ACID, `HA hArr H^(+) + A^(-)`
`K_(a) = ([H^(+)]^(2))/([HA])=((1.413xx10^(-3))^(2))/(0.08) = 2.4957xx10^(-5)`.
Discussion
43.

pH, ionization constant K_(a) and concentration c of the solution of the salt of a weak acid and strong base (like CH_(3)CO Ona) are related as ............

Answer»


Answer :`pH=7+(1)/(2) [ pK_(a) + LOG C ]`
Discussion
44.

Ph-CH=CH_2overset(IC l)to P, Identify major product 'P' is :

Answer»

`Ph-undersetunderset(I)(|)CH-CH_2-CL`
`Ph-undersetunderset(Cl)(|)CH-CH_2-I`
`Ph-undersetunderset(Cl)(|)CH-CH_2-Cl`
`Ph-undersetunderset(I)(|)CH-CH_2-I`

ANSWER :B
Discussion
45.

Ph-CH=CH-CH_3 overset(HOCl)to X,X is :

Answer»

`Ph-undersetunderset(CL)(|)CH-oversetoverset(OH)(|)CH-CH_3`
`Ph-undersetunderset(OH)(|)CH-oversetoverset(Cl)(|)CH-CH_3`
`Ph-CH=CH-CH_2Cl`

ANSWER :B
Discussion
46.

Ph-CH_(2)-overset(CH_(3))overset(|)(C)H-CH_(3)underset(("monochlorination"))overset(Cl_(2)//hv)to which statements is/are correct about protochemical chlorination of the above compound

Answer»

The MAJOR PRODUCT will have chiral carbon atom but is optically inacive.
the intermediate free RADICAL of the major product is resonance STABILISED
the intermediate free radical is tertiary for major product.
The intermediate free radical is plannar, stabilised by only hyperconjugation.

Answer :A::B
Discussion
47.

Ph-CH_2 - underset(D)underset(|)CH-CH_3 overset(Br_2// hv)to Product of the above reaction will be :

Answer»

DIASTEREOMERS
RACEMIC mixture
Meso
Constitutional isomers

Solution :Diastereomers
Discussion
48.

Ph-CH_2-CH=CH_2overset(dil.H_2SO_4)to X, 'X' is :

Answer»

`Ph-CH_2-CH_2-CH_2-OH`
`Ph-CH_2-undersetunderset(OH)(|)CH-CH_3`
`Ph-undersetunderset(OH)(|)CH-CH_2-CH_3`
`Ph-CH_2-OH`

ANSWER :C
Discussion
49.

Ph-C -=CH underset(MeOH)overset(MeO)to major product of the reaction is

Answer»



`PH-C -= - OME`
`Ph- UNDERSET(OMe)underset(|)(C)=CH_2`

SOLUTION :
Discussion
50.

Ph-C-=C-CH_3overset(Hg^(2+)//H^(+))toA,A is :

Answer»




ANSWER :A
Discussion