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.

Excessive accumulation of organic matter in water bodies leads to

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Decrease in species diversity
Increase in species diversity
Green house effect
No effect on species diversity

Solution :Organic matter (organic wastes) contains a NUMBER of pathogen, SECONDARY pollutant, pesticides etc. Biological oxygen DEMAND becomes high, and therefore the dissolved oxygen reduced. Hence, planktons Mollusca and fishes will be eliminated due to reduced dissolved oxygen and presence of secondary pollutants. Some species like annelid worm Tubifex and some insect larvae (Chironomus) tolerate POLLUTION
2.

Excess of silver nitrate solution is added to 100 ml of 0.01 M pentaaqua chloro chromium (III) chloride solution .The mass of silver chloride obtained in grams is [Atomic mass of silver is 108]

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`287xx10^(-3)`
`143.5xx10^(-3)`
`143.5xx10^(2)`
`287xx10^(-2)`

Solution :`[Cr(H_(2)O_(5))_(5)Cl]Cl_(2)underset(AgNO_(3))overset(Excess)to`
`[Cr(H_(2)O)_(5)Cl]^(+)NO_(3)^(-)+2AgCldarr`
Thus Meq. Of `Cl^(-)` ions outside=Meq. Of AGCL FORMED
`THEREFORE` NUMBER of moles of `Cl^(-)` ions present
`=(2xx100xx0.01)/(1000)=0.002 `moles
`therefore` Mass of AgCl=`0.002xx143.5`
=0.287 g=`287xx10^(-3)g`
3.

Excess of NH_3 combine with sodium hypochlorite solution gives -

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`NH_(4)`
`NH_(2)NH_(2)`
`NaNH_(2)`
`NH_(3)NCl_(3)`

Answer :B
4.

Excess of PCl_(5) reacts with conc , H_(2)SO_(4) giving

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sulphuryl CHLORIDE
sulphurous ACID
chlorosulphonic acid
thionyl chloride

Solution :Excessof `PCl_(5)` REACTS with conc. `H_(2)SO_(4)`to give sulphurly chloride.
`underset("Sulphurice acid")(SO_(2)(OH)_(2)+2PCl_(5))tounderset("Sulphuryl chloride")(SO_(2)Cl_(2)+2POCl_(3)+2HCl`
5.

Excess of Na^(+) ions in human system causes

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diabetes
anaemia
HIGH PRESSURE
low BLOOD pressure

Solution :High Blood Pressure
6.

Excess of isopropyl alcohol is heated with conc. H_2SO_4 at 413 K, gives

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`(CH_3)_2CHOCH(CH_3)_2`
`CH_3CH=CH_2`
both 'a' and 'B'
none of these

Answer :B
7.

Excess of KI reacts with CuSO_(4) solution and then Na_(2) S_(2) O_(3) solution is added to it . Which of the statements is incorrect for this reaction ?

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`Na_(2) S_(2) O_(3)` is oxidised
`CuI_(2)` is FORMED
`Cu_(2) I_(2)` is formed
Evolved `I_(2)` is reduced .

ANSWER :B
8.

Excess of glycol when dehydrated gives :

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ETHYLENE oxide
Ethanol
Acrolein
1,4-dioxan

Answer :D
9.

Excess of ethanol is heated with conc. H_2SO_4 at 413 K. the compound that distills is

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DIETHYL sulphate
diethyl ether
ethylene HYDROGEN sulphate
ethylene

Answer :B
10.

Excess of copper and iron are removed mainly by which of the following chelating ligands via formation of coordination compounds?

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DESFERRIOXIME-B and EDTA RESPECTIVELY
EDTA and D-penicillamine respectively
Desferrioxime-B and D-penicillamine respectively
D-penicillamine and Desferrioxime B-respectively

Answer :D
11.

Excess of copper in toxic proportions in plants/animals can be

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EDTA
Ethane-1,2- diamine
Oxalate ion
D-penicillamine

SOLUTION :D-penicillamine is used in treating wilson's disease, a RARE GENETIC disorder of copper metabolism, penicillaminetreatment relies on its binding to ACCUMULATED copper and ELIMINATION through urine.
12.

Excess of copper and iron are removed by chelating ligands like ______

Answer»


ANSWER :D-penicillamine
13.

Excess of conc. H_(2)SO_(4) reacts with C_(2)H_(2)OH at 413 K to form.

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`C_(2)H_(5)HSO_(4)`
`(C_(2)H_(5))_(2)SO_(4)`
`CH_(2)=CH_(2)`
`C_(2)H_(6)OC_(2)H_(5)`

Solution :Alcohol is dehydrated to give alkene. It is a `BETA-"elimination"`
`CH_(3)-CH_(2)-OH underset(-H_(2)O)overset(H_(2)SO_(4).443K)to CH_(2)=CH_(2)`
Note: At 413K, ethyl ether is obtained
`C_(2_)H_(5)OH +HCO_(2)H_(5) underset(-H_(2)O) overset(H_(2)SO_(4).443K)to C_(2)H_(5)OC_(2)H_(5)`
14.

Excess of concentrated sodium hydroxide can separate mixture of

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`Al^(3+)` and `Cr^(3+)`
`Cr^(3+)` and `Fe^(3+)`
`Al^(3+)` and `Zn^(3+)`
`Zn^(2+)` and `Pb^(2+)`

Solution :`AI^(2+),Zn^(2+) and Cr^(2+)` form a soluble COMPLEX with excess of `NAOH` WHEREAS `Fe^(2+)` does not
15.

CH_3MgI is made to react with acetone and the addition product formed is hydrolysed will give :

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SEC- butanol
n- butanol
iso- butanol
t- butanol

Answer :D
16.

When CH_3MgI is made to react with acetone and the addition product formed is hydrolysed we get :

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ETHYL HYDROGEN sulphate
Ethylene
Diethyl sulphate
Diethy ether

Answer :D
17.

Excess of carbon dioxide is passed through 50 ml of 0.5 M calcium hydroxide solution. After the completion of the reaction, the solution was evaporated to dryness. The solid calcium carbonate was completely neutralised with 0.1 N hydrochloric acid. The volume of hydrochloric acid required is (At. mass of calcium=40)

Answer»

`200cm^(3)`
`500cm^(3)`
`400cm^(3)`
`300cm^(3)`

Solution :According to the question, the REACTION occurs as
`underset(74(=1mol))(Ca(OH)_(2))+underset(44(=1mol))(CO_(2))to underset(100(=1mol))(CaCO_(3))+underset(18(=1mol))(H_(2)O)`
Given, 50 ml of 0.5 M `Ca(OH)_(2)` reacts with excess of `CO_(2)`
`THEREFORE` No. of millimoles of `Ca(OH)_(2)` reacts with excess of `CO_(2)`
`therefore` No. of millimoles of `Ca(HO)_(2)` reacted=25
`therefore` 1 MOLE of `Ca(OH)_(2)` gives 1 mole of `CaCO_(3)`,
`therefore`No. of millieq. `=("WEIGHT in mg")/("eq. Wt")=(25xx100)/(50)=50`
`implies`No. of milliequivalents of `CaCO_(3)=50`
As, volume of `CaCO_(3)` solution=50 ml
So, normality of `CaCO_(3)` solution=1N(millieq.=`NxxV` in ml)
Normality of HCl=0.1N (given)
Volume of HCl=?
`N_(HCl)xxV_(HCl)=N_(CaCO_(3))xxV_(CaCO_(3))`
`implies0.1xxV_(HCl)=1xx50impliesV_(HCl)=(50)/(0.1)=500cm^(3)`.
18.

Excess of CH_(3)CO OH is reacted with CH-=CH in presence of Hg^(2+), the product is

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`CH_(3)CH(OCOCH_(3))_(2)`
`CH_(2)=CH(OCOCH_(3))`
`(CH_(3)CO O)CH_(2)-CH_(2)(O OC CH_(3))`
None of these

Solution :`CH-=CH underset(Hg^(2+))OVERSET(CH_(3)CO OH)to CH_(2)=CHOCOCH_(3) overset(CH_(3)CO OH)to CH_(3)CH(OCOCH_(3))_(2)`
19.

Excess of carbon dioxide is passed through 50 mL of 0.5 M calcium hydroxide solution. After the completion of the reaction, the solution was evaporated to dryness. The solid calcium carbonate was completely neutralised with 0.1 N hydrochloric acid. The volume of hydrochloric acid required is (At. mass of carbon = 40)

Answer»

200 mL
500 mL
400 mL
300 mL

Solution :`Ca(OH)_(2)+CO_(2)RARR CaCO_(3)+H_(2)O`
50 mL of 0.5 `Ca(OH)_(2)` contain `Ca(OH)_(2)`
`=(0.5)/(1000)xx"50 mol"=0.025"mol"`
1 mol of `Ca(OH)_(2)` PRODUCES `CaCO_(3)` = 1 mol
`therefore 0.25` mole of `Ca(OH)_(2)` will produce `CaO_(3)`
= 0.025 mol
`CaCO_(3)+2HClrarr CaCl_(2)+H_(2)O+CO_(2)`
1 mole of `CaCO_(3)` is NEUTRALISED by HCL = 2 mol
`therefore 0.025` mol of `CaCO_(3)` will be neutralised by HCl
`=2xx0.025" mol" = 0.05 " mol"`
`" 1000 mL of 0.1 N HCl contain HCl = 0.1 g eq"`
`"= 0.1 mol"`
`"or 0.1 of HCl is present in HCl sol = 1000 mL"`
`therefore" 0.05 mol of HCl will be present in HCl sol."`
`=(1000)/(0.1)xx0.05mL="500 mL"`
20.

Excess of calcium orthophosphate is reacted with magnesium to from calcium phosphide (Ca_(3)P_(2)) along with magnesium oxide. Calcium phosphide on reacting with excess of water liberate phosphine gas (PH_(3)) along with calcium hydroxide. Phosphine is burnt in excess of oxygen to from P_(2)O_(5) along with water. oxides of magnesium & phosphorous react to give magnesium metaphosphate. calculate grams of magnesium metaphosphate obtained if 1.92 gm of magnesium is taken .

Answer»


SOLUTION :`Ca_(3)(PO_(4))_(2)+underset(0.08)(8 Mg) to underset("0.01 MOLE")(Ca_(3)P_(2))+underset(0.08)(8 MgO)`
`underset(0.01)(Ca_(3)P_(2))+6H_(2)Oto underset(0.02) (2PH_(3))+3Ca(OH)_(2)`
`underset(0.02)(2PH_(3))+4O_(2) to underset(0.01)(P_(2)O_(5))+3H_(2)O)`
`underset(MgO)(0.08) +underset(0.01)(P_(2)O_(5))to underset("0.01 Mole" ~~ 1.82 GM ~~ 2gm)(MgP_(2)O_(6))`
21.

Excess of carbon dioxide is passed through 50 mL of 0.5 M calcium hydroxide solution. After the completion of the reaction, the solution was evaporated to dryness. The solid calcium carbonate was completely neutralised with 0.1 N hydrochloric acid. The volume of hydrochloric acid required is

Answer»

`200 cm^(3)`
`500 cm^(3)`
`400 cm^(3)`
`300 cm^(3)`

Solution :According to the question the reaction occurs as
`underset((=1"mol"))underset(74)(Ca(OH)_(2))+underset((=1"mol"))underset(44)(CO_(2))tounderset((=1"mol"))underset(100)(CaCO_(3))+underset((=1"mol"))underset(18)underset(H_(2)O`
GIVEN 50 ml of 0.5 M `Ca(OH)_(2)` reacts with excess of `CO_(2)`
`:.` no. of millimoles of `Ca(OH)_(2)` reacted =25
`:'` 1 mile of `Ca(OH)_(2)` gives 1 mile of `CaCO_(3)`,
`:.` no. of millimoles of `CaCO_(3)` formed =25
`:.` no. of millieq `=("weight in mg")/("eq. wt")=(25xx100)/50=50`
`implies` no. of MILLIEQUIVALENTS of `CaCO_(3)=50`
As volume of `CaCO_(3)` solution =50 ml
So, normality of `CaCO_(3)` solution =1N
(millieq `=NxxV` in ml)
Normality of HCl `=0.1N` (given)
Volume of HCl=?
`N_(HCl)xxV_(HCl)=N_(CaCO_(3))xxV_(CaCO_(3))`
`implies0.1xxV_(HCl)=1xx50`
`impliesV_(HCl)=50/0.1=00 cm^(3)`
22.

Excess of ammonia with sodium hypochloride solution in the presence of glue or gelatine gives.

Answer»

`NaNH_(2)`
`NH_(2)NH_(2)`
`N_(2)`
`NH_(4)Cl`

Solution :`NH_(3) + NAOCL to NH_(2)- NH_(2) + NaCl + H_(2)O`
23.

Excess of bromo ethane reactswithalcholicammonia to give,

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ETHYL AMINE
DIETHYL amine
TRIETHYL amine
all of these

ANSWER :D
24.

Excess of ammonia with sodium hypochlorite solution in the presense of gule or gelatin gives

Answer»

`NaBH_(2)`
`NH_(2)NH_(2)`
`N_(2)`
`NH_(4)Cl`

Solution :Excess of ammonia `(NH_(3))` when reacts with sodium HYPOCHLORITE solution (NaOCl) in PRESENCE of glue or GELATIN gives hydrazine `(NH_(2)*NH_(2))`. This process is known as Reschig process
`2NH_(3)+NaOC l toN_(2)H_(4)+NaCl+H_(2)O`
Addition of gelatin or glue suppress the side reaction by complexing with the metal ions and yield of hydrazine is increased.
25.

Excess of AgNO_(3) solution was added to 2.2 g of commercial sample of common salt dissolved in water. The mass of dried precipitate of silver chloride was 2.11 g . Calculate the percent purity of common salt.

Answer»


ANSWER :39.1
26.

Excess of AgCl is added to 0.1 M solution of KBr at 298 K. Calculate the equilibrium concentrations of Br^(-) and Cl^(-) ions. E_(Cl^(-)//AgCl//Ag)^(@)=0.222" volt"E_(Br^(-)//AgBr//Ag)^(@)=0.095 volt

Answer»


Solution :`{:(,AgCl,+,Br^(-),hArr,AGBR,+,Cl^(-)),("At equilibrium,",,,(0.1-X),,,,x):}`
APPLY `log K =E_(cell)^(@)/0.0591`
or `"log"x/((0.1-x))=E_(cell)^(@)/0.0591`
27.

Excess nitratein drinkin watercan cause

Answer»

methemoglobinemia
liver DAMAGE
kindey
damage

Solution :Excess NITRATE in drinking watercan causemethemoglobinemia(bluebabysyndrome)
28.

Excess chlorine,when bubbled through aqueous ferrous bromide will yeild

Answer»

`FeCl_(2)`
`FeBr_(2)`
`FeCl_(3)`
`Br_(2)`

Answer :C::D
29.

Excess H^+ and Cl^+ ions from a colloidal solution of Fe(OH)_3 can be removed by

Answer»

filtration
DIALYSIS
sublimation
chromatography

SOLUTION :Ionic IMPURITIES can be REMOVED from a colloidal solution by dialysis.
30.

Excess chlorine is passed through boiling toluene, how many chloro deriatives would you get?

Answer»


SOLUTION :THREE, i.e., benzyl CHLORIDE, BENZAL dichloride and benzotrichloride.
31.

Except which alcohol, other alcohols are weaker acid than water?

Answer»

ETHANOL
PHENOL
METHANOL
PROPANOL

SOLUTION :Methanol
32.

Except ……………… other alkanes gives a mixture of nitro aklkanes due to C-C cleavage by nitration of alkanes.

Answer»


ANSWER :METHANE
33.

Except……….all other amino acids are optically active.

Answer»

SOLUTION :GLYCINE
34.

Except helium, all the noble gases crystallize with cubic close packed structures. Derive an equation relating atomic radius to the density of cubic close packed solid of given mass and apply it to deduce the relative size of Ar and Xe if their densities are 1.4 g/cm and 2.83 g/cm^(3) respectively.

Answer»

Solution : For a CCP solid : `(rho)=(nM)/(n_(A)a^(3))`
Where,
`rho`=Density,n=No. of atoms/unit cell
M = Molar MASS, `N_(A)` = Avogadro constant,
and a=Edge LENGTH
`implies rho=(4M)/(N_(A)(2sqrt(2R))^(3))=(M)/(4sqrt(2)N_(A)R^(3))`
`implies(rho_(x_(e)))/(rho_(Ar))=(2.83)/(1.40)((r_(Ar))/(r_(Xe)))^(3)`
`implies(r_(Xe))/(R_(Ar))=((131xx1.4)/(40xx2.83))^(1//3)=1.174`
35.

Except glycine all other naturally occurring amino acids are optically active.

Answer»


ANSWER :1
36.

Except for vitamin B_(12) all other vitamins of group B should be regularly supplied in diet. Why?

Answer»

Solution :All the vitamins of group B except `B_(12)` are water SOLUBLE. So they are READILY excreted in urine. Hence they NEED to be RESTORED in the body. `B_(12)` is water insoluble and is not much excreted through urine.
37.

Except for vitamin B_12, all other vitamins of group B, shlould be supplied regularly in diet. Why ?

Answer»

Solution :Vitamin- `B2` does not occur in plants but is exclusively synthesized by the microorganisms PRESENT in the body. THEREFORE, it may not be supplied REGULARLY in die , while all other vitamins of group B should be supplied regularly in DIET.
38.

Which statement correctly the statement ? Except for glycine, which is achiral, all the amino acids present in proteins....

Answer»

Are CHIRAL , but recemic
Have the L CONFIGURATION at their `alpha` CARBON
Have the R configuration at their `alpha` carbon
Have the S configuration at their `alpha` carbon

Answer :B
39.

Except ............ elements, all transition metals are hard and have very high melting point

Answer»

SOLUTION :group-11
40.

Except ………………. all other alcohols are weaker acid than water.

Answer»

SOLUTION :METHANOL
41.

Except_________ all elementfrom Rfto Cnare syntheticallyprearedand have verylowhalflifeperiods .

Answer»

CADMIUM
ACTINIUM
ytrium
cadmium

ANSWER :B
42.

Examples of thermoplastic polymer is/are :

Answer»

UREA FORMALDEHYDE resin
polythene
polystyrene
bakelite

Solution :N//A
43.

Examples for natural polymers are

Answer»

Cotton, Silk, Bakelite and WOOL
CELLULOSE, Polystyrene and Neoprene
Nylon, TERYLENE and PVC
Silk, Cotton and Proteins Natural Rubber

Answer :D
44.

Example the function of H_(2)-O_(2) fuel cell.

Answer»

Solution :In this case, hydrogen act as a fuel and oxygen as an oxidend and the electrolyte is aqueous KOH maintained at `200^(@)`C and 20 - 40 atm. Porous graphite electrode containinng NI and NiO serves as the inert electrodes.
Hydrogen and oxygen gases are bubbled through the anode and cathade, respectively.
Oxidation occurs at the anode:
`2H_2(G) + 4OH_((AQ))^(-) to 4H_2O_((l)) + 4e^(-)`
Reduction occurs at the cathode `O_2(g) + 2H2O(l) + 4e^(-) to 4OH^(-)(aq)`
The overall reaction is `2H_(2(g)) + O_(2)(g) to 2H_2O(l)`
The above reactio is the same as the hydrogen combustion reaction, however, they do not react directly ie. the oxidation and reduction reactions take place separately at the anode and cathode respectively like `H_2 - O_2` fuel CELL. Other fuel cell like propane `-O_2` and methane `O_2` have also been developed .
45.

Examples of regional pollution are

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Smog
Acid rain
Both (a) and (B)
NONE of these

Answer :C
46.

Example of unit cell with crystallographic dimensions a!=b!=c, alpha=beta=gamma=90^(@) is :

Answer»

Calcite
GRAPHITE
RHOMBIC SULPHUR
MONOCLINIC sulphur

ANSWER :B
47.

Example showing ionisation isomerism

Answer»

`[Co(NO_3)(NH_3)_5]SO_4` & `[Co(SO_4)(NH_3)_5]NO_3`
`[Co(NH_3)_4Cl_2]CL` & `[Co(NH_3)_5Cl]Cl_2`
`[Cr(H_2O)_5Cl]Cl_2. H_2O` & `[Cr(H_2O)_4Cl_2]Cl. 2H_2O`
`[PT(NH_3)_4][Pt Cl_4]` & `[Pt(NH_3)_3 Cl]_2 [Pt Cl_4]`

Answer :A
48.

Example of Ring silicate is ………….

Answer»

Olivine
Beryl
Spodumene
Asbestos

Answer :B
49.

Example of neutral complex compound in the following list

Answer»

`CoCl_3. 6NH_3`
`CoCl_3. 5NH_3`
`CoCl_3 . 4NH_3`
`CoCl_3 . 3NH_3`

ANSWER :D
50.

Exampleof intrinsic colloid is

Answer»

glue
sulphur
Fe
`As_(2)S_(3)`

Solution :On shaking with the DISPERSION medium, colloids directly form the colloidal sol. Hence, they are called INTRINSIC colloids. i.e., glue.