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.

In permanganate titrations, potassium pemanganats is used an an oxidizing agent in acidic medium. The medium is maintained acidic by the use of dilute sulphuric acid. Potassium permanganate acts as self indicator . The potential equation, when potassium permanganate acts as an oxidizing agent is 2KMnO_(4) + 3H_(2)SO_(4) to K_(2)SO_(4) + 2MnSO_(4) + 3H_(2)O + 5[O] or MnO_(4)^(-) + 8H^(+) + 5e^(-) to Mn^(2+) + 4H_(2)O Before the end point, the solution remains colourless but after teh equivalence point only one extra drop of KMnO_(4) solution imparts pink colour , i.e. appearance of pink colour indicates end point. These titrations are used for estimation of ferrous salts, oxalic acid, oxalates, hydrogen peroxide, As_(2)O_(3) etc. Mass of KHC_(2)O_(4) (potassium acid oxalate ) required to reduce 100 ml of 0.02 KMnO_(4) in acidic medium (MnO_(4)^(-)" changes to " Mn^(2+)) is x gm and to neutralize 100 ml of 0.05M Ca(OH)_(2) is y gm ,then

Answer»

`X = y`
`2X = y`
`x = 2Y`
NONE of above

SOLUTION :N/A
Discussion
2.

In permanganate titrations, potassium pemanganats is used an an oxidizing agent in acidic medium. The medium is maintained acidic by the use of dilute sulphuric acid. Potassium permanganate acts as self indicator . The potential equation, when potassium permanganate acts as an oxidizing agent is 2KMnO_(4) + 3H_(2)SO_(4) to K_(2)SO_(4) + 2MnSO_(4) + 3H_(2)O + 5[O] or MnO_(4)^(-) + 8H^(+) + 5e^(-) to Mn^(2+) + 4H_(2)O Before the end point, the solution remains colourless but after teh equivalence point only one extra drop of KMnO_(4) solution imparts pink colour , i.e. appearance of pink colour indicates end point. These titrations are used for estimation of ferrous salts, oxalic acid, oxalates, hydrogen peroxide, As_(2)O_(3) etc. For decolourization of 1 mol of KMnO_(4) , the number of moles of H_(2)O_(2) required is

Answer»

`1/2`
`3/2`
`5/2`
`7/2`

SOLUTION :N/A
Discussion
3.

In permanganate titrations, potassium pemanganats is used an an oxidizing agent in acidic medium. The medium is maintained acidic by the use of dilute sulphuric acid. Potassium permanganate acts as self indicator . The potential equation, when potassium permanganate acts as an oxidizing agent is 2KMnO_(4) + 3H_(2)SO_(4) to K_(2)SO_(4) + 2MnSO_(4) + 3H_(2)O + 5[O] or MnO_(4)^(-) + 8H^(+) + 5e^(-) to Mn^(2+) + 4H_(2)O Before the end point, the solution remains colourless but after teh equivalence point only one extra drop of KMnO_(4) solution imparts pink colour , i.e. appearance of pink colour indicates end point. These titrations are used for estimation of ferrous salts, oxalic acid, oxalates, hydrogen peroxide, As_(2)O_(3) etc. In order to prepare one litre of normal solution of KMnO_(4), how many grams of KMnO_(4) is required if solution is to be used in acid medium for oxidation ?

Answer»

`158` GM
`31.6` gm
`62` gm
`790` gm

SOLUTION :N/A
Discussion
4.

In permanganate ion, all the bonds formed between Mn and oxygen are covalent. Give reasons.

Answer»

Solution :Oxidation number of Mn in `MnO_4` IONS is +7. To form `Mn^(7+)`ions, SEVEN electrons are to be removed. However, ENERGETICALLY this is not feasible. HENCE, it forms COVALENT bonds by sharing of electrons.
Discussion
5.

In permanganate ion, all the bonds between Mn and O are covalent. Give reasons.

Answer»

Solution :Oxidation state of Mn in `MnO_(4)^(-)` is +7. It is energetically not possible to lose SEVEN electrons to give ionic SPECIES. It FORMS bonds by sharing of electrons. HENCE covalent bonds are formed.
Discussion
6.

In PCl_(5), phosphorus is in sp^(3)d hybridised state but all its five bonds are not equivalent. Justify your answer with reason.

Answer»

Solution :`PCl_(5)` has trigonal bipyramid geometry. It has three equatorial and two axial bonds. Since each axial P-Cl bond is repelled by three bond pairs and each equatorial P-Ci bond is repelled by only two bond pairs, therefore, axial bonds are longer (240 PM) than equatorial bonds (202 pm) than equatorial bonds (202 pm). Thus, all the five P-Cl bonds in `PCl_(5)` are NT equivalent.
Discussion
7.

In PCl_5, phosphorus is in sffid hybridised state but all its five bonds are not equivalent. Justify your answer with reason.

Answer»

SOLUTION : In `PCl_5`, the three equatorial P-Cl bonds are EQUIVALENT present at 120°. The remaining TWO P - CI bonds are axial which has GREATER bond lengththan equatorialbonds.
Discussion
8.

In passing the electricity through the acidulated H_(2)O , 5.6 dm^(3) of O_(2) liberated at anode . The volume of H_(2) liberated at cathode will be

Answer»

`5.6 DM^(3)`
`22.4 dm^(3)`
`11.2 dm^(3)`
`44.8 dm^(3)`

SOLUTION :By passing 1 F , 1 g equivalent of `H_(2)` (1 g `-= 11.2 dm^(3)`) and 1 g equivalent of `O_(2)` ( 8 g `-= 5.6 dm^(3))` liberated.
Discussion
9.

In PCl_5 molecule, P is :

Answer»

`sp^3`-hybridised
`dsp^2`-hybridised
`ds^3`p-hybridised
`sp^3d`-hybridised

Answer :D
Discussion
10.

In para (iv), if NaOH is replaced by an NH_(3) solution, what will the solution D contain?

Answer»

`[ZN(NH_(3))_(4)]^(2+)`
`[AI(OH)_(4)]^(-)`
`[Pb(NH_(3))_(4)]^(2+)`
`SnO_(2)^(2-)`

Solution :`[Zn(NH_(3))_(4)]^(2+)`
Discussion
11.

In paper chromatography :

Answer»

both PHASES are liquids
mobile phase is LIQUID and STATIONARY phase is solid
mobile phase is solid and stationary phase is liquid
both phases are solids

Answer :B
Discussion
12.

In P_4O_6 the number of oxygen atoms bonded to each P atom is:

Answer»

1.5
2
3
4

Answer :C
Discussion
13.

In P_4O_6 and P_40_10, the numbers of oxygen atoms bonded to each phosphorus atoms are respectively

Answer»


Solution :The STRUCTURE of `P_4O_6 and P_4O_10` are gives as under

In `P_4O_6` each PHOSPHORUS is linked to THREE oxygen atoms, whereas in `P_4O_10,` each phosphorus atoms.
Discussion
14.

In P_(4)O_(10) each P atom is linked with ...... O atom

Answer»

2
3
4
5

Solution :In `P_(4)O_(10)`, each P atom is linked to 4 oxygen ATOMS as can be CONFIRMED by its STRUCTURE. It is linked to three oxygen atoms by single bond and with one oxygen atom by double bond.
Discussion
15.

In P_4 (tetrahedral)

Answer»

Each P is joined to four P
Each p is jonied to three P
Each P is joined to two P
NONE

SOLUTION : Each P - SI Tetrahedral and `Sp^3` - Hybridised - each P - is Joined to three P
Discussion
16.

In P versus V graph, the gorizontal line is found in which _______ exists.

Answer»

Gas
Liquid
Equilibrium between gas and liquid
Super critical temperature.

Solution :
From the graph, at point A, `CO_(2)` exists as a gas, As pressure is INCREASED, the volume of the gas decreases along the CURVE AB. At B liquefaction of the gas starts. Hence, volume decreases rapidly along BC because liquid has much LESS volume than the gas. At point C, liquefaction is complete. Amount of gas decreased = Amount of liquid formed or increased. Because along horizontal line, gas converts into liquid.
Discussion
17.

In oxidation 1 Hydrogen is displaed form a substance 2 an electropostive element is added to or proportion of electropositive element increases in a substance Select the corrcect answer using the code given below

Answer»

1 and 3
2 and 4
2 and 3
1 and 4

Solution :Oxidation and reduction are TWO type of chemical RECTIONS that often work together oxidation and reduction reaction involoves anexchange of ELECTRONS are between REACTANTS oxidation and reduction reactions are common when working with acids and bases and other electrochemical processes
Discussion
18.

In oxidisin roasting of Zn, S porducts are

Answer»

`ZNO + SO_2`
`ZN O + Zn SO_4 +SO_2`
`ZnCl_2`
`Zn +SO_2`

ANSWER :A
Discussion
19.

In Ostwald's process, nitric oxide is prepared by the oxidation of

Answer»

`NH_(3)`
`N_(2)`
AIR
NITROUS oxide

Answer :1
Discussion
20.

In Ostwald's dynamic method, the loss in weight in solventbulb is directly proportional to :

Answer»

<P>P
`P^(0)`
`P^(0)-P`
`(P^(0)-P)/(P^(0))`

ANSWER :C
Discussion
21.

In osmosis, there is a net flow of …………..from…………….to………… .

Answer»

SOLUTION :SOLVENT, solvent, solution
Discussion
22.

In osmosis

Answer»

solvent MOLECULES PASS from high concentration of solute to low concentration
solvent molecules pass from a SOLUTION of low concentration of soluteto a solution of high concentration of solute
solute molecules pass from low concentration to high concentration
solute molecules passfrom high concentration to low concentration

ANSWER :A::B::C
Discussion
23.

In orthorhombic, the values of a,b and c are respectively 4.2 Å and 8.3Å. Giventhe molecular of thesolute of 155 g"mol"^(-1)and density is 3.3 g /cc the number of formula units per unit cell is

Answer»

2
3
4
6

Solution :`Z=(VxxN_(A)xxd)/M`
`=(4.2xx8.6xx8.3xx10^(-24)xx6.023xx10^(23)xx3.3)/155=3.84~~4`
Discussion
24.

In organic reactions, metallic lithium in liquid ammonia behaves as

Answer»

oxidising agent
reducing agent
bleaching agent
DEHYDRATING agent

ANSWER :B
Discussion
25.

In organic reaction, metallic lithium in liquid ammonia behaves as

Answer»

Oxidising agent
Reducing agent
Bleaching agent
Dehydrating agent

SOLUTION :Lithium metal in liquid AMMONIA CONTAINS ammoniated electron, which ACTS as a reducing agent.
Discussion
26.

In organic layer test, CS_(2) or C CI_(4) is added to Lassaigne's extract and then Cl_(2) water or KMnO_(4) is added. This test is used to distinguish between

Answer»

`BR^(-)` and `I^(-)`
`Cl^(-)` and `Br^(-)`
`Cl^(-)` and `I^(-)`
`Cl^(-), Br^(-)` and `I^(-)`

Solution :Appearance of ORANGE colour in organic layer confirms the presence of Br while violet color confirms iodine
`2Br^(-)+Cl_(2) to underset("orange")(Br_(2))+2Cl^(-)`
`2I^(-)+Cl_(2) to underset("(violet)")(I_(2))+2Cl^(-)`
`16H^(o+)+2MnO_(4)^(-)+10Br^(-) to 5Br_(2)+2Mn^(2+)+8H_(2)O`
`16H^(o+)+2MnO_(4)^(-)+10I^(-) to 5I_(2)+2Mn^(2+)+8H_(2)O`
Here `Cl_(2)` (or) `KMnO_(4)` works as oxidizing AGENT
Discussion
27.

In organic compounds, nitrogen is tested in Lassaigne's test as

Answer»

`NaNH_(2)`
`NaCN`
`NaNO_(2)`
`NaNO_(3)`

Answer :B
Discussion
28.

In organic compounds , halogens are estimated by

Answer»

Liebig METHOD
Duma's method
Carius method
Schiff's and PINA method

Solution :Estimation of halogen in an organic COMPOUND can be DONE by Carius method and Schiff.s and Piria method .
Discussion
29.

In organic compound phosphorus is estimated as :

Answer»

MAGNESIUM PYROPHOSPHATE `Mg_2P_2O_7`
`H_3PO_4`
`Mg_3(PO_4)_2`
`P_2O_5`

ANSWER :A
Discussion
30.

Which of the following elements forms oxide easily in air?

Answer»

N
O
H
S

Answer :B
Discussion
31.

In organic compound, phosphorus is estimated as :

Answer»

`Mg(NH_(4))PO_(4)`
`Mg_(3)(PO_(4))_(2)`
`Mg_(2)P_(2)O_(7)`
`H_(3)PO_(4)`

Answer :C
Discussion
32.

In order to wash clothes with water containing dissolved calcium hydrogen carbonate.which cleaning agent will you prefer and why : soaps or synthetic detergents ? Give one advantage of soaps over synthetic detergents .

Answer»

Solution :WATER containing calcium HYDROGEN carbonate is hard water. Detergents are preferred over SOAPS for cleaning clothes in hard water because calcium salts of detergents are soluble in water while calcium salts of soaps are INSOLUBLE . As a result, lot of soap is WASTED.
Discussion
33.

In organic chemistry the element which is estimated by difference :

Answer»

N
O
S
H

Answer :B
Discussion
34.

In ordinary Cl_2 gas Cl^35 and Cl^37 are in the ratio:

Answer»

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

Answer :B
Discussion
35.

In order to wash clothes with water containing dissolved calcium hydrogencarbonate, which cleaning agent would you prefer and why: soaps or synthetic detergents ? Give one advantage of soaps over synthetic detergents.

Answer»

Solution :Synthetic detergent will be preferred because soaps will FORM scum with WATER containing calcium hydrogencarbonate (hard water).
Soaps are BIODEGRADABLE and do not cause water pollution. Detergent MAY not degrade and cause ENVIRONMENTAL problem.
Discussion
36.

In order to wash clothes with water containing dissolved calcium hydrogen carbonate, which cleaning agent will you prefer and why : soaps or synthetic detergents ?

Answer»

SOLUTION :Disadvantage of using detergents. SOAPS are biodegradable while detergents having branched HYDROCARBON chains are not biodegradable and hence cause WATER POLLUTION in rivers and other waterways.
Discussion
37.

In order to wash clothes which cleaning agent what will you prefer and why: soap or synthetic detergents? Give one advantage of soaps and synthetic detergents each.

Answer»

SOLUTION : Soaps have straight hydrocarbon CHAINS and are easily degraded by bacteria PRESENT in the sewage water and hence, do not CAUSE water pollution. Most of the DETERGENTS are non-biodegradable and hence cause water pollution of rivers and waterways. So, one will prefer soap.
Discussion
38.

In order to separate oxygen from one mole of H_(2)O the required quantity of coulomb would be:-

Answer»

`1.93xx10^(5)`
`9.6xx10^(4)`
`1.8`
3.2

Answer :A
Discussion
39.

In order to synthesise acetophenone from acetyl chloride, we can use :

Answer»

WURTZ REACTION
Friedel-Crafts reaction
CANNIZZARO's reaction
Liebermanns reaction

ANSWER :B
Discussion
40.

In order to refine "copper" it is melted in a furnace and it is stirred with green logs of wood. The purpose is :-

Answer»

to expel the dissolved gases in blister copper
to reduce the METALLIC oxide impurities with hydrocarbon gases LIBERATED from the wood
to bring the impurities to surface and OXIDISE them
to increase the CARBON content in copper

Answer :B
Discussion
41.

In order to refine blister copper , it melts in a furnance and is stirred with green logs of wood . The purpose is

Answer»

to expel the dissoved gases in blister copper
to bringthe impuritiesto surface and oxidise them
to increase the CARBON CONTENT of copper
to REDUCE the metallic oxide IMPURITIES with hydrocarbon gases LIBERATED from the wood

Solution :`2Cu_(2)S+3O_(2) to 2Cu_(2)O+2SO_(2) `
`3Cu_(2)O+ underset("From green logs of wood")(CH_(4)) to 6Cu+2H_(2)O+CO`
Discussion
42.

In order to refine blister copper it is melted in a furnance and is strirred with gree longs of woman the purpose is

Answer»

To expel the dissolved gases in the blister
To bring the impurities to surface and oxidise them
To INCREASE the carbon content of copper
To reduce the metallic OXIDE impurities with hydrocarbon gases liberated from the wood

Answer :D
Discussion
43.

In order to refine blister copper it is melted in a furnace and is stirred with green logs of wood. The purpose is

Answer»

to increase the carbon content of copper.
to bring the impurities to SURFACE and oxidize them.
to reduce the metallic OXIDE impurities with HYDROCARBON gasses liberated from the wood.
to expel the dissolved gases in blister copper.

Answer :C
Discussion
44.

In order to refine "blister copper' it is melted in a furnace and is stirred with green logs of wood. The purpose is _______

Answer»

to EXPEL the dissolved gases in blister copper
to BRING the impurities to SURFACE and oxidize them
to inrease the carbon content of copper
to reduce the metallic OXIDE impurities with HYDROCARBON gases liberated from the wood

Answer :D
Discussion
45.

In order to prevent the hot metal flament from getting burnt, when the electric current is switched on, the bulb is filled with:

Answer»

`CH_4`
An INERT GAS
`CO_2`
`Cl_2`

ANSWER :B
Discussion
46.

In order to refine blister copper it is melted in a furnace and is stirred with green logs of wood. The purpose is:

Answer»

To EXPEL the dissolved gases in blister copper
To bring the impurities to SURFACE and oxidise them
To INCREASE CARBON content of copper
To reduce the metallic OXIDE impurities with hydrocarbon gases liberated from the wood

Answer :D
Discussion
47.

In order to prepare one litre normal solution of KMnO_(4), how many gm of KMnO_(4) are required, if the solution is to be used in acid medium for oxidation

Answer»

158 gm
31.60 gm
62.0 gm
790 gm

Answer :B
Discussion
48.

In order to prepare benzonitrile from C_6H_5N_2CI, the reagent required is

Answer»

KCN in presence of ZINC powder
CuCN DISSOLVED in KCN
`Ba(CN)_2` and HCN
Aqueous HCN

Answer :2
Discussion
49.

In order to prepare a primary amine from an alkyl halide with simultaneous addition of one CH_2 group in the carbon chain, the reagent used as source of nitrogen is

Answer»

Sodium amide (`NaNH_2`)
Sodium azide (`NaN_3`)
Potassium cyanide (KCN)
Potassium phthalimide `[C_6 H (CO)_2N^(-) K^(+) ]`.

Solution :Cyanides on REDUCTION gives `1^@`amines with a `CH_2`group, i.e.,
` R - X underset(-KX ) OVERSET("KCN" )toR -CNoverset(Na//C_2 H_5OH) to R -CH_2 NH_2`
Discussion
50.

In order to prepare a 1^(@) amine from an alkyl halide with simultaneous addition of one CH_(2)group in the carbon chain, the reagent used as source of nitrogen is…..

Answer»

Sodium AMIDE (`NaNH_2`)
Sodium azide (`NaN_3`)
POTASSIUM cyanide (KCN)
Potassium PHTHALIMIDE `[C_6 H (CO)_2N^(-) K^(+) ]`.

Solution :Cyanides on reduction gives `1^@`amines with a `CH_2`group, i.e.,
` R - X underset(-KX ) overset("KCN" )TOR -CNoverset(Na//C_2 H_5OH) to R -CH_2 NH_2`
Discussion