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.

Which of the following decreasing order of acid strength of (i) Methanoic acid (ii) Ethanoic acid (iii) Propanoic acid (iv) Butanoic acid is correct ?

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`(i) gt (ii) gt (III) gt (IV)`
`(iv) gt (iii) gt (ii) gt (i)`
`(i) gt (iv) gt (iii) gt (ii)`
`(iv) gt (i) gt (ii) gt (iii)`

Solution :With INCREASE in the `+I` effect of alkyl group ACIDIC NATURE decreases
2.

The values of Delta H and Delta S for the reaction, C_("(graphite)") + CO_(2(g)) to 2CO_((g)) are 170kJ and 170 JK^(-1), respectively. This reaction will be spontaneous at

Answer»

910 K
1110 K
510 K
710 K

Solution :`Delta G = Delta H - T Delta S`
At equilibrium, `Delta G = 0`
`rArr 0 = (170 xx 10^(3) J ) - T(170 "JK"^(-1) )`
`rArr T= 1000K`
For SPONTANEITY, `Delta G ` is -ve, which is possible only if `T gt 1000` K.
3.

The solubilityof alkali metal hydroxide follows the order :

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LiOH `lt` NAOH `lt` KOH `lt` RbOH `lt` CsOH
LiOH `GT` NaOH `gt` KOH `gt` RbOH `gt` CsOH
LiOH `gt` CsOH `gt` RbOH `gt` NaOH `gt` KOH
LiCl `lt` RBCL `lt` KCl `lt ` NaCl

ANSWER :a
4.

X forms an oxide X_(2)O_(3) 0.36 grams of X forms 0.56 grams of X_(2)O_(3). So the atomic weight of X is

Answer»

36
565
28
43.2

Answer :D
5.

Which of the following do not give rearrangement of carbonation in the addition reaction of alkene?

Answer»

`Br_2// C Cl_4`
`HBr`
`HBr_2// H_2O_2,HV`
`OSO_4`

Solution :1. Reaction occur by cyclic bromonium ion by `Br_2`
2. No reaggrangement by FREE radical
3. Cyclic TRANSITION state by `OSO_4`
6.

The wavelength of the wave function of a bonding molecular orbital formed by LCAO is

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EQUAL to the wave function of ATOMIC ORBITAL
LESS than the wave function of atomic orbital
Greater than the wave function of atomic orbital
Double the wave function of atomic orbital

Answer :B
7.

Which of the following is the correct order of solubility in water ?

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`BaSO_(4) GT SrSO_(4) gt CaSO_(4) gt MgSO_(4)`
`BaSO_(4) gt MgSO_(4) gt SrSO_(4) gt CaSO_(4)`
`MgSO_(4) gt CaSO_(4) gt SrSO_(4) gt BaSO_(4)`
`MgSO_(4) gt BaSO_(4) gt CaSO_(4) gt SrSO_(4)`

Answer :C
8.

Which of the following is used as purgative in medicine.

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`MgSO_(4).7H_(2)O`
`CaSO_(4).7H_(2)O`
`ZnSO_(4).2H_(2)O`
`MnSO_(4)`

Solution :`MgSO_(4).7H_(2)O` is USED as PURGATIVE in medicine.
9.

Which of the following pairs are not position isomers ?

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Solution :Position isomers have same structure of the CARBON CHAIN but DIFFER only in the position of the functional group or the heteroatom.
Here, in OPTION (a) position of OXYGEN atom is the same but the structure of the carbon chain is different. Hence, option (a) correct.
10.

The vapour pressure of water at 80^(@)C is 355 torr A 100mL vessel contained water saturated oxygen at 80^(@)C the total gas prssure being 760 torr The contnets of the vessel were pumped into a 50.0mL vessel at the same temperature What were the partial prssures of oxygen and water vapour waht was the total pressures of oxygen and water vapour waht was the total pressure in the first equilibarated state Neglectvolume of any water which might condense .

Answer»


ANSWER :810 TORR, 355 torr 1165 torr
11.

Which of the following compounds is most stable ?

Answer»

LIF
LiCl
LiBr
LiI

Solution :Because of the SMALL size of LI and F , LiF has HIGHEST lattice enthalpy and hence most STABLE .
12.

Which of the followingis /arerecirculated in solvay process

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`CaCl_(2)`
`NH_(3)`
`CO_(2)`
Both (b) and (C)

SOLUTION :`NH_(3)` & `CO_(2)`are recirculated in solvay process
13.

Which pair does represent isomers?

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`CH_(3)COOH` and `HCOOCH_(3)`
`CH_(3)-CHO` and `CH_(2)=CH-OH`
`CH_(3)-CHO` and `CH_(3)-CO-CH_(3)`
`CH_(3)-CO-CH_(3)` and `CH_(3)-CH_(2)-CHO`

Solution :In C Molecular formularare different
14.

The relation between pressure exerted by an ideal gas (p_("ideal")) and observed pressure (p_("real")) is given by the equationP_("ideal")=P_("real")+(an^(2))/(V^(2))(i) If pressure is taken in Nm^(-2), number of moles in mol and volume in m^(3), Calculate the unit of .a..(ii) What will be the unit of .a. when pressure is in atmosphere and volume in dm^(3) ?

Answer»

Solution :Given that, `P_("ideal")=P_("real")+(an^(2))/(V^(2))`
(i) a = If units of `p = NM^(2)`,
units of `V=m^(3)`, units of n = mol
then, units of `a =(Nm^(-2)(m^(3))^(2))/((mol)^(2))=Nm^(4)mol^(-2)`
(II) If units of p = atm, units of `V = DM^(3)`, units of n = mol
then, units of `a = (PV^(2))/(n^(2))=(atm.(dm^(3))^(2))/((mol)^(2))`
= atm `dm^(6)mol^(-2)`
15.

Which of the following is correct regarding the hydrolysis of BX_(3) (X-hylogen)?

Answer»

All `BX_(3)` undergo HYDROLYSIS to produce
`B(OH)_(3)` (aq) and `HX`(aq)
`BF_(3)` does not undergo COMPLETE hydrolysis due to formation of `HBF_(4)`
`BBr_(3)` does not undergo hydrolysis hydrolysis due to formation of `HBF_(4)`
All the above are correct

Answer :B
16.

What gaseous species are present in the mesosphere and thermosphere ?

Answer»

Solution :Gaseous IONS like `NO^(+), O_2^(+), N_2^(+) , O^(+)` and atoms of N and O.
17.

Which of these are biodgradable pollutants ?

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(1) and (ii)
(1) and (III)
(i),(ii)and (IV)
(iii) only

Solution :Only sewage is biodegradable .
18.

What is pK_a, pK_b and pK_w ? Give its relation.

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SOLUTION :`pK_a =-LOG K_a`
`pK_b=-log K_b`
`pK_w=-log K_w`
`pK_w=pK_a + pK_b`=14.0
19.

The shape and hybridisation of some xenon oxyfluorides are given. Choose the wrong set.

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`XeOF_2 to` T-SHAPE `sp^3d`
`XeOF_4 to` Square PYRAMIDAL `sp^3d^2`
`XeO_2F_2 to` Distorted trigonal bipyramidal - `sp^3d`
`XeO_3 F_2 to` OCTAHEDRAL- `sp^3d`

Solution :The structure of `XeO_3F_2` -
No. of lone pair of Xe= 0
No. of BOND pair = 5
Hybridisationof Xe= `sp^3d`
Hence , shape of `XeO_3F_2` should be trigonal bipyramidal and not octahedral .
20.

The standard free energy change, Delta G^(@) is related to equilibrium constant K_(p) as

Answer»

`K_(p) = - RT LN Delta G^(@)`
`K= ((e)/( RT))^(DeltaG^(@))`
`K_(p) = (DeltaG^(@))/( RT)`
`K_(p) = e^(- DeltaG^(@) //RT)`

Solution :`DeltaG^(@)= - RT lnK_(p) ` or `ln K_(p)=- DeltaG^(@) // RT ` or `K_(p) = e^(-DeltaG^(@) //RT )`
21.

What is used to express derived unit obtained from SI unit for smaller or larger quantities ?

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MULTIPLE UNIT
Prefix
Proper power
None of these

Answer :B
22.

What is the strength of 20 volume solution of H_(2)O?

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5g/L
10g/L
30g/L
60g/L

Answer :D
23.

Which can be directly converted into solid state from gaseous state

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CO
`CO_(2)`
`PH_(3)`
`CO+H_(2)`

Solution :`CO_(2)` dry ICE directly goes to GASEOUS STATE (Sublimation)
24.

Which of the following element is not consider as group - II-A ?

Answer»

MG
SR
RA
FR

ANSWER :D
25.

What will be the conjugatebases for the following Bronsted acids ? HF, H_(2)SO_(4) and HCO_(3)^(-)

Answer»

Solution :Conjugate acid `HARR` Conjugate base + `H^(+)`
or Conjugate base = Conjugate acid - `H^(+)`
`:.` Conjugate bases of the given ACIDS will be `F^(-), HSO_(4)^(-), CO_(3)^(2-)`.
26.

Which of the following compounds can be purified by steam distillation

Answer»

SALICYALDEHYDE
BROMOBENZENE
p-hydroxybenzaldehyde
ANILINE

SOLUTION :(A) salicyaldehyde
(D) Aniline
27.

What would happen, if the greenhouse gases were totally missing in the earth's atmosphere?

Answer»

Solution :(i) The primary greenhouse gases in Earth.s atmosphere are water VAPOUR, carbon dioxide, methane, nitrous oxide and ozone
(ii) Naturally occurring greenhouse gases allow solar radiations to reach the earth.s SURFACE. while trapping radiations from the earth on its way back out to space.
(iii) There would be no LIFE on Earth without the warmth PROVIDED by this natural greenhouse gases.
(iv) In the absence of greenhouse gases, the average temperature of the earth will decrease drastically. As a result, life on Eárth would be impossible. .
28.

Which one is the correct order of acidity ?

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`CH-=CH gt CH_3 -C -=CH gt CH_2 =CH_2 gt CH_3 -CH_3`
`CH-=CH gt CH_2 =CH_2 gt CH_3 -C -=CH gt CH_2- CH_3`
`CH_3-CH_3 gt CH_2 =CH_2 gt CH_3 -C -= CH gt CH-=CH`
`CH_2=CH_2 gt CH_3 -CH=CH_2`

Solution :`+I`-Effect of `CH_3` group increases the electron-density in the C-H bond thereby DECREASING its tendency to release a proton. THUS, PROPYNE is LESS acidic than ethyne. Thus, the acidity of ethane, ethene , ethyne and propyne DECREASES in the order : option (a) is correct.
29.

Which of the following does contain the maximum number of gram molecules ? (i) 10 g of H_(2), (ii) 34 of NH_(3) (iii) 44g of CO_(2), (iv) 100 g of H_(2)SO_(4)

Answer»


ANSWER :10 G of `H_(2)`
30.

Which of the followinglaws of chemicalcombination differentiarte a comparedfrom a mixture?

Answer»

LAW of MULTIPLE proportions
Law of definite proportions
Law of reciprocal proportios
Law of COMBINING volumes

Solution :Unlike a mixturre a compoundhas a fixed COMPOSITION by mass.
31.

Which of the following method is used to prevent rusting of iron?

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Galvanisation
Painting
Chrome plating
all the above

Answer :D
32.

The specific test for H_2O_2is

Answer»

It turns starach iodide paper to blue colour
It WHITENS blackened LEAD PAINTINGS
It turns `Ti(SO_4)_2` solution to orange colour
It DECOLOURISES acidified `KMnO_4` solution

Solution :Only `H_2O_2` turns `Ti(SO_4)_2` solution to orange colour.
33.

Why are the melting and boiling points of heavy water more than those of ordinary water?

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Solution :The molecular masss of heavy WATER `(D_(2)0)` is more than that of ordinary water `(H_(2)O)`. As a result there is a grater DEGREE of association of molecules in heavy water than in ordinary water. This results in greater melting and BOILING POINTS as compared to ordinary water.
34.

Which of the followingare amphoteric oxides ?

Answer»

`SnO_(2)`
`Al_(2)O_(3)`
`Ga_(2)O_(3)`
`SiO_(2)`

SOLUTION :Only `SiO_(2)` is ACIDIC while all other are AMPHOTERIC.
35.

Which of these represents the correct order of their increasing bond order.

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`C_(2)ltC_(2)^(2-)ltO_(2)^(2-)ltO_(2)`
`C_(2)^(2-)ltC_(2)^(+)ltO_(2)ltO_(2)^(2-)`
`O_(2)^(2-)ltO_(2)ltC_(2)^(2-)ltC_(2)^(+)`
`O_(2)^(2-)LT C_(2)^(+)ltO_(2)lt C_(2)^(2-)`

SOLUTION :bond order `= (1)/(2)(n_(b)- n_(a))`bond order of `O_(2)^(2-)=(1)/(2)(8 - 6)=1`
bond order of `C_(2)^(+)=(1)/(2)(5 - 2)=1.5`bond order of `O_(2)=(1)/(2)(8 - 4) = 2`
bond order of `C_(2)^(2-)= (1)/(2)(8-2)= 3`
36.

What happen when nitrile undergoes acid hydrlysis?

Answer»

Solution :When alkyl NITRILE undergoes acid HYDROLYSIS to give amide, which on further hydrolysis to give carboxylic acid.
`underset("nitrile") (R - CN)overset(H^(+)|H_(2)O) to underset("Amide") (R -CONH_(2)) overset(H^(+)|H_(2)O) tounderset("carboxylic acid")(R-COOH)`
37.

Which of the following pairs contain equal number of atoms?

Answer»

22.4 L (STP) of nitrous oxide and 22.4 L of nitrifc oxide
1 millimole of HCl and 0.5 millimole of `H_(2)S`
1 MOLE of `H_(2)O_(2)` and 1 mole of `N_(2)O_(4)`
11.2 cc (STP) of nitrogen and 0.015g of NITRIC oxide

Solution :d) Number of atoms in `N_(2)`
`=(11.2 XX 10^(-3) xx 6.023 xx 10^(23) x 2)/(22.4) = (6.023 xx 10^(20)`
Number of atoms in NO = `(0.015 xx 2 xx 6.023 xx 10^(23))/(30)`
`=6.023 xx 10^(20)`
38.

What are the major differences between metals and non-metals?

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Answer :METALS have low IP with strong ELECTROPOSITIVE character. Non metals have high EN and are oxidizing in nature.
39.

Which one among the following gives a dicarbonyl compound with O_3 followed by reduction withzinc and water? a) C_2H_4b) C_6H_6 c) C_2H_2d) C_3H_6

Answer»

a and C
a,B and c
a,b and d
b and c only

SOLUTION :
40.

Thesolubility of AgCl (s) with solubility product 1.6xx10^(-10) in 0.1 MNaCl solution would be

Answer»

`1.26xx10^(-5) M`
`1.6xx10^(-9) M`
`1.6xx10^(-11)M`
zero

Solution :`{:(AgCl(s),OVERSET(+aq)hArr ,AG^(+)(aq),+,CL^(-)(aq)),(S,,S,,S+0.M"(from NaCl)"),(,,,,~=0.1M (" as " S lt lt 0.1)):}`
`K_(sp)=[Ag^(+)][Cl^(-)]`
`1.6xx10^(-10)=S(0.1) "or" S=1.6xx10^(-9)M`
41.

When 2g of a gas A is introduced into an evacuated flask kept at 25°C, the pressure is found to be 1 atmosphere. If 3g of another gas is then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal behaviour, the ratio of their molecular weights M_(A): M_(B) is

Answer»

`3:1` 
`1:3` 
`2:3` 
`3:2`

SOLUTION :`(P_A)/(P_B) = (W_A)/(M_A) XX (M_B)/(W_B) = 1/3`.
42.

The wave lengths of first member of Balmer series of a hydrogen atom is nearly (The value Rydberg constant is 1.08 xx 10^(7)m^(-1))

Answer»

`4400 A^@`
`5500 A^@`
`6600 A^@`
`7700 A^@`

Solution :`1/(lambda)= barv = R ( 1/(n_1^2) - 1/(n_2^2)) n_1 , n_2 = 3`
43.

To a 25 mL H_(2)O_(2) solution, excess of acidified solution of potassium iodide was added. The iodine liberated required 20 mL of 0.3 N solution thiosulphate solution. Calculate the volume strength of H_(2)O_(2) solution.

Answer»

Solution :Step 1. To determine the normality of `H_(2)O_(2)` solution.
Let the normality of the `H_(2)O_(2)` solution be `N_(1)` . According to the question.
25 mL of `H_(1)H_(2)O_(2)-= 20 mL` of 0.3 N `Na_(2)S_(2)O_(3)` solution.
or `25xxN_(1)=20xx0.3` or `N_(1)=(20xx0.3)/(25)=0.24N`
Thus, the normality of the given `H_(2)O_(2)` solution=0.24 N
Step 2. To determine the amount of `H_(2)O_(20`in 25 mL solution
1000 mL of 1 `H_(2)O_(2)` solution CONTAIN `H_(2)O_(2)` =17 g `""` (`because ` Eq. wt. of `H_(2)O_(2)` =17)
`therefore ` 25 mL of 0.24 `H N_(2)O_(2)` solution will contain `H_(2)O_(2)=(17xx25xx24)/(1000xx100)=0.102 g`
Step 3. To determine the volume strength of `H_(2)O_(2)` solution.
Consider the chemical equation `underset(2xx34=68 g)(2H_(2)O_(2))to 2H_(2)O + underset(22.4 litres at N.T.P.)(O_(2))`
68 g of `H_(2)O_(2)` give `O_(2)` =22.4 litres at N.T.P.
`therefore 0.102 g H_(2)O_(2)` will give `O_(2)=(22.4xx1000xx0.102)/(68)=33.6` mL at N.T.P.
Now, 25 mL of `H_(2)O` solution give `O_(2)=33.6 mL` at N.T.P.
`therefore ` 1 mL of `H_(2)O_(2)` solution will give `O_(2)=(33.6)/(25)=1.344`
Thus, the volume strength of the given `H_(2)O_(2)` solution =1.344
44.

What is the major product of the following reaction?

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SOLUTION :
45.

Two containers A and B of equal volume contain 6 g of O_(2) and SO_(2) at 300K. and 1 atm. Then

Answer»

No. of MOLECULES in A is less than that in B
No. of molecules in A is more than that in B
No. of molecules in A and B are same
none of these

Solution :`O_(2)` =Mass=6G
Molar mass=32g
32 g of `O_(2)` contains `6.023xx10^(23)` molecules
`:.` 6g of `O_(2)` will contains=`(6.023xx10^(23))/32xx6=1.129xx10^(23)` molecules.
`SO_(2)`=Mass=6g
Molar mass =64g
64 g of `SO_(2)` contains `6.023xx10^(23)` molecules.
`:.` 6g of `SO_(2)` will contains =`(6.023xx10^(23))/64xx6=0.5646xx10^(23)` molecules
`:.` Number of molecules in A is more than that in B.
46.

Why alcoholic KOH is prefered than alcoholic NaOH in organic chemistry ?

Answer»

Solution :KOH Is much more soluble in alcohol and produces `C_(2)H_(5)O^(-)`ions by the FOLLOWING EQUILIBRIUM.
`C_(2)H_(5)OH+OH^(-)hArrC_(2)H_(5)O^(-)+H_(2)O`
This accounts for the USE of alcoholic KOH in organic chemistry.
47.

Write the formula of relation between equilibrium constant K and DeltaG^(@).

Answer»

Solution :`Delta_(R) G^( @) = -2.303 RT LOG K`
48.

What isscattering ?By whichpropertyelectromageticradiationit can beexplain ?

Answer»

SOLUTION :Derivationin wavedue toobstaclecalledscatteringit isbecauseof wavenatureradiation.
49.

Which of the following ions is smallest in size?

Answer»

`CL^(-)`
`NA^(+)`
`MG^(2+)`
`S^(2-)`

ANSWER :C
50.

Two samples of water A and B having concentration 16.2 ppm of Ca(HCO_(3))_(2) and 13.6 pp m of CaCl_(2) respectively. Then

Answer»

Hardness is more in 'B' sample than in 'A'
Hardness is more in 'A' sample than in 'B'
Hardness is same in both cases
Hardness in sample 'B' is TWICE that of the sample 'A'

Solution :A = 16.2 g of `CaHCO_3`
B = 13.6 g of `CaCl_2`
hardness of `A = 16.2 XX 100/162 = 10` PPM
hardnessof `B = 13.6 xx 100/111=12.25` ppm