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.

Sample containing ""_(88)^(234)Ra, which decays by alpha-particle emission, is observed to disintegrate at the following rate expressed as counts per minute (cpm). Calculate half-life of this nuclide. t=0,1000 cpm, t=1h, 992 cpm, t= 10h, 924 cpm, t=100h, 452cpm, t= 250h, 138 cpm

Answer»


ANSWER :3.63 DAYS
Discussion
2.

Sample of 2, 3-dibromo-3-methylpentane is heated with dust. The resulting product is isolated and heated with HI in presence of phosphorus. Indicate which is the structure represent the final organic product formed in the reaction

Answer»

`CH_3-CH_2-underset(CH_3)underset|CH-CH_2-CH_3`
`CH_2=CH-overset(CH_3)overset|CH-CH_2-CH_3`
`CH_3-CHI-undersetunderset(CH_3)|CH-CH_2-CH_3`
`CH_2-CH-undersetunderset(CH_3)|C(I)-CH_2-CH_3`

SOLUTION :
Discussion
3.

Same quantity of electricity was passed through solutions of salts of elements A,B and C with atomic weights 7, 27 and 48 respectively. The masses of A,B and C deposited were 2.1 g,27.7g and 7.2 respectively. The valencies of A,B and C respectively are

Answer»

3,2 and 1
1,2 and 3
1,3 and 2
2,3 and 2

Solution :Suppose valencies of A,B and C are a,b, and c respectively. Then their EQUIVALENT weights will be (At. Wt/valance) 7/a,27/b and 48/c. when same quantity of electricity is passed, masses DEPOSITED are in the ratio of their equivalent weights, i.e.,
`("Mass of A")/("Eq. wt. of A")=("Mass of B")/("Eq. wt of B")=("Mass of C")/("Eq. wt of C")`
`therefore(2.1)/(7//a)=(2.7)/(27//b)=(7.2)/(48//c)` or 0.3 a=0.1, b=0.15c, or 3a=b=1.5c. thus, if b=1, then `a=(1)/(3)`,
`c=(1)/(1.5)=(2)/(3)`
`therefore`Ratio `a:b:c=(1)/(3):1:(2)/(3)=1:3:2`
Discussion
4.

Same mass of CH_(4) and H_(2)is taken in container. The partial pressure caused by H_(2) is

Answer»

`8 //9`
`1//9`
`1//2`
1

Solution :`N_(CH_(4))=` NUMBER of MOLES of `CH_(4)=(m)/(16)`
`N_(H_(2))= ` number of moles of `H_(2) = (m)/(2)`
fractionpartial pressure of `H_(2)` is
`H_(2)=(n_(H_(2)))/(n_(H_(2))+n_(CH_(4)))=((m)/(2))/((m)/(2)+(m)/(16))=((m)/(2))/((9M)/(16))=(8)/(9)`
Discussion
5.

Same quantity of electrical charge that deposited0.583 g of silver was passed through a solution of gold salt. If 0.335 g of gold is deposited, calculate the oxidation state of gold in the given salt. At wt of Au = 197.

Answer»


ANSWER :`+3`
Discussion
6.

Same mass of CH_(4) and H_(2) is taken in container. The partial pressure caused by H_(2) is

Answer»

`(8)/(9)`
`(1)/(9)`
`(1)/(2)`
1

Answer :A
Discussion
7.

Same amount of electricity is passed through solution of ferrous chloride and ferric chloride . Ratio of weight of iron deposited in the two solution is

Answer»

`1:1`
`1:2`
`2:3`
`3:2`

Solution :`UNDERSET("Ferrous")(FE^(++) + e^(-) to Fe) ""underset("Ferric")(Fe^(+3) + 3E^(-) to Fe)`
`2E^(-) xx 3 : 3e^(-) xx 2 therefore 3:2`
Discussion
8.

Same amount of electric current is passed through solutions of AgNO_(3) and HCl. If 1.08g of silver is obtained in the first case, the amount of hydrogen liberated at S.T.P. in the second case is

Answer»

`224cm^(3)`
`1.008g`
`112cm^(3)`
`22400cm^(3)`

Solution :According to Faraday.s second law
`("Mass of Ag")/("Mass of H")=("EQUIVALENT mass of SILVER")/("Equivalent mass of hydrogen")`
or `(1.08)/("Mass of H")=(108)/(1)` or mass, `H=0.01g`
`because2.00` gm of `H_(2)` occupies 22400 ml at STP
`therefore 0.01`gm of `H_(2)` will OCCUPY`=(22400xx0.01)/(2)`
`=112cm^(3)`.
Discussion
9.

Samarium shows the oxidation state

Answer»

`+3 and +4`
`+2,+3 and +4`
`+2,and+3`
only `+3`

ANSWER :C
Discussion
10.

Same amount of electric current is passed through solutions of AgNO_3and HCl. If 1.08 g of silver is obtained in the first case, the volume of hydrogen liberated at S.T.P. in the second case is

Answer»

`112 cm^3`
`22400 cm^3`
`224 cm^3`
1.008g

ANSWER :A
Discussion
11.

Salvarsan is arsenic containing drug which was first used for the treatment of.________.

Answer»

SYPHILIS
TYPHOID
Meningites
DYSENTRY

ANSWER :A
Discussion
12.

Salvarsan is arsentic containing drug which was first usedfor the treatmentof "……."

Answer»

syphilis
typhoid
meningitis
dysentry

Solution :Syphills is an acute and chronic infections disease CAUSED by the bacterium treponaema pallidum, it is a sexually transmitted infection. The primary route of transmission is through sexual contact but it may also be transmitted frommother to foetus during pregnancy or at birth.
Paul EHRLICH investigated arsenic based structure in order to produce less toxic substances for the treatement of syphillis as he got Nobel PRIZE for medicine in `1908` for this discovery. Salvarsan was the firsteffective medicine discoveredfor syphilis. ALTHOUGH salversan is toxic to human beings its effect on bacteria is much greater than on human being. Strucutre of salvarsanis -
Discussion
13.

Salvarsan is having similarity in structure as that of

Answer»

azodye
penicillin
OFLOXACIN
CHLORAMPHENICOL

ANSWER :A
Discussion
14.

Salvarsanis arsenic containing drug which was firstused for the treatment of ___________

Answer»

SYPHILIS
typhoid
meningitis
dysentry

Solution : SALVARSAN, an arsenic containing compound, was FIRST USED for the treatment of syphilic (a S.T.D.).
Discussion
15.

Salvarsan is arsenic containing drug which was first used the treatment of

Answer»

syphilis
TYPHOID
meningitis
dysentry

Solution : Salvarsan, an arsenic CONTAINING compound, was first used for the TREATMENT of syphilic (a S.T.D.).
Discussion
16.

Salvarsan is arsenic containing drug which was first used for the treatment of ..

Answer»

SYPHILIS
typhoid 
meningitis
dysentry 

SOLUTION :syphilis
Discussion
17.

Salvarsan is arsenic containing drug which was first used for the treatment of

Answer»

syphilis
typhoid
meningitis
dysentry

Answer :A
Discussion
18.

Salvarsan is arsenic containing drug which was first used for the treatment of _____

Answer»

syphilis
typhoid
MENINGITIS
DYSENTRY

SOLUTION :SALVARSAN was the first effective TREATMENT discovered for syphilis.
Discussion
19.

Salts of which of the following elements are mostly blue in colour?

Answer»

Cu
ZN
Fe
Cr

Solution :SALTS of Cu ELEMENT are mostly blue in colour .
Discussion
20.

Salvarsan is arsenic containing drug which was first used for the treatment of ________

Answer»

syphilis
typhoid
meningitis
DYSENTRY

ANSWER :A
Discussion
21.

Salts of sorbic acid and propionic acid are used as:

Answer»

ANTIOXIDANTS
flavouring AGENTS
food PRESERVATIVES
nutrional supplement

Answer :C
Discussion
22.

Salts of metals X, Y and Z are electrolysed under identical condition using same quantity of electricity . Ifwas observed that 4.2 g of X, 5.4 g of Y and 19.2 g of Z were deposited at respetive cathode . Ifthe atomic weights of X , Y, Z are 7, 27 and 64 respectively , then their ratio of valencies is

Answer»

`1:2:3`
`1 : 3 : 2`
`2: 3 : 1`
`3: 2:2`

ANSWER :B
Discussion
23.

Salts of sorbic acid and propionic acid are used as

Answer»

Antioxidants
Flavouring Agent
Food preservatives
Nutiritional SUPPLEMENTS 

ANSWER :C
Discussion
24.

Salts can be obtained from a concentrated sea water by :

Answer»

Catalyst
Evaporation
Hydrolysis
Crystallisation

Answer :D
Discussion
25.

Salt 'X' + excess NaOH solution overset("warm")rarr'Y' ("gas")+'Z' ("solution") 'Z'("solution")+BaCl_(2)rarr"white ppt" Which of the following salt follow above reaction.

Answer»

`NA(NH_(4))HPO_(4).4H_(2)O`
`K_(2)SO_(4).Cr_(2)(SO_(4))_(3)24H_(2)O`
`(NH_(4))_(2)Cr_(2)O_(7)`
`KCl.MgCl_(2).6H_(2)O`

Solution :`3^(@)` alcoholcannot be oxidised.
Discussion
26.

Salts of A (atomic weight 7), B (atomic weight 27) and C (atomic weight 48) were electrolysed under identical condition using the same quantity of electricity. It was found that when 2.1 g of A was deposited, the eight of B and C deposited were 2.7 and 7.2 g. The valencies of A,B and C are respectively :

Answer»

`3,1 and 2`
`1,3 and 2`
`3,1 and 3`
`2,3 and 2`

ANSWER :B
Discussion
27.

Salt with which one of the following anion produce deep blue colour with starch and Kl is

Answer»

CHLORIDE
NITRITE
acetate
bromide

ANSWER :B
Discussion
28.

Salt used as a purgative is:

Answer»

NACL
`MgSO_4.7H_2O`
`Ca_3Al_2O_6`
`MgCl_2.6H_2O`

ANSWER :B
Discussion
29.

Salt of the following is used as a water softner

Answer»

`H_(4)P_(2)O_(6)`
`H_(4)P_(2)O_(7)`
`HPO_(3)`
`HPO_(2)`

ANSWER :3
Discussion
30.

Saltmixture overset(dil. HCI) to Saltis consistingof cation

Answer»

`Pb^(2+)` and `HG^(2+)`
`Pb^(2+)` and `Hg_(2)^(2+)`
`Pb^(2+)` and `Ag^(o+)`
`Pb^(2+), Hg_(2)^(2+)` and `Ag^(o+)`

ANSWER :C
Discussion
31.

Salt cake is

Answer»

SODIUM SULPHATE
sodium chloride
Sodium bisulphide
Sodium sulphate and sodium chloride

Answer :A
Discussion
32.

Salt bridge is used to

Answer»

CONNECT two half cells
maintain ELECTRICAL neutrally
reduced junction POTENTIAL
all

Solution :These are the FUNCTIONS of SALT bridge .
Discussion
33.

Salt and water is formed by acid-base neutralisation reaction. If ratio of moles of acid & base taken is not similar to the ratio of their stoichiometric coefficient, then one of the component is limiting reagent.Assume no dissociation of water in following reactions. (Base is 80% pure only, take impurity present as inert & non electrolytic ) Select the correct combination for the resulting acidic solution.

Answer»

<P>(I)(ii)(S)
(I)(iv)(S)
(I)(ii)(P)
(II)(i)(R)

SOLUTION :(i)Mass of pure NaOH`=(10xx80)/100=8 gm`
`{:(NaOH+,HCl to ,NaCl,+H_2O),(8/40=0.2 "mole",(8xx500)/1000,-,-):}`
=0.4 mole
`=0.2 "mole" " " 0.2"mole"`
BASE in L.R.,`[H^+]=0.2 M " " [Na^+]=0.2 M [Cl^(-)]=0.4 M`
(ii)KOH pure `=(28xx80)/100=22.4 gm`
`{:(KOH+,HNO_3to,KNO_3,+H_2O),(22.4/56=0.4 "mole",(2.5xx500)/1000,-,-):}`
=0.1 mole
`=0.3 "mole" " " 0.1"mole"`
Acid in L.R.,`[OH^(-)]=0.3 M " " [K]=0.4 M [NO_3^(-)]=0.1 M`
(iii) `{:(Ca(OH)_2+,H_2SO_4to,CaSO_4,+2H_2O["Pure" Ca(OH)_2=37xx0.8=29.6 gm]),(29.6/74=0.4 ,(0.8xx500)/1000=0.4,-,-):}`
`[Ca^(2+)]=[SO_4^(2-)]=0.4M`
(iv) `{:(Ba(OH)_2+,2HBrto,BaBr_2,+2H_2O["Pure" Ba(OH)_2=342xx0.8=,273.6 gm]),(273.6/171=1.6 ,(6.4xx500)/1000=3.2,-,-),(-,-, ,1.6 "mole"):}`
`[Ba^(2+)]=1.6 M " " [Br^(-)]=3.2M`
Discussion
34.

Salt and water is formed by acid-base neutralisation reaction. If ratio of moles of acid & base taken is not similar to the ratio of their stoichiometric coefficient, then one of the component is limiting reagent.Assume no dissociation of water in following reactions. (Base is 80% pure only, take impurity present as inert & non electrolytic ) Select incorrect combination

Answer»

(I)(II)(P)<BR>(II)(i)(R)
(IV)(iv)(R)
(III)(ii)(S)

Solution :(i)Mass of pure NaOH`=(10xx80)/100=8 gm`
`{:(NaOH+,HCl to ,NaCl,+H_2O),(8/40=0.2 "mole",(8xx500)/1000,-,-):}`
=0.4 mole
`=0.2 "mole" " " 0.2"mole"`
Base in L.R.,`[H^+]=0.2 M " " [Na^+]=0.2 M [Cl^(-)]=0.4 M`
(ii)KOH pure `=(28xx80)/100=22.4 gm`
`{:(KOH+,HNO_3to,KNO_3,+H_2O),(22.4/56=0.4 "mole",(2.5xx500)/1000,-,-):}`
=0.1 mole
`=0.3 "mole" " " 0.1"mole"`
Acid in L.R.,`[OH^(-)]=0.3 M " " [K]=0.4 M [NO_3^(-)]=0.1 M`
(iii) `{:(Ca(OH)_2+,H_2SO_4to,CaSO_4,+2H_2O["Pure" Ca(OH)_2=37xx0.8=29.6 gm]),(29.6/74=0.4 ,(0.8xx500)/1000=0.4,-,-):}`
`[Ca^(2+)]=[SO_4^(2-)]=0.4M`
(iv) `{:(Ba(OH)_2+,2HBrto,BaBr_2,+2H_2O["Pure" Ba(OH)_2=342xx0.8=,273.6 gm]),(273.6/171=1.6 ,(6.4xx500)/1000=3.2,-,-),(-,-, ,1.6 "mole"):}`
`[Ba^(2+)]=1.6 M " " [Br^(-)]=3.2M`
Discussion
35.

Salt and water is formed by acid-base neutralisation reaction. If ratio of moles of acid & base taken is not similar to the ratio of their stoichiometric coefficient, then one of the component is limiting reagent.Assume no dissociation of water in following reactions. (Base is 80% pure only, take impurity present as inert & non electrolytic ) Select the correct combination for the resulting basic solution.

Answer»

(I)(iii)(S)
(I)(iv)(R)
(II)(i)(Q)
(III)(ii)(S)

SOLUTION :(i)Mass of pure NaOH`=(10xx80)/100=8 gm`
`{:(NaOH+,HCL to ,NaCl,+H_2O),(8/40=0.2 "mole",(8xx500)/1000,-,-):}`
=0.4 mole
`=0.2 "mole" " " 0.2"mole"`
Base in L.R.,`[H^+]=0.2 M " " [Na^+]=0.2 M [CL^(-)]=0.4 M`
(ii)KOH pure `=(28xx80)/100=22.4 gm`
`{:(KOH+,HNO_3to,KNO_3,+H_2O),(22.4/56=0.4 "mole",(2.5xx500)/1000,-,-):}`
=0.1 mole
`=0.3 "mole" " " 0.1"mole"`
Acid in L.R.,`[OH^(-)]=0.3 M " " [K]=0.4 M [NO_3^(-)]=0.1 M`
(iii) `{:(Ca(OH)_2+,H_2SO_4to,CaSO_4,+2H_2O["Pure" Ca(OH)_2=37xx0.8=29.6 gm]),(29.6/74=0.4 ,(0.8xx500)/1000=0.4,-,-):}`
`[Ca^(2+)]=[SO_4^(2-)]=0.4M`
(iv) `{:(BA(OH)_2+,2HBrto,BaBr_2,+2H_2O["Pure" Ba(OH)_2=342xx0.8=,273.6 gm]),(273.6/171=1.6 ,(6.4xx500)/1000=3.2,-,-),(-,-, ,1.6 "mole"):}`
`[Ba^(2+)]=1.6 M " " [Br^(-)]=3.2M`
Discussion
36.

Salt bridge is used because it

Answer»

maintains the electrical neutrality of the TWO half cells
It enhances the FLOW of electrons by OVERCOMING liquid jnction potential
It provides the lost ions
It forms insoluble precipitates with the electrolytes in the two half cells

Solution :FUNCTION of salt bridge.
Discussion
37.

Salt A overset("Layer test") to If reddish brownlayer come first , then

Answer»

`BR^(Θ)` PRESENT
`Br^(Θ)` absent
`CI^(Θ)` present
`I^(Θ)` present

Solution :`2Br^(Theta) overset(CI_(2))underset(water) to Br_(2)`: `C CI`layer BECOMES red or reddish brown
`I^(Theta)`ion is oxidised preferably COMPOUND to `Br^(Theta)`,Hence
`I^(Theta)` ID absent

Discussion
38.

Salt (A) makes part of electrode and is insoluble in water (A) is blackened by NH_(3) forming (B),(B) is soluble is equa regia forming (C ),(C ) gives orange ppt with KI but ppt dissolves in extcess of KIforming (D ) , identify (A),(B),(C ) and (D)

Answer»

Solution :(A) `Hg_(2)CI_(2)` (CALOMEL) (B ) `(HgNH_(2)CI+ Hg)`
(C ) `HgCI_(2)`(D) `H_(2)HgI_(4)`
Discussion
39.

Salt (A) is a white deliquescent solid. On heating the solid, brown fumes were evolved. On adding NaOH to the salt solution of (A), a white gelatinous ppt. was obtained. On adding, a drop of blue litmus solution to the ppt., a blue lake was obtained. On adding DPA to the salt solution, a deep blue colouration was obtained. Identify the salt and give the equations involved.

Answer»

Solution :(i) `2AL(NO_(3))_(3)OVERSET(Delta)toAl_(2)O_(3)+6NO_(2)uarr+(3)/(2)O_(2)uarr`
`(ii) `Al(NO_(3))_(3)+3NaOHtoAl(OH)_(3)darr+3NaNO_(3)`
`Al(OH)_(3)+"Blue litmus solution" to"Blue lake"`
[Note: A lake is a ppt. floating in a clear liquid.]
(iii) `NO_(3)^(-)+(C_(6)H_(5))_(2)NHtoNO_(2)^(-)+(C_(6)H_(3))_(2)N-N(C_(6)H_(5))_(2)+H_(2)O`
Discussion
40.

Salol (phenyl salicylate) is used as an :

Answer»

Insecticide
Analgesic
Ointment
Intestinal antiseptic

Answer :D
Discussion
41.

Salol can be used as:

Answer»

ANTISEPTIC
ANTIPYRETIC
both
none.

Answer :A
Discussion
42.

Salol can be used as :

Answer»

ANTISEPTIC
ANTIPYRETIC
Analgesic
None of these

Answer :A
Discussion
43.

Saliva contains:

Answer»

Amylases
Bile
Vitamins
Trypsin

Answer :A
Discussion
44.

Saline soil contains :

Answer»

HIGH CONCENTRATION of salt
Lot of moisture
Hard rocks
None of the above

Answer :A
Discussion
45.

Salicylic acid on brornination gives:

Answer»

2,4,6-tribromosalicylic ACID
2,4,6-tribromophenol
2,4,6-tribromobenzoic acid
p-bromosalicylic acid

ANSWER :B
Discussion
46.

Saline hydrides are :

Answer»

STRONG oxidants
Strong reductants
Strong dehydrading agents
Strong BLEACHING agents

Answer :B
Discussion
47.

Salicylic acid on heating with sodalime forms:

Answer»

PHENOL
Benzene
Benzyl alcohol
Benzoic acid

Answer :A
Discussion
48.

Salicylic acid on heating with soda lime (CaO+NaOH) gives :

Answer»

Benzene
Benzoic acid
Phenol
Toluene.

Solution :
Discussion
49.

Salicylic acid is produced when phenol in alcoholic KOH is treated with

Answer»

`CH_(3)CL`
`CHCl_(3)`
`CH_(2)Cl_(2)`
`"CCL"_(4)`

Answer :D
Discussion
50.

Salicylic acid is used to get

Answer»

OIL of WINTER GREEN
Salol
Aspirin
Anisole.

Answer :A::B::C
Discussion