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.

If the bond dissociation energies of XY,X_(2) and Y_(2) (all diatomic molecules) are in the ratio of 1:1:0.5 and DeltaH_(f) for the formation of XY is -200 kJ mol^(-1). The bond dissociation energy of X_(2) will be

Answer»

400 kJ `MOL^(-1)`
300 kJ `mol^(-1)`
20 kJ `mol^(-1)`
none of these

Solution :Formation of XY is known as
`X_(2)+Y_(2) to 2XY`
`DeltaH=(BE)_(X-X)+(BE)_(Y-Y)-2(BE)_(X-Y)` ,brgt If `(BE)` of `X-Y=a`
then (B E) of (X-X)=a
and (B E) of (Y-Y)=`(a)/(2)`
`therefore DeltaH_(f)(X-Y)=-200KJ`
`therefore-400` (for 2 mol XY)`=a+(a)/(2)-2a`
`-400=-(a)/(2)impliesa=+800kJ`
The bond DISSOCIATION ENERGY of `X_(2)=800" kJ "mol^(-1)`
2.

Impure copper containing Fe, Au and Ag as impurities is electrolytically refined. A current of 140 A for 482.5 s decreased the mass of the anode by 22.26 g and increased the mass of the cathode by 22.011g. Calculate the percentage of iron in impure copper. (Given molar mass of Fe=55.5g mol^(-1), molar mass of Cu=63.54 g mol^(-1)).

Answer»

Solution :In the purification of copper, impure copper is anode and pure copper is cathode. Increase in the mass of cathode is due to Cu deposited on it. Decrease in the mass of anode is due to `CutoCu^(2+)`(deposited on cathode), `FetoFe^(2+)` (remain dissolved int he solution) (Fe being more reactive than Cu), AG and Au (being less reactive than Cu do not undergo oxidation) and fall below anode as anode mud.
Cu deposited on cathode (Pure Cu) can be calculated as follows:
`Cu^(2+)+2e^(-)toCu`
Quantity of ELECTRICITY passed=`140xx482.5`coulombs=67550C ltbr `2xx96500C` deposit Cu=63.54g
`therefore67550C` will deposit Cu`=(63.54)/(2xx96500)xx67550g=22.239g`
Actual increase in the mass of cathode=22.011g
Cu not deposited due to Fe passing into the solution`=22.239-22011=0.228g`
Equivalent AMOUNT of Fe that passed into the solution`=(55.5)/(63.54)xx0.228g=0.199g`
Thus, Fe PRESENT as impurity=0.119g
Decrease in the mass of anode=Total impure `Cu(Cu+Fe+Au+Ag)=22.26g`
`therefore%` of Fe in impure `Cu=(0.199)/(22.26)xx100=0.89cong0.90`
3.

If the bond dissociation energies of XY, X_(2) and Y_(2) (all diatomic molecules) are in the ratio of 1 : 1 : 0.5 and Delta_(f) H for the formation of XY is - 200 kJ " mole"^(-1). The bond dissociation energy of X_(2) will be

Answer»

100 KJ `mol^(-1)`
800 kJ `mol^(-1)`
300 kJ `mol^(-1)`
400 kJ `mol^(-1)`

SOLUTION :`XYrarrX_((g))+Y_((g)), DeltaH=+"a kJ/mole"^(-1)".........(i)"`
`X_(2)rarr2X, DeltaH=+" a kJ/mole.....(ii)"`
`Y_(2)rarr2Y, DeltaH=+0.5" a kJ/mole.......(III)"`
`(1)/(2)XX(ii)+(1)/(2)xx(iii)-(i)`, gives
`(1)/(2)X_(2)+(1)/(2)Y_(2)rarrXY,`
`DeltaH=(+(a)/(2)+(0.5)/(2)a-a)"kJ/mole"`
`+(a)/(2)+(0.5a)/(2)-a=-200`
a = 800
4.

If the bond dissociation energies of XY, X_(2) and Y_(2) (all diatomic molecules) are in the ratio of 1 : 1 : 0.5 and Delta_(f)H for the formation of XY is –200 kJ mol^(-1). The bond dissociation energy of X_(2) will be

Answer»

200 kJ `MOL^(-1)`
100 kJ `mol^(-1)`
800 kJ `mol^(-1)`
300 kJ `mol^(-1)`

Answer :C
5.

Impure coppercontaining Fe,Au ,Ag as impuritiesis elecrolyticallyrefined A currentof 140 A for 482.25 s decreasedthe massof theanodeby 22.26 g and increased the mass of cathode by 22.011 g percentageof iron in impure copper is (Given molar mass of Fe =55.5 mol^(-1) molar mass of Cu =63.54 g mol^(-1))

Answer»

0.95
0.85
0.97
0.9

Solution :AMOUNT pf impurity
=decreased mass of anode - increased mass of cathode
amountof pure Cu DEPOSITED
`W=Zit =(C )/(96500)xxit`
`=(63.54)/(2xx96500)xx140 xx482.5`=22.239 g
`therefore`amountof impurity (Fe) = 22.239 -22.011 = 0.228 g
Now from FARADAY SECOND law of electrolysis
`("weightof fe deposited")/(0.228) =(27.55)/(31.77)`
`therefore` Weight of Fe `=(27.75 xx0.288)/(22.26) xx100=0.88 =0.09`
6.

If the boiling point of ethanol (molecular weight = 46) is 78^(@)C, the boiling point of diethyl ether (molecular weight = 74) is

Answer»

`100^(@)`C
`78^(@)`C
`86^(@)`C
`34^(@)`C

Answer :D
7.

Impure copper containing Fe, Au, Ag as impurities is electrolytically refined. A current of 140 A for482.5 s decreased the mass of the anode by 22.26 g and increased the mass of cathode by 22.011 g. Percentage of iron in impure copper is (Given molar mass Fe= 55.5 g "mol"^(-1), molar mass Cu = 63.54 g "mol"^(-1))

Answer»

0.95
0.85
0.97
`0.90`

Solution :`Cu^(2+)+2e^(-)RARR Cu`,
EQUIVALENT wt. `=(63.54)/(2)=31.77`
`Fe^(2+)+2e^(-)rarr Fe`, equivalent wt.`=(55.5)/(2)=27.75`
Mass inceased at cathode is due to deposition of Cu.
Hence, no. of gram equivalents of Cu deposited `=(22.011)/(31.77)=0.6928`
Now, using Faraday's FIRST law,
`Q=it = 140 A xx 482.2 s = 67550C`
`96,500C-31.77 g Cu`
`67550 C - ?`
`(31.77)/(96, 500)xx(67550)/(1)=22.239` g of pure Cu deposited
But, according to the question, the mass of cathode only increases `= 22.011g`
Hence, `22.239-22.011=0.228g`
Now, `31.77g Cu -= 0.228` g deposited
`27.75 g Fe -= (0.228)/(31.77)xx27.75=0.199` g Fe deposited
% of Fe `= ("Mass of Fe")/("Mass of impurities (at anode)")xx100`
`= (0.199)/(22.26)xx100=0.894~~0.90%`
8.

if the atoms touching one of the reactangular planes of symmetry and one of the 4-fold axis of symmetry are removed , then formula of compound is ______

Answer»

Solution :Here, two cases are possible . Rectangular plane and 4-fold AXIS are
(i)Either coinciding each other
(ii)or PERPENDICULAR to each other . Hence, two FORMULAE are possible

(i)Contribution of A =`8xx1/8=1`
Contribution of B =`(6-4)xx1/2`=1
Contribution of C =1-1=0
Contribution of D =`(12-4)xx1/4`=2
Formula = `ABD_2`
(ii)Contribution of A=`8xx1/8`=1
Contribution of B =`(6-6)xx1/2`=0
Contribution of C=1-1=0
Contribution of D =`(12-4)xx1/4=2`
Formula =`AD_2`
9.

If the aufbau principle had not been followed, Ca (Z=20) would have been placed in the:

Answer»

s-block
p-block
d-block
f-block

Answer :C
10.

Impossible orbital among the following is

Answer»

2s
3F
2p
4d

Solution :3f ORBITAL is IMPOSSIBLE because for n=3,l=0 can BOT be 3.
11.

If the atomic mass of M is x , the electrochemical equivalent of M in the solution of M_(2)(SO_(4))_(3) will be

Answer»

`(3x)/(F)`
`(X)/(3F)`
`(2X)/(F)`
`(x)/(F)`

Solution :Valency of M in `M_(2) (SO_(4))_(3) ` is 3
Eq. mass of M = `("ATOMIC mass")/("valency") = (x)/(3)`
`THEREFORE` ELECTROCHEMICAL equivalent = `("Equivalent mass")/(96500) = (E)/(F) = (x)/(3F)`
12.

Important ore of Mg is:

Answer»

Dolomite
Sylvine
Amblygonite
Triphylite

Answer :A
13.

If the atomic number of an element is 33, it will be placed in Mendeleev's periodic table in the :

Answer»

FIRST group
third group
fifth group
seventh group

Answer :C
14.

If the anions (X) form hexagonal closed packing and cations (Y) occupy only 3/8th of octahedral voids in it, then the general formula of the compound is

Answer»

`XY`
`YX_2`
`X_8 Y_3`
`X_3 Y_4`

ANSWER :C
15.

Important ore of boron is ………..

Answer»

bauxite
borosilicate
borax
`BETA`- tetragonal

Answer :C
16.

Ifthe anthalpy of vaporization for water is 186.5 kJ mol^(-1), the entropy of its vaporization will be

Answer»

0.5 `JK^(-1)mol^(-1)`
1.0 `JK^(-1)mol^(-1)`
1.5 `JK^(-1)mol^(-1)`
2.0 `JK^(-1)mol^(-1)`

SOLUTION :`DeltaS_("vap")=186.5//373=0.5 JK^(-1) mol^(-1)`
17.

If the amount of radioactive substance is increased three times, the number of atoms disintegrated per unit time would

Answer»

Be double
Be triple
Remain ONE third
Not change

Solution :`R = LAMDA N`
18.

Important iron ore is iron pyrite

Answer»

SOLUTION :HAEMATITE
19.

If the amount of radioactive substance is increased three times, the number of atoms distintegrated per gram per unit time would be

Answer»

doubel
remain one third
TRIPLE
not changed

Solution :The no. of atoms disintegrating per GRAM per UNIT TIME does not CHANGE.
20.

If the amino group of Glycine and carboxylic acid group of Alanine undergo elimination of wter molecule, the name of the compound thus formed is

Answer»

Alanylglycine(DIPEPTIDE)
Glycyl ALANINE(TRI PEPTIDE)
Glycyl alanine(dipeptide)
Alanineglycine (dipeptide)

ANSWER :A
21.

If the adsorbate is held on a surface by weak Vander Waal’s force, the adsorption process is called:

Answer»

PHYSICAL adsorption
Chemical adsorption
Heat of adsorption
Enthalpy of adsorption

Answer :A
22.

Important allotropic forms of phosphorus are white phosphorus, red phosphorus and black phosphorus. Among these which allotropic form is more reactive ? Why ?

Answer»

Solution :White phosphorus is the most reactive because of the following two reasons :
(i) White phosphorus consists of `P_(4)` molecules which are HELD together by WEAK VAN der Walls forces of attraction.
(ii) The `P_(4)` molecules have considerable ANGLE strain because in them PPP angle is just `60^(@)`.
Therefore, due to considerable angle strain and weaker forces of attraction, white phosphorus is most reactive.
23.

If the activation of forward reaction In a simple chemical reaction A to B is E_(a),then what will be activation energy for reverse reaction?

Answer»

`-E_(a)`
ALWAYS LESS than `E_(a)`
Always TWICE of `E_(a)`
More or less than `E_(a)`

ANSWER :D
24.

Important iron ore is _____.

Answer»

SOLUTION :HAEMATITE
25.

If the alignment of magnetic moments in a substance is in a compensatory way so as to give zero net magnetic moment, then the substance is said to be

Answer»

FERROMAGNETIC
Anti-ferromagnetic
FERRIMAGNETIC
DIAMAGNETIC

ANSWER :B
26.

If the activation energy is high, then the rate of the reaction is

Answer»

high
moderate
LOW
cannot be PREDICTED

Answer :C
27.

If the activation energy is decreased by (5x) % by using a catalyst (300K) to achieve the same effectas of increasing the temperature from 300K-400K in a reaction, then find x.

Answer»


SOLUTION :`r_(1)=r_(2),A.E^(-E_(A1)//RT_(1))=A.e^(-epsilon_(a_(2))//RT),((E_(a_(1)))/(T_(1)))_("cat")=((E_(a_(2)))/(T_(2)))_("No cat"),(E_(a)(100-y))/(300xx100)=(E_(a))/400,y=25%=5x%impliesx=5`
28.

Imprortant iron ore is ______ .

Answer»

SOLUTION :HAEMATITE
29.

If the activation energy for the forward reaction is 150kJ mol^(-1) and that of the reverse reaction is 260kJ mol^(-1). What is the enthalpy change for the reaction

Answer»

`410kJ mol^(-1)`
`-110 KJ mol^(-1)`
`110kJ mol^(-1)`
`-410kJ mol^(-1)`

Answer :B
30.

Important constituent of cell wall is

Answer»

LIPID
CELLULOSE
PROTEIN
vitamin

Answer :B
31.

If the activation energy for the forward reaction is 150 kJ mol^(-1) and that of the reverse reaction is 260 kJ mol^(-1) , what is the enthalpy change for the reaction

Answer»

410 kJ `mol^(-1)`
`-110kJ mol^(-1)`
110 kJ `mol^(-1)`
`-410 kJ mol^(-1)`

Solution :For a reversible reaction ,
`DELTA H = E_(a)` (forward) `- E_(a)` (backward)
`DeltaH = 150 - 260 = -110 kJ mol^(-1)`.
32.

Important allotropic forms of phosphorus are white phosphorus, red phosphorus and black phosphorus. Among these, which allotropic' form is more reactive? Why

Answer»

Solution :White phosphorus. It is less STABLE than other ALLOTROPES because of the ANGULAR strain ;in `P_4` molecules.. So it is more reactive. X
33.

If the activation energy for the forward reaction is 150 "kJ mol"^(-1)and that of the reverse reaction is 260 "kJ mol"^(-1), what is the enthalpy change for the reaction ?

Answer»

410 `"kJ mol"^(-1)`
`-110" kJ mol"^(-1)`
`110" kJ mol"^(-1)`
`-410" kJ mol"^(-1)`

SOLUTION :(b) : `DeltaH=E_(f)-E_(b)=150-260" kJ"=-110" kJ"`
34.

Imino group is present in

Answer»

`1^(0)` amins
`2^(0)` AMINE
`3^(0)` amine
quaternaryammoniumsalt

Answer :B
35.

If the absolute temperature of an ideal gas become double and pressure become half , the volume of gas would be

Answer»

Remain unchange
Will be double
Will be four times
Will be half

Solution :`P(1)=P,V_(1)=V,T_(1)=T`
`P_(2)=(P)/(2),V_(2)=? , T_(2)=2T`
According to gas EQUATION
`(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`or `(PV)/(T)=(PV_(2))/(4T)IMPLIES V_(2)=4V`
36.

Imine formation using an aldehyde/ketone and primary amine is acid - catalyzed, yet the rate drops below pH 4.5. Why does the rate drop below this pH?

Answer»

The CARBINOLAMINE intermediate is stable at LOW pH
The IMINE product is hydrolyzed at low pH
Protonation of the amine DECREASES its nucleophilicity
The amine is hydrolyzed at low pH

Answer :C
37.

If the above plot is replotted at 373 K, then which of the following plots may show the correct behaviour at 373 K ?

Answer»




ANSWER :C
38.

Imines or enamines are selectively reduced to 1^(@) or 2^(@) amines with:

Answer»

a.`NaBH_(4)`
b.`LAH`
c.`NaCNBH_(3)`
d.`NaAiH_(4)`

Solution :`NaCNBH_(3)` reduces IMINES or enamines `(RCH=NH, or RCH=NH-R^(1)) or R_(2)C=NH or R_(2)C=NHR')` to `1^(@)` or `2^(@)` amines. So the ANSWER is `(c )`.
39.

If the 0.05 molar solution of M^(+) is replaced by a 0.0025 molar M^(+) solution, then the magnitude of the cell potential would be

Answer»

35 mV
70 mV
140 mV
700 mV

Solution :From the above EQUATION,
`(2.303RT)/(F)"LOG(1)/(0.05)=70mV`
`=70xx10^(-3)V=0.07V`
`therefore(2.303RT)/(F)(1.301)=0.07`
or `(2.303RT)/(F)=(0.07)/(1.301)=0.0538`
For the NEW concentration,
`E_(CELL)=(2.303RT)/(F)"log"(1)/(0.0025)`
`=0.0538log400=0.0538xx2.6021`
`=0.140V=140mV`
40.

Imine derivatives of aldehyde and ketone is called as

Answer»

Schiff's reagent
FEHLING's reagent
Schiff's base
Schiff's acid

ANSWER :C
41.

(i)Magnetic moment of [Ag(CN)_(2)] is zero .How many unpaired electrons arepresent in this compound?(ii) 'Silver nitrate is usually supplied in brown coloured bottles. Explain.

Answer»

Solution :(i) ZERO
(II) It DECOMPOSES on exposure to LIGHT and thus stored in brown coloured bottles.
42.

If temperature remains contant during a reaction the process is called:

Answer»

Isothermal
Isochoric
Isobaric
Adiabatic

Answer :A
43.

Im Kjedahl's method, the nitrogen present in the organic compound is quantitatively converted into

Answer»

Gaseous ammonia
Ammonium sulphate
Ammonium phosphate
Ammonia

Solution :In KJELDAHL's method, the nitrogen is ESTIMATED in the form of ammonia , which is obtained by heating COMPOUNDS with `NaOH`.
`CH_(3)CONH_(3) +NaOH overset(Delta)rarr CH_(3)COONa + H_(2)O + NH_(3)`
44.

If temperature remains constant during the process. It is called an :

Answer»

ISOTHERMAL process
Adiabatic process
Isobaric process
Isochoric process

Answer :A
45.

If temperature is 10^(@)C, what is temperature in Fahrenheit?

Answer»

Solution :`.^(@)F=(9)/(5)(.^(@)C)+32=(9)/(5)xx10+32=50^(@)`
46.

Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers.

Answer»

Solution :`to` The Williamson synthesis is not a SUITABLE method for the preparation of unsymmetrical ETHERS where compound contains secondary or tertiary ALKYL groups. For example, the reaction between tert-butyl bromide and sodium methoxide yield alkenes. This is because of sodium methoxide is strong base apart from a nucleophile and as a result, the elimination dominates over substitution reaction.
`CH_(3)- underset(CH_(3))underset(|)OVERSET(CH_(3))overset(|)C- Br+ overset(+)Nabar(underset(* *)overset(* *)O)-CH_(3) to CH_(3)- underset(CH_(3))underset(|)C= CH_(2)+NaBr+ CH_(3)OH`
47.

If te solubility product of BaSO_(4) is 1.5 xx 10^(-9) in water, its solubility in moles per litre, is

Answer»

`1.5 xx 10^(-9)`
`3.9 xx 10^(-5)`
`7.5 xx 10^(-5)`
`1.5 xx 10^(-5)`

SOLUTION :`BaSO_(4) hArr BA^(2+) + SO_(4)^(--)`
Solubility constant `= S xx S`
`1.5 xx 10^(-19) = S^(2), S = sqrt(1.5 xx 10^(-19)), S = 3.9 xx 10^(-5)`.
48.

Illustrate: Wurtz fitting reaction

Answer»

SOLUTION :Haloarenes REACTS with etheral solution of an ALKYL halide in the presence of SODIUM from alkyl derivatives of benzene.
49.

Illustrate with examples - (i) Lyophilic and Lyophobic sols (ii) Homogeneous and Heterogeneous catalysis.

Answer»

SOLUTION :(i) The substances such as proteins, starch, RUBBER, etc. directly passes into the colloidal statewhen brought in contact with the solvent. Such colloids are known as lyophillic sols. The substances like metals, their sulphides, hydroxides, etc. do not form colloidal sol READILY when mixed with dispersion medium. The colloidal sols can only be PREPARED by some special methods. Such sols are called lyophobic sols.
(ii) Homogeneous catalysis: Here the reactants and catalyst are present in the same phase. For example, lead chamber process for the manufacture of `H_2SO_4` .
`2SO_(2)(g)+ O_2 (g) overset(NO_((g)))to 2SO_(3)(g)`
Heterogeneous catalysis: Here the reactants and catalyst are present in different PHASES. For example, contact process for manufacture of `H_2SO_4` .
`2SO_(2)(g) + O_(2)(g) overset(V_2O_(5(s)))to 2SO_(3)(g)`
50.

If T_(1)andT_(2) are the temperatures of the heat source and sink respectively, the efficiency of a heat engine may be expressed as

Answer»

`(T_(2)-T_(1))/T_(2)`
`T_(1)/T_(2)`
`(T_(1)-T_(2))/T_(1)`
`T_(2)/T_(1)`

ANSWER :C