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 a closed flask of 5 litres, 1.0 g of H_(2) is heated from 300 to 600 K. Which statement is/are correct?

Answer»

Pressure of the GAS increases
The rate of COLLISION increases
The number of MOLES of gas increases
The energy of GASEOUS molecules increases

Answer :A::B::D
2.

If the initial concentration of reactants in a certain reaction is doubled the half life period of the reaction doubles . The order of the reaction is :

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ZERO
FIRST
Second
Third

ANSWER :A
3.

In a closed insulated container, a liquid is stirred with a paddle to increase the temperature. Which of the following is true ?

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`Delta U = W NE 0, Q = 0`
`Delta U = w, q ne 0`
`Delta U = 0, w = q ne 0`
`w = 0, Delta = q ne 0`

Answer :A
4.

In the initial cocnentration is reduced to 1/4th of the initial value of a zero order reaction the half life of the reaction

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REMAIN CONSTANT
BECOMES 1/4 th
becomes double
Becomes fourfold

Answer :B
5.

In a closed containerr, a liquid is stirred with a paddle to increse the temperature. Which of the following is true?

Answer»

`DeltaE=W ne0,q=0`
`DeltaE=W=qne0`
`DeltaE=0,W=qne0`
`W=0,DeltaE=qne0`

Solution :`DeltaE=q+W`
q=0 (`because` TEMPERATURE is to be increased, no heat should ENTER or leave the SYSTEM)
`DeltaE=q+W=0+W` or `DeltaE=W`
`therefore W ne 0,q=0`.
6.

If the hydrogen ion concentration of a given solution is 5.5 xx 10^(-3) " mol litre"^(-1), the pH of the solution will be

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2.26
3.4
3.75
2.76

Solution :`[H^(+)] = 5.5 xx 10^(-3)` mole/litre
`pH = -LOG [H^(+)] , pH = -log [ 5.5 xx 10^(-3)] , pH = 2.26`.
7.

In a closed insulated container, a liquid is stirred with a paddle to increase its temperature. In this process, which of the following is true

Answer»

`DeltaE=W=Q=0`
`DeltaE ne0,Q=W=0`
`DeltaE=W NE 0, Q=0`
`DeltaE = Qne0, W=0`

SOLUTION :As the system is closed and INSULATED no heat ENTER or leave the system, i.e. q = 0 , `thereforeDeltaE = Q+W=W`.
8.

In a closed container of capacity V with N_(2)(g) and wate vapour at 298 k, the pressure observed is 1.8 atm. If the volume is squeezed to V/3, then the preesue of the container is….., if aq. Tension of H_(2)O(g) is 0.2 atm at 298 K.

Answer»

Solution :5 atm
WATER VAPOURS do not FOLLOW Boyle.s LAW.
9.

If the hydrogen ion concentration of a given solution is 5.5 xx 10^-3 M. Find the pH of the solution.

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2.26
3.4
3.75
4.76

Answer :A
10.

In a closed container NO_(2) gas is getting dimerised into N_(2)O_(4) gas with first order kinetics. Pressure after 10 hours of reaction is 5 atm and pressure after completion of reaction is 4atm. What is half life in hours?

Answer»

<P>

SOLUTION :`2NO_(2(g))hArrN_(2)O_(4(g))` So `P_(0)=8` atm `(P_(0)-y/2)=5`
`(P_(0)-y)=(8-6)=2` atm` Kxxt=2.303xxlog((P_(0))/(Pt))`
`0.693/((t_(1//2)))xx10=2.303xxlog(8/2),(0.3010xx10)/((t_(1/2)))=2xxlog2,(0.3010x10)/((t_(1/2)))=2xx0.3010`,
`(t_(1/2))=10/2=5,t_(1/2)=5`
11.

If the highest oxidation states shown by any Lanthanide and any Actinide are +X and+Y, the sum of X+Y is euqal to?

Answer»


ANSWER :`11.00`
12.

if the heat of neutralisation for a strong acid- base reaction is -57.1 kJ, would be the heat released when 350 cm^(3)of 0.20 M ofa dibasic strong acid is mixed with 650 cm^(3) of 0.10 M monoacidic base ?

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57.1 kJ
3.71 kJ
`-57.1 kJ`
0.317 kJ

Solution :millimoles of dibasic strong ACID
` M XX V=0.20xx350=70 MMOL`
Amount of `H^(+)` IONS in the acid = `2xx70=140 mmol`
SIMILARLY, amount of `OH^(-)`ions in the acid `0.10xx650xx =65 mmol`
[ Here, `OH^(-)` is the limiting reactant]
`65 xx10^(-3)` mole of `OH^(-)` ions will produce
`=57.1xx65xx10^(-3)=3.71kJ`
13.

If the heat of formation of CO_(2) is -393 kJ. The amount of heat evolved in the formation of 0.156 kg of CO_(2) is

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`-1357.9`kJ
`-1275.9` kJ
`-1572.0` kJ
`-1165.5` kJ

Solution :Heat of formation of a SUBSTANCE is the heat exchanged when one MOLE of that substance is FORMED by its constituent elements.
`because` Heat evolved when 1 mole (44g) `CO_(2)` is formed
`=(393xx156)/(44)`
`therefore DELTAH` for the process = 1572 kJ =-1572.0 kJ`.
14.

If the heat of combustion of carbon monoxide at constant volume and at 17^(@)C is - 283.3 kJ, then its heat of combustion at constant pressure (R=8.314 J degree^(-1) mol^(-1)

Answer»

`- 284.5` KJ
284.5 kJ
384.5 kJ
`- 384.5 kJ`

Solution :`CO_((g))+(1)/(2)O_(2(g))rarrCO_(2(g)), Deltan=1-1(1)/(2)=-(1)/(2)`
`DeltaH=DeltaE+DeltanRT`
`DeltaH=-283.3-(1)/(2)XX(8.314)/(1000)xx290=-284.5 kJ`
15.

In a close packing of atoms A of radius r_(a), the radius of atom B that can be fitted in tetrahedral void is

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`0.225 r_(a)`
`0.155 r_(a)`
`0.414 r_(a)`
`0.732r_(a)`

ANSWER :A
16.

If the half-life period of a first order reaction is 138.6 minutes, then the value of decay constant for the reaction will be

Answer»

`5 "miniute"^(-1)`
`0.5 "MINUTE"^(-1)`
`0.05 "minute"^(-1)`
`0.005 "minute"^(-1)`

Solution :`K = (0.693)/(t_(1//2)) = (0.693)/(138.6 "MIN") = 0.005 "min"^(-1)`
17.

In a close-packed structure, if r is the radius of the spherical void and R is the radius of the spheres forming voids, the critical r/R value for the tetrahedral void is equal to

Answer»

0.155
0.225
0.414
0.732

Answer :B
18.

If the half-life of an isotope X is 10 years, its decay constant is

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`6.932 YR^(-1)`
`0.6932 yr^(-1)`
`0.06932 yr^(-1)`
`0.006932 yr^(-1)`

SOLUTION :`K = (0.693)/(t_(1//2)), K = (0.693)/(10) == 0.0693 yr^(-1)`
19.

In a close packed structure of mixed oxides , the lattice is composed of oxide ions , one eighth of the tetrahedrai voids are occupied by divalent cations ( A ) while half of the octahedral voids are occupied by trivalent cations ( B ) . What is the formula of the oxide ?

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`A_2B_3O`
`A_4B_2O_3`
`AB_2O_4`
`A_3B_2O_2`

ANSWER :C
20.

If the half life period for a reaction in A is 100 min . How long will it take [A] to reach 25 % of its initial concentration ?

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50 MIN
250 min
200 min
500 min

Solution :( C) K `= (0.693)/(100) = 6.93 xx10^(-3)`
` t= (2.303)/(6.93xx10^(-3)) "LOG" ([A]_(0))/(0.25[A]_(0))`
`= (2.303)/(6.93xx10^(-3))"log"4`
`= (2.303)/(6.93xx10^(-3))xx0.6020`
=200 min
21.

In a close packed structure of mixed oxide, the lattice is composed of O^(2-) ions. One eighth of tetrahedral voids are occupied by divalent cations A^(2+) and one half of octahedral voids by trivalent cations B^(3+) What is the formula of oxide ?

Answer»

SOLUTION :`AB_(2)O_(4)`
22.

If the half-life of a radioisotope is 4 days, then how long would it take for 75% disitegration of the sample?

Answer»

8 days
10 days
12 days
14 days

Solution :`t_(1//2) = 4 "days", lambda = (0.693)/4 "days"^(-1)`
`N_0 = 100, N_t = 25`
Now `t = (2.303)/(lambda) log (N_0)/(N_t)`
`implies t = (2.303 xx 4)/(0.693) log (100)/(25) = (18.424)/(0.693) xx 0.301 = 8 ` days
It WOULD TAKE 8 days for 75% distintegration of the SAMPLE.
23.

In a close packed centred cubic lattice of potassium the corrcet relation between the atomic radius (r)of potassium and the edge length (a) of the cube is

Answer»

`R=a/(sqrt(2))`
`r=a/(sqrt(3))`
`r=(sqrt(3))/2a`
`r=(sqrt(3))/4a`

SOLUTION :For BCC `4r=sqrt(3)aimpliesr=(sqrt(3))/4a`
24.

If the half cell reactions are given as (i) Fe_((aq))^(2+)+2etoFe_((s)),E^(o)=0.44V (ii) 2H_((aq))^(+)+(1)/(2)O_(2(g))+2etoH_(2)O_((l)),E^(o)=+1.23V The E^(o) for the reaction Fe_((s))+2H^(+)+(1)/(2)O_(2(g))toFe_((aq))^(2+)+H_(2)O_((l))

Answer»

`+1.67V`
`-1.67V`
`+0.79V`
`-0.79V`

Solution :Subtracting EQUATION (i) from (II) we will get equation (iii)
`Fe_((s))+2H^(+)+(1)/(2)O_(2)toFe_((aq))^(2+)+H_(2)O_((l))`. . .(iii)
So, `E^(o)=E_(2)^(o)--E_(1)^(0)`
`=+1.23-(-0.44)=+1.6V`
25.

In a close packed lattice containing 'n' particles, the number of tetrahedral and octahedral voids respectively

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N and 2N
n and n
2n and n
2n and n/2

Answer :C
26.

In a close packed arrangement of N spheres, how many i) tetrahedral and (ii) octahedral sites are present ?

Answer»

SOLUTION :(i) TETRAHEDRAL SITES = 2N
(ii) OCTAHEDRAL sites = N.
27.

If the half life of a first order reaction is 60 min, the approximate time in min, required to complete 90% of the reaction is (log2=0.3)

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`200`
`240`
`50`
`100`

ANSWER :A
28.

In a classroom, the teacher has explained the quantitative aspects of electrolysis by stating the Faraday's laws of electrolysis. Calculate the quantity of electricity required to deposit 0.09 g of aluminium during the following electrode reaction: Al^3 + 3e^(-) to Al ( Al=27u)

Answer»

Solution :Quantity of electricity required to deposit
27 G (1 mole ) of Al = 3F = `3xx96500` coulombs
`THEREFORE .Quantity of electricity required to deposit
(0.09g of Al `3xx96500xx0.09)/27`=965 coulombs
29.

If the half cell reaction A+e^(-)toA^(-) has a large negative reduction potential,it follows that

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A is readily reduced
A is readily oxidised
`A^(-)` is readily reduced
`A^(-)` is readily oxidised

Solution :Since `E_(A//A^(-))^(@)` has LARGE NEGATIVE VALUE, the tendency of A to be reduced to `A^(-)` is very SMALL. In other words tendency of `A^-` to be oxidized to A is very large.
30.

In a close pack array of n spheres, the number of tetrahedral holes are :

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4N
N/2
2N
N

ANSWER :C
31.

If the half cell reaction A to e^(-) to A^(-) has a larger negative potential , it follows that :

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A is not READILY redused
A is readily oxidised
`A^(-)` is readily REDUCED
`A^(-)` is readily oxidised

SOLUTION :`A + E^(-) to A^(-) , `A is not readily reduced , A is readily oxidised
32.

If the half-cell reaction A +e rarr A^(-) has a large negative reduction potential, it follows that

Answer»

`A^(-)` is READILY oxidized
A is readily oxidised
`A^(-)` is readily REDUCED
A is readily reduced

ANSWER :A
33.

If the half cell reaction A +e rarr A^- has a large negative reduction potential, it follows that :

Answer»

A is READILY REDUCED
A is readily oxidised
`A^-` is readily reduced
`A^-` is readily oxidised

Answer :D
34.

If the greenhouse effect or global warming remains unchecked, it alter :

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SEA levels
Temperature
Rainfall
All of these

Answer :D
35.

In a classroom discussion about order and moleculrity of a reaction,Ramu argued that ''there are reactions which appear to be of higher order but actually follows first order kinetics''How far is his statement true? Give your opinion in this regard. Justify your answer using suitable example.

Answer»

Solution :It is a true statement. For example, hydrolysis of ethyl ACETATE in presence of acid APPEARS to be seconds order, but actually the order of the REACTION is one. Such reaction are CALLED pseudo FIRST order reaction.`CH_3COOC_2H_5+H_2O overset(Acid)oversettolarrCH_3COOH+C_2H_5OH`
36.

State the Faraday's laws of electrolysis

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Solution :I law: The amount of a substances deposited or liberated during dectrolysis is directly proportional to the quantity of electricity FLOWING through the electrolyte.
II law :If the same quantity of electricity is PASSED through different ELECTROLYTES the amount of SUBSTANCE formed are directly proportional to their equivalent WEIGHT.
37.

If the gold number of potato starch is 25, how many grams of it is required to prevent coagulating of 100mL of gold sol on adding 10mL of 10% NaCl solution?

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0.3
0.5
0.25
0.4

Answer :C
38.

In a classroom, the teacher has explained the quantitative aspects of electrolysis by stating the Faraday's laws of electrolysis.Explain the term electrochemical equivalent

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Solution : It is the quantity of a substance formed at the electrode, when ONE ampere CURRENT is PASSED through an ELECTROLYTE for one second.
39.

If the given four electronic configurations (i) n = 4, l = 1 (ii) n = 4, I= 0 (iii) n = 3, l = 2 (iv) n = 3, I= 1 are arranged in order of increasing energy, then the order will be

Answer»

(iv)lt (ii)lt (iii) lt (i)
(ii)lt (iv)lt (i) lt (iii)
(i) lt (iii) lt (ii) lt (iv)
(iii) lt (i) lt (iv) lt (ii)

SOLUTION :(i) n = 4, l = 1 `rArr` 4p-orbital
(ii) n = 4, l = 0 `rArr` 4s-orbital
(iii) n = 3, l = 2 `rArr` 3D-orbital
(iv) n = 3, I= I `rArr` 3P-orbital
Increasing order of energy as PER (n +l)rule :
3p < 4s < 3d < 4p
(iv) (ii) (iii) (i)
40.

In a class room, a teacher was teaching about batteries in the chapter of 'electrochemistry' to 12th class. As the use of batteries is very common now a days for invertors and cellphones, students were very curious to know which battery they should use for their invertors at home and for their cellphones as a number of cheap as well as costly batteries are available in the market. the teacher advised them that they should use those batteries which do not cause any pollution or health hazard though these may be a little constlier. thus, he suggested the use of cadmium battery in place of lead storage battery for invertor and lithium ion battery in place of Ni-Cd battery low celphone. After reading the above paragraph, answer the followng questions: (i) What values are expressed by the teacher? (ii) How is lead storage battery harmful in the long run? (iii) How is Ni-Cd battery harmful? What are other advantages of using Li-Ion battery over Ni-Cd battery?

Answer»

Solution :(i) The teacher has expressed concern that we should not use those materials which cause health hazard or pollution in the ATMOSPHERE.
(ii) Lead is poisonous. It has harmful effects on health as well as environment.
(iii). Cd is TOXIC heavy metal and requires special care for disposal. other adavantages of Li-ion battery are: (a) It has a voltage of 3.6-3.7V whereas Ni-Cd battery has 1.2 V
(B) It does not need periodic discharge whereas Ni-Cd battery is charged only after it is fully discharged. (C) Li-ion batteries do not LOSE their charge even after months of storage whereas Ni-Cd batteries lose 1-5% of their charge per day.
41.

If the geometry of [PtCl_(4)]^(2-) is square planar, which orbitals of Pt are involved in the bonding ?

Answer»

SOLUTION :`dsp^(2)` HYBRIDISATION INVOLVING 5d, 6s and 6P ORBITALS.
42.

In a chemical reaction K_(2)Cr_(2)O_(7)+xH_(2)SO_(4)+ySO_(2)toK_(2)SO_(4)+Cr_(2)(CO_(4))_(3)+zH_(2)O the values of x,y,z are

Answer»

`1,3,1`
`4,1,4`
`3,2,3`
`2,1,2`

SOLUTION :BASED on BALANCING.
43.

If the freezing point of a 0.01 molal aqueous solution of a cobalt(III) chloride-ammonia complex (which behaves as a strong electrolytes) is -0.0558^(@)C, the number of chloride(s) in the coordination sphere of the complex is [K_(f) of water =1.86 Kkgmol^(-1)]

Answer»


SOLUTION :`DELTA T_(f)=K_(f)xx ixx m rArr 0.0558=1.86xx ixx0.01`
i=3
Given COMPLEX BEHAVES as a STRONG electrolyte `alpha = 100%`
n = 3(no. of particles given by complex)
`therefore` complex is `[Co(NH_(3))_(5)Cl]Cl_(2)`
no. of `Cl^(-)` ions in the co-ordination sphere of the complex = 1
`Delta T_(f)=K_(f)xx ixx m`
`0.0558=1.86xx ixx0.01`
i = 3
44.

In a chemical reaction triangleH=150kJ and triangleS= 100JK^-1 at 300K. The triangleG for the reaction is :

Answer»

Zero
300 kJ
330kJ
120 kJ

Answer :D
45.

If the freezing point of 0.01 molal aqueous solution of cobalt (III) chloride-ammonia complex (which behaves as a strong electrolyte) is -0.0558^(@)C, the number of chloride(s) in the coordination sphere of the complex is (K_(f)" of water "="1.86 K kg mol"^(-1))

Answer»

Solution :`DeltaT_(f)=iK_(f)m`
Given `m=0.01 " molal, "T_(f)=-0.0558^(@)C`,
`K_(f)="1.86 K kg mol"^(-1)`
`DeltaT_(f)=T_(f)^(@)-T_(f)=0-(-0.0558)=0.0558^(@)`
`i=(DeltaT_(f))/(K_(f)m)=(0.0588)/(1.86xx0.01)=3`
As 3 ions are produced on dissociation of the complex, the molecular FORMULA of the complex is `[Co(NH_(3))_(5)Cl]Cl_(2)`. Thus, only one chloride ion is present in the coordination sphere.
46.

In a chemical reaction, the rate constant for the backward reaction is 7.5xx10^(-4) and the equilibrium constant is 1.5.The rate constant for the forward reaction is :

Answer»

`5xx10^(-4)`
`2xx10^(-3)`
`1.125xx10^(-3)`
`9.0xx10^(-4)`

SOLUTION :`K=(k_(F))/(k_(b))`
or `k_(f)=K xx k_(b)`
`=1.5xx7.5xx10^(-4)`
`=1.125xx10^(-3)`
47.

If the freezing point of a 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as a strong electrolyte) is -0.0558^(@)C, the number of chloride (s) in the coordination sphere of the complex if [K_(f) of water =1.86 K kg mol^(-1)]

Answer»

Solution :`DeltaT_(F)=ixxK_(f)xxm`
`i=(DeltaT_(f))/(K_(f)xxm)=((0.0558K))/((1.86 KM^(-1))xx(0.01m))=3`
The value of I indicates that in aqueous solution, the complex dissociates to give ions.
`[Co(NH_(3))_(5)CI]CI_(2)OVERSET(("aq"))to[Co(NH_(3))_(5)CI]^(+)2CI^(-)`
This means that one chlorine ATOM is present in the cooedination sphere. For more details, consult Unit-9 (Co-ordination compounds).
48.

If the freezing point of 0.01 molal aqueous solution of a cobalt (III) chloride-ammonia complex (which behaves as strong electrolyte) is -0.0558^(@)C, the number of chloride (s) in the coordination sphere of the complex is(k_(f) of water =1/86K kg mol^(-1))

Answer»


Solution :`((DeltaT_(f))_(obs))/((DeltaT_(f))_(CAL))=i=3` which shows
`[Co(NH_(3))_(5)Cl]Cl_(2) to [Co(NH_(3))_(5)Cl]^(2+)+2Cl^(-)`
49.

If the four tubes of a car are fitted to the same pressure with N_(2), O_(2), H_(2) and Ne separately then which one will be filled first?

Answer»

`N_(2)`
`O_(2)`
`H_(2)`
Ne

Answer :C
50.

If the fragment of one strand in DNA molecule has the base sequence CCATGCATG, what is the base sequence of complementary strand ?

Answer»

Solution :Base sequence of COMPLEMENTARY STRAND of DNA is GGTACGTAC.