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.

From the following thermochemical equations {:(C_(2)H_(4)+H_(2) to C_(2)H_(6),,DeltaH=-32.7 kcal),(C_(6)H_(6)+3H_(2) to C_(6)H_(12),, DeltaH=-49.2 kcal):} Calculate the resonance energy of benzene.

Answer»

Solution :`C_(6)H_(6)` has three isolated carbon-carbon DOUBLE bonds. `DeltaH` for hydrogenation of `C_(2)H_(4)`, with one double bond, but the given value is `-49.2 "kcal"`
`:.` resonance ENERGY `=3xx(-32.7)-(-49.2)`
`=-48.9 "kcal"`.
2.

The empirical formula of a compound is CH. Its molecular weight is 78. The molecular formula of the compound will be :

Answer»

`C_nh(_2n)O`
`C_Nh(_2n+2)O`
`C_Nh(2N+1)O`
`C_NH_(2n+2)O_2`

ANSWER :A
3.

From the following substances, which carbohydrate has the maximum sweetness ?

Answer»

Saccharin
Alitame
Aspartame
Sucrolose 

ANSWER :D
4.

General formula of an amine is :

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`C_nH_(2N+1)N`
`C_nH_(2n+2)N`
`C_NH_(2n+3)N`
`C_NH_(2n)N`

ANSWER :C
5.

From the followingstatementregarding H_(2)O_(2) choose the incorrect statement

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it can act only as an oxidizing agent
it decomposes on exposure to light
it has to be stroed in plastic or wax linedgalss bottles in dark
it has to be kept AWAY from dust

SOLUTION :`H_(2)O_(2)` can be REDUCED or oxidised HENCE it can act as an oxidising agent as well as a REDUCING agent
6.

General formula of aldehyde is :

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`C_nH_(2N)O`
`C_nH_(2n+2)O`
`C_nH_(2n+1)O`
`C_nH_(2n+2)O_2`

ANSWER :A
7.

Fromthe followingseries of reactions, {:(Cl_(2)+2KOH to KCl+KClO+H_(2)O),("" 2KClO to 2KCl+KClO_(3)),(""4KClO_(3) to 3 KClO_(4) + KCl):} calculate the mass of chlorineneeded to produce100 g of KCiO_(4)

Answer»

Solution :Wt, of `Cl_(2) = 100 g KCIO_(4) XX ((" 1 mole " KCIO_(4))/( 139 g KCIO_(3)))`
`xx (("4 mole " KCIO_(3))/("3mole " KCIO_(4))) xx ((3 "mole " KCIO)/( 1 " mole " KCIO_(3)))`
`xx (("1 mole " Cl_(2))/( "1 mole KCIO")) xx ((71 g Cl_(2))/( " 1 mole " Cl_(2)))`
204.5 g
8.

General formula of alkyl group is

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`c_nH_(2N)`
`c_nH_(2n+2)`
`c_nH_(2n+1)`
`c_nH_(2n-1)`

ANSWER :C
9.

From the followingreaction sequence,{:(CaC_(2) + H_(2) O to CaO+ C_(2)H_(2)),(""C_(2)H_(2)toC_(2)H_(4)),(""nC_(2)H_(4) to (C_(2)H_(4))_(n)):} calculate the mass of polyethylenewhichcan be produced from 10 kg of CaC_(2)

Answer»

Solution :Wt. of `(C_(2)H_(4))_(n) = (1000 g CaC_(2)) xx ((1 "mole CA"C_(2))/(64 "g Ca"C_(2))) `
`((1 "mole "C_(2)H_(2))/(1"mole Ca" C_(2)))xx ((" 1 mole " C_(2) H_(4))/( " 1 mole "(C_(2)H_(2))))`
`(("1 mole " (C_(2)H_(4))_(n))/( "n more " C_(2) H_(4))) xx (( " 28 n g " (C_(2) H_(4))_(n))/("1 mole " (C_(2) H_(4))_(n)))`
= 4375 g
(note that n cancels)
Second Method APPLY POACfor C atoms.
10.

General formula for carboxylic acid is

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`C_(n) H_(n)O_(2)`
`C_(n)H_(2N)O_(2)`
`C_(n)H_(2n)O`
`C_(n)H_(2n^(+2))O_(2)`

Answer :B
11.

Fromthe followingreaction , 2CoF_(2) + F_(2) to 2 CoF_(3)(CH_(2))_(n) + 4 n CoF_(3) to (CF_(2))_(g) _(n) + 2n HF + 4n CoF_(2)calculate how much F_(1)will be consumedto produce 1 kgof (CF_(2))_(n)

Answer»


ANSWER :1 . 52 KG
12.

General formula for carbohydrate is

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`C_(N)H_(2n)O_(2n+2)`
`C_(x)(H_(2)O)_(2x)`
`C_(x)(H_(2)O)_(y)`
none of these

Solution :CARBOHYDRATES are hydrates of CARBON. Their general formula is `C_(x)(H_(2)O)_(y)`
13.

From the following reaction sequences 2A = B + C 5B + D = 2E + F E + G = 4H + J calculate moles of H produced by 10 moles of A

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20
10
5
8

Answer :D
14.

General formula for alkenes is

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`C_(N)H_(2N)`
`C_(2n)H_(2n)`
`C_(n)H_(2n+2)`
`C_(n)H_(2n-2)`

ANSWER :A
15.

From the following reaction of H_(2) and X_(2) ,in which reaction catalyst is necessary?

Answer»

`H_(2)+Br_(2)to2HBr`<BR>`H_(2)+I_(2)to2HI`
`H_(2)+Cl_(2)to2HCl`
`H_(2)+F_(2)to2HF`

SOLUTION :The REACTIVITY decreases from up to down because the reactivity of `I_(2)` is LESS,the catalysis is necessary.
16.

From the following plot, predict the order of the reaction.

Answer»

SOLUTION :FIRST ORDER.
17.

General electronic configurationof actinoids is (n-2)f^(1-14) ( n-1)d^(0-2) ns^(2) . Which of the followingactinoids have one electron in 6d orbital ?

Answer»

U ( ATOMIC no. 92)
Np ( Atomic no . 93)
Pu ( Atomic no. 94)
Am ( Atomic no. 95)

Answer :a,b
18.

From the following metallurgical flow chart, identify steps A, B and C. "Sulphide" overset((A))rarr "Oxide" overset((B))rarr "Impure metal" overset((C))rarr "Pure metal"

Answer»

Solution :(A) Roasting
(B) Reduction by CARBON or more ELECTROPOSITIVE METAL
(C) Electrolysis
19.

From the following …………………….. is the example of biopolymer.

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teflon
neoprene
nylon 66
DNA

Solution :DNA
NATURAL polymers LIKE polysaccharide, protein, nuclic ACID is necessary for human life is called biopolymer substance.
Two TYPES of Nuclic acid (1) DNA & (2) RNA
20.

General empirical formula of silicone is ………………. .

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`(R_(2)SIO)`
`(RSiO)`
`(R_(2)CO)`
`(RSIH)`

Answer :A
21.

General formula for alcohols is :

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`rarrCOH`
`GT CHOH`
`-CH_2OH`
All

Answer :D
22.

General electronic configuration of transition elements is

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`(n-1)d^(1-10)NS^(1-2)`
`(n-2)^(1-10)ns^(1-2)`
`(n-1)d^(1-8)ns^(2)`
`(n-2)^(1-10)ns^(2)`

ANSWER :A
23.

From the following half cells a galvanic cell is made, A^(2+)+2e=A" "..." "E_(1)^(0)=0.8V B=B^(3+)+3e" "..." "E_(2)^(0)=-0.3V E^(0) cell is

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`E_(1)^(0)-E_(2)^(0)`
`E_(1)^(0)+E_(2)^(0)`
`3E_(1)^(0)-2E_(2)^(0)`
`3E_(1)^(0)+2E_(2)^(0)`

ANSWER :B
24.

General electronic configuration of lanthanide is :

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`(n-2)F^(1-14), (n-1) s^2p^6d^(0-1) , ns^2`
`(n-2)f^(10-14), (n-1) d^(10-1) , ns^2`
`(n-2)f^(0-14) , (n-1) d^10 , ns^2`
`(n-2)d^(0-1) , (n-1)f^(1-14) , ns^2`

ANSWER :A
25.

From the following ions magnetic moment is maximum for

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`Mn^(2+)`
`FE^(2+)`
`CR^(2+)`
`Zn^(2+)`

ANSWER :A
26.

General electronic configuration of actionoids is (n-2) f^(1-14) (n-1) d^(0-2) ns^(2). Which of the following actinoids have one electron in 6d orbital?

Answer»

U(atomic no92)
Np (Atomic no.93)
Pu (Atomic no.94)
Am (Atomic no.95)

Solution :`""_(92)U : [Rn] 5f^(3) 6d^(1) 7s^(2)`
`""_(93)NP: [Rn]5f^(4) 6d^(1) 7s^(2)`
27.

From the following graph, identify order of reaction and mention the unit of its rate constant.

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SOLUTION :FIRST ORDER reaction
Unit - `s^(-1)`
28.

General electronic configuration of d-block elements is

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`(n-1)d^(1-10) NS^(1-2)`
`ns^(2) NP^(1-6)`
`(n-2)F^(0-14) (n-1)^(1-2) ns^(2)`
`(n-1)d^(1-5) ns^(1-2)`

ANSWER :A
29.

From the following graph , identify order of reaction and mention the unit of its rate constant.

Answer»

Solution :FIRST ORDER Reaction , Unit `RARR s^(-1)`
30.

General electronic configuration of group 16 elements is

Answer»

`ns^(2)NP^(3)`
`ns^(2)np^(4)`
`ns^(2)np^(2)`
`ns^(2)np^(5)`

Solution :(x) (B) `ns^(2)np^(4)`
31.

From the following equations, calculate the standard molar heat of formation of AgCl at 25^(@)C. Ag_(2)O(s)+2HCl (g) to 2AgCl(s)+H_(2)O(l), DeltaH^(0)=-77.61 kcal 2Ag(s)+(1)/(2)O_(2)(g) to Ag_(2)O(s), DeltaH^(0)=-73.1 kcal (1)/(2)H_(2)(g)+(1)/(2)Cl_(2)(g) to HCl (g), DeltaH^(0)=-22.06 kcal H_(2)(g)+(1)/(2)O(g) to H_(2)O(l),DeltaH^(0)=-68.32 kcal

Answer»


ANSWER :(-63.20 KCAL)
32.

From the following E^(o) value of helf cells: (i) A+e^(-)rarr A ^(-) , E^(o) = -0.824 V (ii) B^(-)+e^(-)rarr B ^(2-) , E^(o) = +1.25V (iii) C^(-)+2e^(-)rarr C ^(3-) , E^(o) = -1.25V (iv) D^(-)+2e^(-)rarr D ^(2-) , E^(o) = +0.68V What combination of two half cells would result is a cell with the largest cell potential?

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(II)&(III)
(ii)&(IV)
(i)&(iii)
(i)&(iv)

ANSWER :A
33.

General electronic configuration of element of Group 16 is :

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`ns^2np^6`
`ns^2np^4`
`ns^2np^5`
`ns^2np^2`

ANSWER :B
34.

From the following equations pick up the possible fusion reactions :

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`""_(6)^(13) C + ""_(1)^(1) H + ""_(0)^(1) e + 4.3 MeV`
`""_(6)^(12) C + ""_(1)^(1) H to ""_(7)^(13)N + 2 MeV`
`""_(7)^(14) N + ""_(1)^(1) H to ""_(8) ^(15) O + 7.3 MeV `
`""_(92)^(235) U + ""_(0) ^(1) n to ""_(54)^(140) Xe + ""_(38)^(94) Sr + ""_(0)^(1) n + gamma + 200 MeV`

Solution :`""_(6) C^(12) + ""_(1)H^(1) to ""_(7) N^(13) + Q , ""_(7) N^(14) + ""_(1)H^(1) to ""_(8) O^(15) + Q`
35.

Gemstone is the impure form of which compound?

Answer»

`Al_2O_3`
`Cu_2O`
`Cr_2O_3`
`Mn_2O_3`

ANSWER :A
36.

From the following data, the activation energy for the reaction is (cal/mol):H_2 + I_2---->2HI

Answer»

`4xx10^4`
`2xx10^4`
`8xx10^4`
`3xx10^4`

ANSWER :A
37.

From the following data , show that the decomposition of hydrogen peroxide is a reaction of the first order : {:(t(min),0,10,20),(V(ml),46.1,29.8,19.3):} where t is the time in minutes and V is the volume of standard KMnO_4 solution required for titrating the same volume of the reaction mixture.

Answer»

Solution :Volume of `KMnO_4` , solution used `prop ` AMOUNT of `H_2O_2` present . Hence if the given reaction is of the first ORDER , it must obey the equation.
`k=(2.303)/tlog.(a)/((a-x))`
`k = (2.303)/tlog.V_o/V_t`
In this case, `V_o=46.1` ml
The value of k at each INSTNAT can be calculated as follows :

THUS, the value of k comes out to be nearly constant . Hence it is a reaction of the first order.
38.

Gelatine is mostly used in making ice creams in order to

Answer»

prevent forming the COLLOIDAL sol
enrich the fragrance
prevent crystallisation and STABILISE the mix
modify the teste

Solution :Gelatine is a PROTECTIVE colloid. It stabilises the mix and PREVENTS carystallisation.
39.

From the following data, the activation energy for the reaction (cal/mol):

Answer»

`4xx10^4`
`2xx10^4`
`8xx10^4`
`3xx10^4

Answer :A
40.

Gemstone is an impure form of ..........

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`Cu_2O`
`Cr_2O_3`
`Mn_2O_3`
`Al_2O_3`

ANSWER :D
41.

From the following data, show that the decomposition of hydrogen peroxide is a reaction of the order : {:("t (min)",0,10,20),("V (ml)",46.1,29.8,19.3):} Where t is the time in minutes and V is the volume of standard KMnO_(4) solution required for titrating the same volume of the reaction mixture.

Answer»

SOLUTION :`k=(2.303)/(t)"log"([A_(0)])/([A])`
`k=((2.303)/(t))log((V_(0))/(V_(t)))`
In the PRESENT CASE, V0 = 46.1 ml.
The value of k at each instant can be calculated as follows :

Thus, the value of k comes out to be NEARLY constant. HENCE it is a reaction of the first order.
42.

Gem dihalides on hydrolysis give :

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Acetone
Aldehydes
Ketone
Ketone and aldehyde

Answer :D
43.

From the following data of DeltaH of the following reactions, C(s)+(1)/(2)O_(2)(g) to CO(g),DeltaH=-110 kJ and C(s)+H_(2)O(g) to CO(g)+H_(2)(g),DeltaH=132 kJ Calculate the mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, keeping the reaction temperature constant.

Answer»

Solution :From the QUESTION we see that the first reaction is exothermic and the second ONE is endothermic. Thus, if a mixture of oxygen and steam `(H_(2)O)` is passed over coke and at the same time, the temperature does not change, the composition should be such that `DeltaH` of both the reactions are numerically EQUAL.
In the first reaction, consumption of `1//2` "mole" of `O_(2)` evolves 110 kJ of energy, while in the second reaction, for 1 mole of steam `(H_(2)O)`, 132 kJ of energy is absorbed.
`:. "mole"` of `O_(2)` needed to EVOLVE `132kJ=(0.5)/(110)xx132=0.6`
`:. "mole"` ratio of `O_(2)` and steam `(H_(2)O)=0.6:1`.
44.

Gelatin which is a peptide is added in ice-creams. What can be its role ?

Answer»

Solution : Ice-creams are EMULSIONS which get STABILISED by emulsifying agents like GELATIN.
45.

From the following data, show that the decomposition of hydrogen peroxide is a reaction of the first order : {:(t,,,0,,,10,,,20),(x,,,46.1,,,29.8,,,19.3):} where t is the time in minutes and x is the volume of standard KMnO_(4) solution in "cm"^(3) required for titrating the same volume of the reaction mixture.

Answer»

Solution :Volume of `"KMnO"_(4)` solution used `prop`amount of `H_(2)O_(2)` present. Hence, ifthe given reaction is of the first order, it must obey the equation : `k=(2.303)/(t)log""(a)/(a-x)=(2.303)/(t)log""(V_(0))/(V_(t))`
(Note that the SYMBOL x in the numerical PROBLEM isin FACT `a-x, i.e., V_(t)`)
In the present CASE, `V_(0)=46.1" cm"^(3)`
The value of k at each instant can be calculated as follows:
`{:("t (min)",,,V_(t),,,""k=(2.303)/(t)log""(V_(0))/(V_(t))),(10,,,29.8,,,k=(2.303)/(10min)log""(46.1)/(29.8)=0.0436min^(-1)),(20,,,19.3,,,k=(2.303)/(20min)log""(46.1)/(19.3)=0.0435min^(-1)):}`
Thus, the value of k comesout to be nearly constant. Hence, it is a reaction of the first order.
46.

From the following data show that the decomposition of H_(2)O_(2) is a reaction of first order. Also calculate the value of the rate constant.

Answer»

Solution :Let us substitute the values in the rate EQUATION for FIRST order reaction and calculate the value of rate constant(k)
`k=2.303/tlog(a/(a-x)) = (2.303)/tlogV_(0)/V_(t)`
CASE I. `V_(0) = 22.8 mL, V_(t) = 13.8 mL, t=10 min`
`k=(2.303)/(10 min) log(22.8)/(13.8) = (2.303)/(10 min) (log 22.8 - log13.8)`
`=(2.303)/(10 min) (1.358-1.140) = (2.303 XX 0.128)/(10 min) = 0.0502 min^(-1)`
Case II. `V_(0) =22.8 mL, V_(t)=8.3, t=20min`
`therefore k=(2.303)/(20 min)(1.388 - 0.919) = (2.303 xx 0.439)/(20 min) = 0.0505 min^(-1)`
Since, the value of k comes out to be almost constant, the reaction is of first order.
47.

Gelatin which is a peptide is added is ice-creams. What can be its role ?

Answer»

SOLUTION :Ice-creams are emulsions. These are STABILIZED by emulsifying agents LIKE GELATIN.
48.

From the following data show that decomposition of H_(2)O_(2) in aqueous solution is first order. Time (in minutes) 0 10 20 Volume (in c.c. of KMnO_(4)) 22.8 13.3 8.25

Answer»

SOLUTION :FIRST ORDER
49.

Gelatin which is a peptide is added in ice creams. What can be its role?

Answer»

SOLUTION :GELATIN ACTS as an emulsifying AGENT. It STABILIZES the ice-cream.
50.

From the following data of DeltaH of the following reactions C (s) + (1)/(2) O_(2) (g) , Delta H = -110 KJ and C(s) + H_(2) O (s) to CO_(g) + H_(2) (g) , Delta H = 132 KJ Calculate the mole composition of the mixture of steam and oxygen on being passed over coke at 1273 K, keeping the reaction temperature constant.

Answer»

SOLUTION :MOLE % `O_(2) (G) = 37.5 , H_(2) O (g) = 62.5`