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.

Quantitative estimation of C, H, and extra elements (e.g., N,S,P and halogens) is carried out by Liebig.s combustion, Carius, Dumas, and Kjeldahl.s method. Carius is the name of

Answer»

A chemist
A biologist
A SEALED CAPILLARY tube
A long-necked round-bottom FLASK

Answer :C
Discussion
2.

Quantitative estimation of C, H, and extra elements (e.g., N,S,P and halogens) is carried out by Liebig.s combustion, Carius, Dumas, and Kjeldahl.s method.In the quantitative estimation of phosphorous by using magnesia misture, the formula used is:

Answer»

<P>PERCENTAGE of `P =62/222 xx (W xx 100)/(w)`
Percentage of `P = 31/222 xx (W xx 100)/(w)`
Percentage of `P = 62/222 xx (w xx 100)/(W)`
Percentage of `P = 31//222 xx (w xx 100)/(W)`

Answer :A
Discussion
3.

Quantitative estimation of C, H, and extra elements (e.g., N,S,P and halogens) is carried out by Liebig.s combustion, Carius, Dumas, and Kjeldahl.s method.Kjeldahl.s is the name of

Answer»

A chemist
A BIOLOGIST
A SEALED capillary TUBE
A long-necked round-bottom flask

ANSWER :D
Discussion
4.

Quantitative estimation of C, H, and extra elements (e.g., N,S,P and halogens) is carried out by Liebig.s combustion, Carius, Dumas, and Kjeldahl.s method. Carius method is used for the quantitative estimation of:

Answer»

C and H
HALOGENS, S, and P
N
All

Solution :Halogens, S and P are QUALITATIVELY ESTIMATED by carius METHOD.
Discussion
5.

Quantitative estimation of C, H, and extra elements (e.g., N,S,P and halogens) is carried out by Liebig.s combustion, Carius, Dumas, and Kjeldahl.s method.Dumas and Kjeldahl.s methods are used for quantative estimation of:

Answer»

C and H
Halogens, S, and P
N
All

Solution :NITROGEN is quantitatively estimated by Delmas & KJELDAHL METHOD.
Discussion
6.

Quantitative estimation of C, H, and extra elements (e.g., N,S,P and halogens) is carried out by Liebig.s combustion, Carius, Dumas, and Kjeldahl.s method.Liebig.s combustion method is used for the quantitive estimation of:

Answer»

<P>C and H
Halogens
S and P
N

SOLUTION :Lie big.s combustion method is USED for the quantitative estimation of C and H.
Discussion
7.

Quantitative analysis of calcium is carried out by ........

Answer»

SOLUTION :FLAME PHOTOMETRY
Discussion
8.

Quantitaivemeasurementof nitrogenin anorganiccompoundis doneby themethod_____.

Answer»

BERTHELOT method
Belisteinmethod
LASSAIGNE TEST
Kjeldahl'smethod

ANSWER :D
Discussion
9.

Qualitative analysis of calcium can be done by using .......

Answer»

Microscope
Flame photometer
Electron microscope
Spectro photometer

Answer :B
Discussion
10.

Q.12, Q.13 and Q.14 by appropriately matching the information given in the three columns of the following table. In which amongs the following product is formed by free radical mechanism?

Answer»

<P>I) II) P)
IV) ii) Q)
II) iii) Q)
III) iii) P)

SOLUTION :N//A
Discussion
11.

Q.12, Q.13 and Q.14 by appropriately matching the information given in the three columns of the following table. In which amongs the following more than one products are not formed?

Answer»

IV) iv) R)
III) ii) Q)
II) iii) R)
I) i) P)

SOLUTION :N//A
Discussion
12.

Q.12, Q.13 and Q.14 by appropriately matching the information given in the three columns of the following table. Match the correct combination?

Answer»

(IV) iii) R
I) II) R)
III) iv) P)
II) ii) R)

Solution :N//A
Discussion
13.

Q. The molecule C_(3)H_(4)O_(3) is sequential reaction is

Answer»

`HO-CH=CH-COOH`
`CH_(2)=underset(OH)underset(|)(C)-COOH`
`CH_(2)=underset(O)underset(||)(C)-underset(O)underset(||)(C)-OH`
`H-underset(O)underset(||)(C)-CH_(2)-underset(O)underset(||)(C)-OH`

ANSWER :D
Discussion
14.

Q. The conversion of (D) to (F) involves

Answer»

ELECTROPHILIC addition
Nucleophilic addition
Electrophilic substitution
Nucleophilic substitution

Answer :D
Discussion
15.

Q. Match the properties of column-I with the formulae in Column-II .

Answer»


ANSWER :A::B::C::D
Discussion
16.

Q. In the sequential reaction, if instead of CO_(2),(CH_(3))_(3)C-Br is used. Product will be

Answer»

`HC-=HC`
`(CH_(3))_(3)CH`
`HC-=C-C(CH_(3))_(3)`
`BrCH=CH-C(CH_(3))_(3)`

ANSWER :A
Discussion
17.

Q. Match the column-I with Column-II .

Answer»


ANSWER :A::B::C::D
Discussion
18.

Q. C_(3)H_(2)O_(2)underset(KMnO_(4))overset("Hot alkaline")toProduct. One of the products is

Answer»

`underset(CHO)OVERSET(CHO)(|)`
`underset(COOH)overset(COOH)(|)`
`O=CH-CH_(2)-COOH`
`CH_(3)-underset(O)underset(||)(C)-COOH`

ANSWER :B
Discussion
19.

Q. 'B' in the above passage is

Answer»

`Cu-C-=C-CH_(2)CH_(2)CH_(2)CH_(3)`
`CH_(3)CH_(2)C-=C-CH_(2)CH_(3)`
`CH_(3)-CH_(2)-C-=C-Cu`
All of these

Answer :A
Discussion
20.

Q and S can be

Answer»

`NaCl,Na_(2)S_(2)O_(3)`
`NAH,Na_(2)S_(2)O_(4)`
`NaH,Na_(2)S_(2)O_(7)`
`NaOH,Na_(2)S_(2)O_(4)`

Answer :B
Discussion
21.

Pyrolusite, MnO_(2), is the main ore from which manganese is produced. The manganese content of the ore may be determined by reducing the MnO_(2) under acetic conditions to Mn^(2+) with the oxalate ion, C_(2)O_(4)^(2-), the oxalate ion being oxidised to carbon dioxide during the reaction. The analytical determination is carried out by adding a known excess volume of oxalate solution to a suspension of the pyrolusite and digesting the mixture on a hot water bath until all the MnO_(2) has been reduced. The excess, unreacted oxalate solution is then titrated with standardised potassium permanganate, KMnO_(4) solution after which the manganese content of the ore can be calculated. A student prepared a standard solution of sodium oxalate by weighing 3.2 g of the dry anhydrous salt, dissolving it in distilled water and making the solution up to 500 mL. 25 mL of the oxalate solution required 24.76 mL of KMnO_(4) solution. Role of KMnO_(4) in the given titration can be described as :

Answer»

OXIDISING AGENT
reducing agent
indicator
oxidising agent and indicator

Solution :N//A
Discussion
22.

Pyrolusite, MnO_(2), is the main ore from which manganese is produced. The manganese content of the ore may be determined by reducing the MnO_(2) under acetic conditions to Mn^(2+) with the oxalate ion, C_(2)O_(4)^(2-), the oxalate ion being oxidised to carbon dioxide during the reaction. The analytical determination is carried out by adding a known excess volume of oxalate solution to a suspension of the pyrolusite and digesting the mixture on a hot water bath until all the MnO_(2) has been reduced. The excess, unreacted oxalate solution is then titrated with standardised potassium permanganate, KMnO_(4) solution after which the manganese content of the ore can be calculated. A student prepared a standard solution of sodium oxalate by weighing 3.2 g of the dry anhydrous salt, dissolving it in distilled water and making the solution up to 500 mL. 25 mL of the oxalate solution required 24.76 mL of KMnO_(4) solution. What is the molarity of KMnO_(4) solution ?

Answer»

0.04776
0.01929
0.038
0.028

Solution :N//A
Discussion
23.

Pyrolysis of Methane and ethane respectively are

Answer»

EXOTHERMIC & ENDOTHERMIC
Endothermic & exothermic
Both are endothermic
Both are Exothermic

Answer :C
Discussion
24.

Pyrolusite, MnO_(2), is the main ore from which manganese is produced. The manganese content of the ore may be determined by reducing the MnO_(2) under acetic conditions to Mn^(2+) with the oxalate ion, C_(2)O_(4)^(2-), the oxalate ion being oxidised to carbon dioxide during the reaction. The analytical determination is carried out by adding a known excess volume of oxalate solution to a suspension of the pyrolusite and digesting the mixture on a hot water bath until all the MnO_(2) has been reduced. The excess, unreacted oxalate solution is then titrated with standardised potassium permanganate, KMnO_(4) solution after which the manganese content of the ore can be calculated. A student prepared a standard solution of sodium oxalate by weighing 3.2 g of the dry anhydrous salt, dissolving it in distilled water and making the solution up to 500 mL. 25 mL of the oxalate solution required 24.76 mL of KMnO_(4) solution. How many moles of C_(2)O_(4)^(2-) ions will be oxidised by 1 mole MnO_(4)^(-) ?

Answer»

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

SOLUTION :N//A
Discussion
25.

Pyrolusite, MnO_(2), is the main ore from which manganese is produced. The manganese content of the ore may be determined by reducing the MnO_(2) under acetic conditions to Mn^(2+) with the oxalate ion, C_(2)O_(4)^(2-), the oxalate ion being oxidised to carbon dioxide during the reaction. The analytical determination is carried out by adding a known excess volume of oxalate solution to a suspension of the pyrolusite and digesting the mixture on a hot water bath until all the MnO_(2) has been reduced. The excess, unreacted oxalate solution is then titrated with standardised potassium permanganate, KMnO_(4) solution after which the manganese content of the ore can be calculated. A student prepared a standard solution of sodium oxalate by weighing 3.2 g of the dry anhydrous salt, dissolving it in distilled water and making the solution up to 500 mL. 25 mL of the oxalate solution required 24.76 mL of KMnO_(4) solution. Molarity of the sodium oxalate solution is ..

Answer»

0.04776
0.07446
0.06447
0.07644

Solution :N//A
Discussion
26.

Pyrolusite, MnO_(2), is the main ore from which manganese is produced. The manganese content of the ore may be determined by reducing the MnO_(2) under acetic conditions to Mn^(2+) with the oxalate ion, C_(2)O_(4)^(2-), the oxalate ion being oxidised to carbon dioxide during the reaction. The analytical determination is carried out by adding a known excess volume of oxalate solution to a suspension of the pyrolusite and digesting the mixture on a hot water bath until all the MnO_(2) has been reduced. The excess, unreacted oxalate solution is then titrated with standardised potassium permanganate, KMnO_(4) solution after which the manganese content of the ore can be calculated. A student prepared a standard solution of sodium oxalate by weighing 3.2 g of the dry anhydrous salt, dissolving it in distilled water and making the solution up to 500 mL. 25 mL of the oxalate solution required 24.76 mL of KMnO_(4) solution. What is the equivalent mass of MnO_(2) in the present titration ?

Answer»

`(M. W.)/(1)`
`(M.w.)/(2)`
`(M.w.)/(3)`
`(2M.w.)/(3)`

SOLUTION :N//A
Discussion
27.

Pyridine like benzene, has six pi -electrons in delocalized pi -orbitals, but unlike benzene the orbitals will be deformed as electrons are more attracted towards the nitrogen atom because of its high electro negativity. This is reflected in the dipole of pyridine which has the negative end on N and the positive end on the nucleus. We can compare the respective properties of benzene, pyridine, pyrrole, furan and thiophene, which very frequently give aromatic electrophilic and nucleophilic substitution reactions in different positions. Which of the following compounds are aromatic.

Answer»




Solution :
Is aromatic because planar, CONJUGATION, it follows Huckels rule(4M + 2)`PI` ELECTRONS.
Discussion
28.

Pyridine like benzene, has six pi -electrons in delocalized pi -orbitals, but unlike benzene the orbitals will be deformed as electrons are more attracted towards the nitrogen atom because of its high electro negativity. This is reflected in the dipole of pyridine which has the negative end on N and the positive end on the nucleus. We can compare the respective properties of benzene, pyridine, pyrrole, furan and thiophene, which very frequently give aromatic electrophilic and nucleophilic substitution reactions in different positions. Nitration, chlorination, bromination of pyridine is very difficult at normal conditions. Why?

Answer»

High aromaticity of pyridine than benzene
Pyridine has an LONE PAIR of e-s on NITROGEN so it protonates
pyridine has 5 CARBON atoms
pyridine is cyclic compound.

Solution :In pyridine .N. is highly EN. So it protonates and can behave as LEWIS base.
Discussion
29.

(pV^(2)T^(2))/(n) What would be the SI unit for the quantity.

Answer»

Solution :SI UNIT of `(PV^(2)T^(2))/(n)`
`= ((Pa)(m^(3))^(2)(K)^(2))/(mol)`
`= Pa XX m^(6)xx K^(2)xx mol^(-1)`
where p = Pascal Pa
V = Volume `= m^(3)`
T = Absolute Temp. K
n = mol.
Discussion
30.

Purpose of 'U' shaped perforated vessel in Nelson cell is mainly to prevent the reaction

Answer»

`2NaOH+Cl_(2)RARRNACL+NaOCl+H_(2)O`
`Na+1//2Cl_(2)rarrNaCl`
`NaOH+HClrarrNaCl+H_(2)O`

ANSWER :A
Discussion
31.

Purple of cassius is

Answer»

COLLIDAL SOLUTION of GOLD
COLLOIDAL solution of silver
colloidal solution of platinum
oxyacids of gold.

Solution :It is a colloidal solution of gold
Discussion
32.

Purification of silicon element used in semiconductors is done by

Answer»

Zone refining
Heating
Froth floatation
Heating in VACUUM .

Solution :PURIFICATION of SI USED in semi-conductors is done by zone refining.
Discussion
33.

Purest form of hydrogen is obtained by

Answer»

PETROLEUM industry
Electrolysis of water
Reaction of METALS with acids
Steam with coke

Answer :B
Discussion
34.

Pure will can be obtained from sea water by

Answer»

centrifugation
plasmolysis
REVERSE osmosis
sedimentation

SOLUTION :PURE WATER can be OBTAINEDFORM sea water by reverse osmosis
Discussion
35.

Pure water can be obtained from water by

Answer»

cetrifugation
plasmolysis
reverse osmosis
sedimentation

Answer :C
Discussion
36.

Pure substances which show conductivity similar to that of silicon and germanium are called __________Conductors

Answer»


ANSWER :INTRINSIC
Discussion
37.

Pure sodium chloride is not hydroscopic but common salt gets wet in rainy season ? Why so ?

Answer»

Solution :PURE `NaCl` is not hygroscopic but table salt is impure NaCl containing impurities of `MgSO_(4) , CaSO_(4) , MgCl_(2) and CaCl_(2)` , all of these being hygroscopic absorb MOISTURE from air in rainy season . As a result , table salts gets wet .
Discussion
38.

Pure methane can be prepared by:

Answer»

WURTZ reaction
Kolbe ELECTROLYSIS METHOD
soda-lime de-carboxylation
REDUCTION with `H_2`

SOLUTION :Sodaline decarboxylaisation method.
Discussion
39.

Pure hydrogen is obtained by carrying electrolysis of

Answer»

water CONTAINING `H_(2)SO_(4)`
water containing NaOH
`Ba(OH)_(2)` SOLUTION
KOH solution

Solution :Pure HYDROGEN is obtained by the electrolysis of `Ba(OH)_(2)` solution in a U-tube using nickel electrode. The GAS is liberated at the cathode and is passed over heated platinum gauze to remove oxygen if present as impurity.
Discussion
40.

Pure H_(2)O_(2)is not very stable and decomposes in the presence of light or on standing or heating overset((-1))(H_(2)O_(2l))+overset((2-)) (H_(2)O_((l)))+overset((0))(O_(2(g))) The oxidation state of oxygen in H_(2)O_(2) is 1 it is oxidistion state of zero in O_(2) and is reduced to an oxidation state of -2 in H_(2)O. Thus, this decomposition is an example of auto oxidation reduction reaction or disproportionation reaction. 30-volume, hydrogen peroxide means

Answer»

30% of `H_2O_2` by VOLUME
30g of `H_2O_2` solution contains 1g of `H_2O`
`1 cm^3`of solution liberates `30 cm^3`of DIOXYGEN gas at STP
`30cm^3` of the solution contains ONE mole of `H_2O_2`

Solution :30 -volume,HYDROGEN PEROXIDE means `1 cm^3` of solution liberates `30 cm^3` of dioxygen gas at STP
Discussion
41.

Pure H_(2)O_(2)is not very stable and decomposes in the presence of light or on standing or heating overset((-1))(H_(2)O_(2l))+overset((2-)) (H_(2)O_((l)))+overset((0))(O_(2(g))) The oxidation state of oxygen in H_(2)O_(2) is 1 it is oxidistion state of zero in O_(2) and is reduced to an oxidation state of -2 in H_(2)O. Thus, this decomposition is an example of auto oxidation reduction reaction or disproportionation reaction. The compound which gives H_(2)O_(2) on treatment with dilute acid is

Answer»

`PbO_(2)`
`MnO_(4)`
`TiO_(4)`
`KO_(2)`

SOLUTION :The compound wich GIVES `H_(2)O_(2)` on tratement with acid is `KO_(2)`
Discussion
42.

Pure H_(2)O_(2)is not very stable and decomposes in the presence of light or on standing or heating overset((-1))(H_(2)O_(2l))+overset((2-)) (H_(2)O_((l)))+overset((0))(O_(2(g))) The oxidation state of oxygen in H_(2)O_(2) is 1 it is oxidistion state of zero in O_(2) and is reduced to an oxidation state of -2 in H_(2)O. Thus, this decomposition is an example of auto oxidation reduction reaction or disproportionation reaction. Pure H_(2)O is

Answer»

Semi-Solid
Liquid
Solid
Gas

Solution :Pure `H_(2)O_(2)` is thick SYRUPY pale BLUE liquid
Discussion
43.

Pure benzne freezes at 5.3^(@)C. A solution of 0.223g of phenyl acetic acid (C_(6)H_(5)CH_(2)COOH) in 4.4g of benzene (K_(f)=5.12 "K kg mol"^(-1)) freezes at 4.47^(@)C. From the observation one can conclude that :

Answer»

phenyl ACETIC ACID exists as such in benzene
phenyl acetic acid UNDERGOES partial ionization in benzene
phenyl acetic acid undergoes complete ionization in benzene
phenyl acetic acid dimerizes in benzene

Answer :D
Discussion
44.

Pure ammonia is placed in a vessel at a temperature where its degree of dissociation (alpha) is appreciable. At equilibrium 2NH_(3) = N_(2) + 3H_(2) .

Answer»

`K_p` does not change with PRESSURE
`ALPHA` changes with pressure
concentration of `NH_3` does not change with pressure
concentration of `H_2` is more than that of `N_2`

Solution :Le-chatelier.s principle
Discussion
45.

Psi^(2) = 0 represents

Answer»

a node
an ORBITAL
angular wave function
wave function

Solution :`Psi^(2) = 0` MEANS probability of finding the electron in the orbital is zero, i.e., it represents a node
Discussion
46.

Psi^2 (r , theta , phi) represents (for schrodinger wave mechanical model)

Answer»

Amplitude of electron wave
Probability density of electron
product with `4pi r^2dr` GIVES TOTAL probability density of electron
orbit

Solution :`Psi^2` is MEASURE of probability density of `e^-` .
Discussion
47.

Prussian blue colour is obtained by mixing together aqueous solution of Fe^(3+) salt with:

Answer»

ferricyanide
ferrocyanide
hydrogen cyanide
sodium cyanide

Answer :B
Discussion
48.

Prove the relationship between C_(p) and C_(p) for an ideal gas.

Answer»

Solution :At constant volume, the heat capacity, C is denoted by `C_v` and at constant pressure, this is denoted by `C_p`.
At constant volume as `q_(V) = DELTA U = C_(v) Delta T`
At constant pressure as `q_(p) = Delta H = C_(p) Delta T`
The DIFFERENCE between `C_(p) and C_(v)` can be derived for an ideal gas as :
For a mole of an ideal gas,
`Delta H = Delta U + Delta (pV)`
`= Delta U + Delta (RT)`
`= Delta U+ R Delta T`
`therefore Delta H = Delta U+ R Delta T ""...(1)`
bn putting the values of `Delta H and Delta U` we have
`C_(p) Delta T = C_(v) Delta T + R Delta T`
`C_(p) = C_(v) + R`
`C_(p) - C_(v) = R`
Discussion
49.

Prove that the pressure necessary to obtain 50% dissociation of PCl_(5)at 500 K is numerically equal to three times the value of the equilibrium constant, K_(p).

Answer»

Solution :` {: (,PCl_(5),hArr,PCl_(3),+,Cl_(2),) , (" Intial moles",1,,0,,0,) , ("Moles at EQM".,1-0*5=0*5,,0*5,,0*5,"TOTAL " 1*5 "moles") :}`
IfP isthe total requiredpressure, then
` p_(PCl_(5)) = (0*5)/(1*5) XX P=P/3, p_(PCl_(3)) = (0*5)/(1*5) xx P = P/3, p_(Cl_(2)) = (0*5)/(1*5) xx P= P/3`
` K_(p) = (p_(pCl_(3)) xx p_(cl_(2)))/(p_(PCl_(5)))= ((P//3)(P//3))/(P//3) = P/3 or P = 3 K_(P).`
Discussion
50.

Prove that valency is a periodic property.

Answer»

SOLUTION :Variation in period: The number of valence electrons INCREASES from 1 to 8 on moving across a period. The valency of the elements with respect to hydrogen and CHLORINE increases from 1 to 4 and then decreases from 4 to zero.
Variation in group: On moving down a group, the number of valence electrons remains same. All the elements in a group exhibit same valency. For example, all the elements of group 1 have valency equal to 1.
Hence, valency is a periodic property.
Discussion