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.

Lactic acid, CH_3CH(OH)COOH molecule shows:

Answer»

GEOMETRICAL isomerism
Metamerism
Optical isomerism
Tautomerisms

Answer :C
Discussion
2.

Lactic acid and extracted from muscles is :

Answer»

Laevorotatory
Dextrorotatory
similar with syntheticlactic ACID
NONE of these

Answer :B
Discussion
3.

Lactam from which nylon-4 is synthesised is

Answer»




SOLUTION :(LACTAM ) is USED in the SYNTHESIS of nylon-4
Discussion
4.

Lactic acid which extracted from muscles is :

Answer»

Laevorotatory
Dextrorotatory
similar with syntheticlactic ACID
NONE of these

Answer :B
Discussion
5.

Lactase enzyme hydrolyses the lactose intoitsconstituent as …………. .

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GLUCOSE , FRUCTOSE
glucose , GALACTOSE
fructose only
glucoseonly

ANSWER :B
Discussion
6.

Lacrymator or tear gas is:

Answer»

`C_6H_5COCl`
`C_6H_5OC_6H_5`
`C_6H_5COCH_2Cl`
`C_6H_5COCH_3`

ANSWER :C
Discussion
7.

Lack of vit-P causes

Answer»

beri-beri
weakness of muscles
Hemorrhage
Scurvy

Answer :C
Discussion
8.

Lack of vitamin B_1 causes scurvy.(True/False)

Answer»


ANSWER :beri-beri
Discussion
9.

Labourer.s working with phosphorus suffer from a diseses in which bones decay. It is known as:

Answer»

Arthritis
Phossy jaw
Rickets
Cancer

Answer :B
Discussion
10.

Laboratory preparation of Nitro ethane from ethyl bromide follows……………….. Mechanism.

Answer»


ANSWER :`SN^(2)`
Discussion
11.

Laboratory method for the preperation of diborane involves the oxidation of

Answer»

SODIUM BOROHYDRIDE with iodine
Sodium borohydride with fluorine
Sodium borohydride with chlorine
Sodium borohydride with `KMnO_4`

ANSWER :A
Discussion
12.

Laboratory method for the preparation of acetyl chloride is

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`CH_(3)COOH+SOCl_(2)toCH_(3)COCL`
`CH_(3)COOH+PCl_(3)toCH_(3)COCl`
`CH_(3)COONa+PCl_(3)toCH_(3)COCl`
All of these

Solution :In laboratory it is prepared by `SOCl_(2)` because byproduct `SO_(2) and HCl` are gases which can EASILY ESCAPE and leave behind `CH_(3)COCl` as a RESIDUE.
`CH_(3)COOH+SOCl_(2)toCH_(3)COCl+HCl uarr+SO_(2)uarr`
Discussion
13.

Label the hydrophilic and hydrophobic parts in the following compounds. (i) CH_(3)(CH_(2))_(10)CH_(2)OSoverset(-)(O_(2))overset(+)Na (ii) CH_(3)(CH_(2))_(15)overset(+)N(CH_(3))_(3)overset(-)Br (iii) CH_(3)(CH_(2))_(16)COO(CH_(2)CH_(2)O)_(n)CH_(2)CH_(2)OH

Answer»

SOLUTION :
Discussion
14.

Label the hydrophilic and hydrophobic parts in the following compounds : (i) CH_(3)(CH_(2))_(10)CH_(2)OSO_(3)^(-)Na^(+) (ii) CH_(3)(CH_(2))_(15)-overset(+)N(CH_(3))_(3)Br^(-) (iii) CH_(3)(CH_(2))_(16)COO(CH_(2)CH_(2)O)_(n)CH_(2)CH_(2)OH

Answer»

Solution :(i) `underset("HYDROPHOBIC part")ubrace(CH_(3)(CH_(2))_(10))-underset("Hydrophilic part")ubrace(OSO_(3)^(-)Na^(+))`
(II) `underset("Hydrophobic part")ubrace(CH_(3)(CH_(2))_(15))-underset("Hydrophilic part")ubrace(overset(+)N(CH_(3))Br^(-))`
(III) `underset("Hydrophobic part")(ubrace(CH_(3)(CH_(2))_(6)))-underset("Hydrophilic part")ubrace(COO(CH_(2)CH_(2)O)_(n)CH_(2)CH_(2)OH)`
Discussion
15.

Label the hydrophillic and hydrophobic parts in the following compounds: (a) CH_(3)(CH_(2))_(10)CH_(2)OSO_(3)""^(-)Na^(+) (b) CH_3 (CH_2)_(15) N^(+) (CH_3)_2 Br^(-) (c) CH_3 (CH_2)_(16) COO (CH_2 CH_2 O)_n CH_2 CH_2 OH

Answer»

SOLUTION :(a) `underset("Hydrophobic part")(CH_3 (CH_2)_(10)) underset("HYDROPHILIC part")(CH_2- OSO_3 NA)`
(B) `underset("Hydrophobic part")(CH_3 (CH_2)_(15))- underset("Hydrophilic part")(N^(+) (CH_3)_3 Br^(-))`
(c) `underset("Hydrophobic part")(CH_3 (CH_2)_(16)) - underset("Hydrophilic part")(COO (CH_2 CH_2 O)_n CH_2 CH_2 OH)`
Discussion
16.

Label the sterogenic center in each biologically active compound as R orS.

Answer»


ANSWER :(a) S (B) S ( C) R
Discussion
17.

Label the hydrophilic and hydrophobic parts in the following compounds. (a)CH_3(CH_2)_10CH_2OSO_3^(-)Na^+ , (b)CH_3(CH_2)_15-overset+N(CH_3)_3Br^- , (c ) CH_3(CH_2)_16COO(CH_2CH_2O)_n CH_2CH_2OH

Answer»

Solution :(a)`ubrace(CH_3(CH_2)_10CH_2)_"Hydrophobic"-ubrace(OSO_3^(-)Na^(+))_"Hydrophilic PART"`
(b)`ubrace(CH_3(CH_2)_15)_"Hydrophobic part"-ubrace(overset+N(CH_3)_3Br^-)_"Hydrophilic part"`
(c )`ubrace(CH_3(CH_2)_16)_"Hydrophobic part"-ubrace(COO(CH_2CH_2O)_nCH_2CH_2OH)_"Hydrophilic ACID"`
Discussion
18.

Label the following as, E, Z isomers.

Answer»





SOLUTION :ECLIPSED < PARTIALLY eclipsed < SKEW < staggard.
Discussion
19.

(a) Complete the following chemical equations : (i) MnO_(4)^(-) (aq) + S_(2)O_(3)^(2-) (aq) + H_(2)O(l) to (ii)Cr_(2)O_(7)^(2-)(aq)+Fe^(2+) (aq) + H^(+) (aq) to (b) Explain the following observatons : (i) La^(3+) (Z = 57) and Lu^(3+) (Z = 71) do not show any colour in solutions. (ii) Among the divalent cations in the first series of transition elements, manganese exhibits the maximum paramagnetism. (iii) Cu^(+) ion is not known in aqueous solutions.

Answer»

SOLUTION :Configurationof `La^(3+) is 4F^(0)` and that of `Lu^(3+) is 4f^(14)` . Due to the absence of partially FILLED f-orbitals, `La^(3+)` and `Lu^(3+)` do not SHOW any colour in solution.
Discussion
20.

Label each stereogenbic center as R or S.

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ANSWER :(a) S (B) R ( C) R (d) R `(##GRB_CHM_ORG_HP_C02_E01_313_A01##)`
`(##GRB_CHM_ORG_HP_C02_E01_313_A02##)` (G) S `(##GRB_CHM_ORG_HP_C02_E01_313_A03##)`
Discussion
21.

La^(3+)(Z=57) and Lu^(3+)(Z=71) do not show any colour in solutions.

Answer»

Solution :CONFIGURATIONOF `La^(3+) is 4f^(0)` and that of `LU^(3+) is 4f^(14)` . Due to the ABSENCE of partially filled f-orbitals, `La^(3+)` and `Lu^(3+)` do not show any colour in solution.
Discussion
22.

La (lanthanum) having atomic number 57 is a member of :

Answer»

s-block elements
p-block elements
d-block elements
f-block elements

Answer :C
Discussion
23.

La^(3+) and Lu^(3+) ions exhibit ________ property.

Answer»

SOLUTION :DIAMAGNETIC
Discussion
24.

La ^(3+) is isoelectronic with

Answer»

`TB ^(3+)`
`Eu^(3+)`
`CE ^(4+)`
`Sm^(3+)`

Answer :C
Discussion
25.

l high temperature the amount of sodium formed in the Castner process decreases due to

Answer»

Sodium metal HIGH densit
Sodium metal dissolves in fused NaOH
High electrode potential
Low JONISATION potential

ANSWER :B
Discussion
26.

Kwashiorkor is caused by the deficiency of-

Answer»

vitamins
hormones
AMINO acids
essential amino acids

Answer :D
Discussion
27.

K_(w) of H_(2)O at 373 K is 1 xx 10^(-12). Identify which of the following is/are correct

Answer»

`pK_(w)` of `H_(2)O` is 12
pH of `H_(2)O` is 6
`H_(2)O` is neutral
`H_(2)O` is acidic

Solution :`pK_(w) = -logK_(w) - log 1 xx 10^(-12) = 12`
`RARR K_(w) = [H^(+)][OH^(-)] = 10^(-12)`
`[H^(+)] = [OH^(-)] rArr`
`[H^(+)]^(2) = 10^(-12), [H^(+)] = 10^(-6), pH = -log [H^(+)] = -log 10^(-6) = 6`
`H_(2)O` is neutral because `[H^(+)] = [OH^(-)]` at 373 K even when pH = 6
(d) Is not correct at 373 K. Water cannot become acidic.
Discussion
28.

Kwashiorkar is caused by the deficiency of:

Answer»

Vitamins
Hormones
AMINO acids
Essential amino acids

Answer :C
Discussion
29.

K_w=1 times 10^-14 at 25^@C Justify that statement.

Answer»

Solution :(i) Experimentally found that the concentration of `H_3O^+` in pure water is `1 times 10^-7` at `25^@C`.
(ii) SINCE the dissociation of water PRODUCES equal number of `H_3O^+ and OH^-`, the concentration of `OH^-` is also equal to `1 times 10^-7` at `25^@C`.
(ii) `therefore` The IONIC PRODUCT of water at `25^@C` is
`K_w=[H_3O^+][OH^-]`
`=[1 times 10^-7][1 times 10^-7]`
`K_w=[1 times 10^-14]`
Discussion
30.

K_w represents

Answer»

IONIC PRODUCT CONSTANT of water
Solubility product of water
Equilibrium constant of water
Buffer index

Answer :A
Discussion
31.

What are the possible products.

Answer»

SOLUTION :` (##KSV_CHM_ORG_P2_C15_E01_007_S01.png" WIDTH="80%">.
Discussion
32.

NaNO_(2) +Dil. HCI. What are the possible products?

Answer»

SOLUTION :` (##KSV_CHM_ORG_P2_C15_E01_006_S01.png" WIDTH="80%">,
Discussion
33.

K_(sp) value of Al(OH)_(3) and Zn(OH)_(2) are 8.5 xx 10^(-23) and 1.8 xx 10^(-14) respectively. If NH_(4)OH is added in a solution of Al^(3+) and Zn^(2+), which will precipitate earlier

Answer»

`AL(OH)_(3)`
`ZN(OH)_(2)`
Both together
None

Solution :Solubility of `Al(OH)_(3)` is lesser than `Zn(OH)_(2)`.
Discussion
34.

K_(sp) of an electrolyte AB is 1 xx 10^(-10).[A^(+)] = 10^(-5) m, which concentration of B^(-) will not give precipitate of AB

Answer»

`5 XX 10^(-6)`
`1 xx 10^(-5)`
`2 xx 10^(-5)`
`5 xx 10^(-5)`

Solution :`K_(sp) gt` IONIC PRODUCT `1 xx 10^(-10) gt 5 xx 10^(-11)`.
Discussion
35.

K_(sp) of Al(OH)_3 is 1 times10^-15 At what pH does 1.0 times 10^-3 M Al^(3+) precipitate on the addition of buffer of NH_4Cl and NH_4OH solution?

Answer»

Solution :`Al(OH)_3 LEFTRIGHTARROW Al^(3+)(aq)+3OH^- (aq)`
`K_(sp)=[Al^(3+)][OH^-]^3`
`Al(OH)_3` PRECIPITATES when
`[Al^(3+)][OH]^3 gtK_(sp)`
`(1 times10^-3)[OH^-]^3gt1 times10^-15`
`[OH]^3 GT 1 times 10^-12`
`[OH] gt1 times10^-4M`
`[OH^-]=1 times10^-4M`
`pOH=-log_10[OH^-]=-LOG( 1 times 10^-4)=4`
`pH=14-4=10`
Thus `Al(OH)_3` precipitates at a pH of 10
Discussion
36.

K_(sp)of AgCl is 1.8 times 10^-10. Calculate molar solubility in 1M AgNO_3

Answer»

Solution :`AGCL(s) leftrightarrow Ag^+(aq)+Cl^-(aq)`
X= SOLUBILITY of AgCl in 1 M `AgNO_3`
`AgNO_3(s) leftrightarrow Ag^+(aq)+NO_3^- (aq)`
`[Ag^+]=x+1 approx 1M (because x lt lt 1)`
`[Cl^-]=x`
`K_(SP)=[Ag^+][Cl^-]`
`1.8 times 10^-10 =(1) (x)`
`x=1.8 times 10^-10 M`
Discussion
37.

K_(sp) of AgCl at 25^(@)C is 1.782 xx10^(-10). At 35^(@)C, K_(sp) is 4.159xx10^(10). What are DeltaH^(0) and DeltaS^(0) for the reaction : AgCl(s)=Ag^(+)(aq)+Cl^(-)(aq) ?

Answer»


ANSWER :`(DeltaH^(0)=64.6555 KJ, DELTA S^(0)=30.3 J)`
Discussion
38.

K_(sp) of AI(OH)_3 is 1xx10^(-15) M. At what pH does 1.0xx10^(-3) M AI^(3+)precipitate on the addition of buffer of NH_4 CI and NH_4 OH solution.

Answer»

Solution :`AI(OH)_3 hArr AI+_((aq))^(3+)+3OH_((aq))^(-)`
`[K_(sp)=pAI^(3+)][OH^-]^3`
`AI(OH)_3` precipitates when
`[AI^(3+)][OH^(-)]^3 GT K_(sp)`
`[1xx10^(-3)][OH^(-)]^3 gt 1xx10^(-15)`
`[OH]^3 gt 1xx10^(-12)`
`[OH^-] gt 1xx10^(-4) M`
`[OH^-]=1xx10^(-4) M`
`POH=-log_(10)[OH^-]`
`=-LOG[1xx10^(-4)]=4`
`PH=14-4=10`
Thus, `AI(OH)_3` precipitates at a `pH of 10.`
Discussion
39.

K_sp of AgCl at 18^@C is 1.8 xx 10^-10. If Ag^+ of solution is 4 xx 10^-3 mol/litre. The Cl^- that must exceed before AgCl is precipitated would be:

Answer»

`4.5xx10^(-8)` mol/lit
`7.2xx10^(-13)` mol/lit
`4XX10^(-3)` mol/lit
`4.5xx10^(-7)` mol/lit

Answer :A
Discussion
40.

K_(sp) of Ag_2CrO_4 is 1.1 times 10^-12. What is solubility of Ag_2CrO_4 in 0.1 M K_2CrO_4.

Answer»

Solution :`Ag_2CrO_4 leftrightarrow 2Ag^+ +CrO_4^(2-)`
x 2X x
x is the solubility of `Ag_2CrO_4` in 0.1 M `K_2CrO_4`
`K_2CrO_4 leftrightarrow 2K^+ + CrO_4^(2-)`
0.1 M 0.2 M 0.1 M
`[Ag^+]=2x`
`[CrO_4^(2-)]=(x+0.1) APPROX 0.1 therefore x lt lt 0.1`
`K_(sp)=[Ag^+]^2 [CrO_4^(2-)]`
`1.1 TIMES 10^-12=(2x)^2 (0.1)`
`1.1 times 10^-12=0.4x^2`
`x^2=(1.1 times10^-12)/0.4`
`x=sqrt((1.1 times10^-12)/0.4)`
`x=sqrt(2.75 times10^-12)`
`x=1.65 times 10^-6 M`
Discussion
41.

K_(sp) of AgCI is1.8xx10^(-10). Calculate molar solubility in 1M AgNO_3.

Answer»

SOLUTION :`AgCI_((s))HARR Ag_((aq))^(+)+CI_((aq))^(-)`
`x="solubility of in " 1M AgNO_3`
`AgNO_(3(aq)) hArr underset(1M)(Ag_((aq))^(+))+underset(1M)(NO_(3(aq))^-)`
`[AG^+]=x+1 cong 1M (because x ltlt 1)`
`[CI^-]=x`
`K_(sp)=[Ag^+][CI^-]`
`1.8xx10^(-10)=(1)(x)`
`x=1.8xx10^(-10)M`.
Discussion
42.

K_(sp) of Ag_2 CrO_4 is 1.1xx10^(-12). What is solubility of Ag_2 CrO_4 in ox.1M K_2 CrO_4.

Answer»

Solution :`Ag_2 CrO_4 hArr 2AG^(+)+CrO_(4)^(2-)`
x is the solubility of `Ag_2 CrO_4` in `0.1 M K_2 CrO_4`
`UNDERSET(0.1M)(K_2 CrO_4) hArr underset(0.2 M)( 2K^+)+underset(0.1 M)(CrO_(4)^(2-))`
`[Ag^+]=2x`
`[CrO_(4)^(2-)]=(x+0.1)=0.1""because xlt LT 0.1`
`K_(sp)=[Ag^(+)^(2)[CrO_(4)^(2-)]`
`1.1xx10^(-12)=(2x)^2(0.1)`
`1.1xx10^(-12)=0.4x^2`
`x^2=(1.1xx10^(-12))/(0.4)`
`x=sqrt((1.1xx10^(-12))/(0.4))`
`x=sqrt(2.75xx10^(-12))`
`x=1.65xx10^-5M`.
Discussion
43.

K_(sp) for sodium chloride is 36 mol^(2)//"litre"^(2). The solubility of sodium chloride is

Answer»

`(1)/(36)`
`(1)/(6)`
6
3600

Solution :`{:(NaCl,hArr,NA^(+),+,Cl^(-)),(,,S,,S):}`
`K_(sp) = S^(2), S = sqrt(K_(sp)) = sqrt(36) = 6`.
Discussion
44.

K_sp for sodium chloride is 36 mol^2/liter^2. The solubility of sodium chloride is :

Answer»

1/36 M
1/6 M
6 M
3600 M

Answer :C
Discussion
45.

K_(sp) for sodium chloride is 36 "mol"^(2)//"litre"^(2)". The solubility of sodium chloride is

Answer»

6M
`(1)/(6)M`
`3600M`
`(1)/(36)`M

Answer :A
Discussion
46.

K_(sp) for Cr(OH)_(3) is 2.7 xx 10^(-31). What is its solubility in moles/litre

Answer»

`1 XX 10^(-8)`
`8 xx 10^(-8)`
`1.1 xx 10^(-8)`
`0.18 xx 10^(-8)`

SOLUTION :`{:(Cr(OH)_(3),rarr,Cr^(+3)+,3OH^(-)),(,,x,3x):}`
`K_(sp) = x.(3x)^(3) = 27 x^(4)`
`x = 4sqrt((K_(sp))/(27)) , x = 4sqrt((2.7 xx 10^(-31))/(27))`
`x = 1 xx 10^(-8)` mole/litre.
Discussion
47.

K_(sp) for Cr(OH)_3 is 2.7 times 10^-3. What is the solubility in moles/litre?

Answer»

`1 TIMES 10^-8`
`8 times 10^-8`
`1.1 times 10^-8`
`0.18 times 10^-8`

Solution :`Cr(OH)_3 to Cr^(3+)+3OH^-`
x 3x
`K_(SP)=x.(3x)^3=27x^4`
`therefore x=4sqrt(K_(sp)/27)=4sqrt((2.7 times 10^-31)/27)`
`therefore x=1 times 10^-8`
Discussion
48.

K_(sp) = [A]^3 [B]^2 for the salt where A and B are the cation and anion as the case may be stands true for

Answer»

`As_2S_3`
`Ca_3(PO_4)_2`
`Bi_2S_3`
All are CORRECT

ANSWER :D
Discussion
49.

K_sp = 1.2 xx 10^-5 of M_2SO_4 (M^+ is monovalent metal ion) at 298K find the maximum concentration of M^+ ions that could be attaiined in a saturated solution of this solid at 298 K.

Answer»

`3.46xx10^(-3)` M
`7XX10^(-3)` M
`2.88xx10^(-2)` M
14.4 XX 10^(-3) M

Answer :C
Discussion
50.

K_s for the acid HA is 1 xx 10^-6. The value of K fo the reaction A + H_3O^+ ⇌ HA + H_2O is:

Answer»

`1xx 10^(-6)`
`1xx10^(12)`
`1xx10^(-12)`
`1xx10^(6)`

ANSWER :D
Discussion