94420 + Interview Questions in Chemistry in Class 12

1.	Half life period of a reaction is directly proportional to initial concentration of the reactant. What is the order of this reaction?
Answer» SOLUTION :Half LIFE of the zero order REACTION is directly proportional to INITIAL concentration of the reactant. `t_(1//2)=([R]_(0))/(2K)""rArr""t_(1//2)alpha[R]_(0)` The order of the reaction is zero order.

Discussion

2.	Half-life period of a zero order reaction is
Answer» PROPORTIONAL to INITIAL CONCENTRATIONS of REACTANTS independent of initial concentrations of reactants inversely proportional to initial concentrations of reactions inversely proportional to the square of intioal concentrations of reactants . Answer :A

Discussion

3.	Half-life period of a radioactive element is 100 seconds. Calculate the disintegration constant and average life period. How much time will it take for 90% decay?
Answer» Solution :`t^(1//2)=100 SEC` average LIFE period `tau=(1)/(lambda)=(1)/(0.00693)=1.44 sec^(-1)` `=(2.303)/(0.00693) LOG (100)/(10)` `=(2.303)/(0.00693) log 10=332.3 sec`

Discussion

4.	Half life period of a radio active element is 1500 years. Find the value of disintegration constant interms of second.
Answer» SOLUTION :`LAMBDA=(0.693)/(t^(1//2)) =(0.693)/(1500 yrs)` `=(0.693)/(1500xx365xx24xx60xx60 sec)` `=0.1465 xx10^(-10) sec^(-1)`

Discussion

5.	Half life period of a first-order reaction is 1386 seconds . The specific rate constant of the reaction is
Answer» `5. 0 xx 10^(-3) s^(-1)` `0.5 xx 10^(-2) s^(-1)` `0.5 xx 10^(-3) s^(-1)` `5.0 xx 10^(-2) s^(-1)` SOLUTION :`t_(1//2) = (0.693)/(k) , k = (0.693)/(1386) = 0.5 xx 10^(-3) s^(-1)`.

Discussion

6.	Half-life period of a first order reaction is
Answer» directly proportional to initial concentration a. inversely proportional to a. INDEPENDENT of a. independent of RATE constant of the REACTION. ANSWER :C

Discussion

7.	Half-life period of a first - order reaction is 1386 seconds. The specific rate constant of the reaction is
Answer» `5.0xx10^(-2)s^(-1)` `5.0xx10^(-3)s^(-1)` `0.5xx10^(-2)s^(-1)` `0.5xx10^(-3)s^(-1)` ANSWER :D

Discussion

8.	Half life period of a first order reaction is 10min. Starting with initial concentration 12 M. The rate after 20 min is
Answer» `0.0693 xx3 MIN^(-1)` `0.0693 M min^(-1)` `0.0693 xx4 "M min"^(-1)` `0.693xx3 "M min"^(-1)` Solution :(A) `K=(0.693)/(t_(1//2))= (0.693)/(10)` =0.0693 `min^(-1)` `t= (2.303)/(k) "LOG" ([R]_(0))/([R])` `20 = (2.303)/(0.0693)"log" (12)/([R])` or `"log"(12)/([R]) = 0.601` `(12)/([R])` = Antilog (0.601)=4 . or `[R] =(12)/(4) =3` RATE after 20 min Rate =`k[R]^(1)` `=0.0693xx3` M `min^(-1)`

Discussion

9.	Half life ofreaction is found to be inversely proportional to the cube of its initial concentration.The order of reaction is
Answer» 2 5 3 4 Answer :D

Discussion

10.	Half-life of radius is 1580 yr. Its average life will be
Answer» `2.5 xx 10^(3) YR` `1.832 xx 10^(3) yr` `2.275 xx 10^(3) yr` `8.825xx 10^(2) yr` Solution :Average life PERIOD `= 1.44 xx t_(1//2)` `1.44 xx 1580 = 2275.2 = 2.275 xx 10^(3) yr`

Discussion

11.	Half life period is independent of initial concentration of reactant for
Answer» FIRST order REACTION second order reaction Zero order reaction Third order reaction. Solution :a) it is the CORRECT ANSWER.

Discussion

12.	Half-life of a reaction is found to be inversely proportional to the cube of its initial concentration. The order of reaction is
Answer» 2 5 3 4 Solution :Half-life of a reaction is found to be inversely proportional to the CUBE of initial concentration . The ORDER of REACTIONS is 4 .

Discussion

13.	Half - life of polonium is of
Answer» 138 days 56 days 13.8 days 110 days SOLUTION :Half-life of POLONIUM is 13.8 days

Discussion

14.	half life of a radioactive substance which disintegrates by 75% in 60 minutes will be
Answer» 120 MIN 30 min 45 min 20 min SOLUTION :75% of the substance disintegrates into two half lives. `:.` 2 half - lives = 60 min `:. t_(1//2) = 30` min.

Discussion

15.	Half-life of a reaction is found to be inversely proportional to the cube of initial concentration. The order of reaction is
Answer» 5 2 4 3 Solution :The half-life period of `N^(TH)` ORDER reaction is `t_(1//2)=(K)/(a^(n-1))` a is the initial concentration of reactant. `t_(1//2)prop(1)/(a^(3))""thereforet_(1//2)=(K)/(a^(3))` `(K)/(a^(n-1))=(K)/(a^(3))impliesa^(3)=a^(n-1)` `impliesn-1=3impliesn=4`. `therefore`The order of the reaction is 4.

Discussion

16.	Half-life of a radioactive substance is 120 days. After 480 days, 4 g will be reduced to
Answer» 2 1 0.5 0.25 Solution :`n = (480)/(120) = 4,N = (N_(0))/(2^(n)), N = (4)/(2^(4)) = (4)/(16) = 0.25 g`

Discussion

17.	Half-life of a radioactive element is 100 yr. The time in which it disintegrates to 50% of its mass, will be
Answer» 50 yr 200 yr 100 yr 25 yr Solution :`t = 100` YEARS

Discussion

18.	Half-life of a radioactive disintegration (A rarr B) having rate constant 231 s^(-1) is
Answer» `3.0 XX 10^(-2) s` `3.0 xx 10^(-3) s` `3.3 xx 10^(-2) s` `3.3 xx 10^(-3) s` SOLUTION :`t_(1//2) = (0.693)/(LAMDA) = (0.693)/(231 s^(-1)) = 3.0 xx 10^(-3) s`

Discussion

19.	Half-life of a first order reaction is 5xx10^(4)s. What percentage of the initial reactant will react in 2 hours? Calculate.
Answer» Solution :`k=(0.693)/(t_(1//2)) and k=(2.303)/(t)"log"([R]_(0))/([R])` From these two EQUATIONS, we have `(0.693)/(t_(1//2))=(2.303)/(t)"log"([R]_(0))/([R])` Substituting the VALUES, wehave `(0.693)/(5XX10^(4))=(2.303)/(7200)"log"([R]_(0))/([R])` or `"log" ([R]_(0))/([R])=(0.693xx7200)/(5xx10^(4)xx2.303)=(4989.6)/(11.515xx10^(4))=0.0433` or `([R]_(0))/([R])=1.105 or ([R])/([R]_(0))=(1)/(1.105)` PERCENTAGE `=(1)/(1.105)xx100=90.49%`

Discussion

20.	Half life of a first order reaction completes in 5 minutes. What persent of reactant reacts after 40 minutes ?
Answer» Solution :`t_(½) = 15 min` `K = (0.693)/(t_(½))` `= (0.693)/(15)` `K = 0.0462 min^(-1)` For first order REACTION, `K = (2.303)/(t) log_(10)((a)/(a-x))` `a = 100 K =0.0462 min^(-1), x = ? T = 40 min` `0.0462 = (2.303)/(40) log_(10)((100)/(a-x))` `log_(10)((100)/(a-x)) = 0.8024` `(100)/(a-x)` antilog (0.8024) `a-x = (100)/(6.345)` `a-x = 15.76` `x = 100 - 15.76` `x = 100 - 15.76` `:. x = 84.24%`

Discussion

21.	Half life of a certain zero order reaction, A rightarrow P is 2 hour when the initial concentration of the reactant, 'A' is 4 mol L^(-1). The time required for its concentration to change from 0.40 to 0.20 mol L^(-1) is
Answer» 0.75 min 2 min 12 min 8 min Answer :C

Discussion

22.	Half-lifeof ""^(210)Pb is 22 years. 2gram of lead is allowed decay for 11 years. (a) How much lead is left?and (b) Whatis the percentage decay ?
Answer» Solution :Radioactive decay follows FIRST ORDER kinetics Number of half - lives `"(n)"=("Total - TIME")/("Half - life")=(11y)/(22y)=1/2=0.5`. (a) AMOUNT of lead left undecayed `=("Amount of lead taken")/(2^(n))=(2)/(sqrt(2))=1.414g` (b) Amount of lead decayed = Amount taken - left amount `=2-1.414=0.586g` Percentage pf lead decayed `=(0.586xx100)/(2)=29.3%`.

Discussion

23.	Half-life is the time in which 50% of radioactive element disintegrates. Carbon -14 disintegrates 50% in 5770 years. Find the half-life of carbon -14
Answer» 5770 YEARS 11540 years `SQRT(5770)` years None of the above Solution :`t_(1//2) = 5770` years

Discussion

24.	Half-cell reactions are given below : Mn^(2+)+2e^(-) to Mn,E_(0)=-1.18V 2(Mn^(3+)+e^(-) to Mn^(2+)),E_(0)=+1.51V then what is E_(0) for 3Mn^(2+) to Mn+2Mn^(3+) ?
Answer» `-0.33V`, no REACTION `-0.33V`, reaction can be possible `-2.69V`, no reaction `-2.69V`, no reaction Solution :For given reactioin : `underset(+3)(3MN^(2+)) to underset(0)(M)n+underset(+3)(2MN)^(3+)` So, half OXIDATION : `2Mn^(2+) to 2Mn^(3+)+2e^(-)("Anode")` `UNDERLINE("and half reduction : "Mn^(2+)+2e^(-) to Mn" (Cathode)")` Total reaction : `3Mn^(2+) to Mn+2Mn^(3+)` So `E_("reaction")^(@)=E_("reduction")^(@)("Cathode")-E_("red")^(@)("Anode")` `=E_(Mn^(2+)\|Mn)^(@)-E_(Mn^(3+)\|Mn^(2+))^(@)` `=(-1.18-1.51)` `=-2.69V` `E_("Reaction")^(@) lt 0` so no reaction is possible and here, `E_(("Reaction"))^(@)=-2.68V=-2.69V`.

Discussion

25.	Half-life for radioactive C^(14) is 5760 years. In how many years 200 mg of C^(14) sample will be reduced to 25 mg
Answer» 11520 years 23040 years 5760 years 17280 years Solution :`25 = [(1)/(2)]^(N) xx 200, [(1)/(2)]^(n) = (25)/(200) = (1)/(8) = [(1)/(2)]^(3)` n = 3, NUMBER of half lives = 3 so time required `= 3xx 5760 = 17280 yr`

Discussion

26.	Half life for radioactive ""^(14)C is years. In how many years, 200 mg of ""^(14)C sample will be reduced to 25 mg ?
Answer» 23040 years 17280 years 115200 years 5760 years. Answer :B

Discussion

27.	Hair shampoos belong to which class of synthetic detergent ?
Answer» Solution :Hair shampoos are made up of cationic detergents. These are quartenary AMMONIUM SALTS of amineswith acetates, chloride or bromidesas anions e.g., cetyltrimethyl ammonium BROMIDE.

Discussion

28.	Hair shampoos belong to which class of detergents ?
Answer» SOLUTION :CATIONIC DETERGENTS.

Discussion

29.	Hair dyes contain:
Answer» COPPER nitrate Gold chloride Silver nitrate Copper sulphate Answer :C

Discussion

30.	Hair dyes contain
Answer» COPPER sulphate Gold chloride SILVER nitrate Copper nitrate Solution :HAIR dye CONTAINS silver nitrate.

Discussion

31.	Hair cream is ..................
Answer» gel EMULSION SOLID sol sol. SOLUTION :Emulsion dispersedphase Dispersionmedium - LIQUID

Discussion

32.	Hair and nail contains
Answer» CELLULOSE FAT KERATIN LIPID SOLUTION :keratin

Discussion

33.	Hair cream is
Answer» gel emulsion SOLID sol sol. Solution :Emulsion dipresed phase DISPERSION MEDIUM - liquid

Discussion

34.	Haemorrhage disease is caused by deficiency of which vitamin ?
Answer» SOLUTION :VITAMIN K.

Discussion

35.	Haemoglobin, the red pigment of blood acts as ______.
Answer» ANSWER :OXYGEN CARRIER

Discussion

36.	Haemophilia is a disease caused by deficiency of:
Answer» RBCs WBCs Thromboplastin Water in plasma Answer :C

Discussion

37.	haemoglobin is a complex of
Answer» `FE^(3+)` `Fe^(2+)` `Fe^(4+)` `Cu^(2+)` ANSWER :B

Discussion

38.	Human haemoglobin is made up of
Answer» 4 HAEME units and one globular protein 4haeme units and four globular protein 2 haeme units and one globular protein 4 haeme units and two globular protein Answer :A

Discussion

39.	haemoglobin (Hb) binded with oxygen and give oxy-haemoglobin (HbO_(2)). This process is partially regulated by the concentration of H_(3)O^(+) annd dissolved CO_(2) in blood as HbO_(2)+H_(3)O^(+)+CO_(2)hArr H^(+)-Hb-CO_(2)+O_(2)+H_(2)O If there is production of Lactic acid and CO_(2) in an muscle exercise then
Answer» More `O_(2)` is release More `HbO_(2)` is formed both (a) and (B) `CO_(2)` will be formed Answer :A

Discussion

40.	Haemoglobinin the example of
Answer» SIMPLE protein derived protein fibrous protein conjugated protein Answer :A::C::D

Discussion

41.	Haemoglobin is ……..
Answer» an ENZYME a globularprotein a VITAMIN CARBOHYDRATES ANSWER :B

Discussion

42.	Haemoglobin is…………………
Answer» an ENZYME a globular protein a VITAMIN CARBOHYDRATE Answer :B

Discussion

43.	Haemoglobin contains 0.334% of Fe by weight. The molecular weight of haemoglobin is approximately 67200. The number of Fe atoms present in one molecule of haemoglobin is
Answer» 1 6 4 2 Answer :C

Discussion

44.	Haemoglobin contains:
Answer» `CU` `Co` `FE` `Ni` ANSWER :C

Discussion

45.	Haemoglobin contains 0.33% of iron by weight. The molecular weight of haemoglobin is approximately 67200. The number of iron atoms (at. wt. of Fe = 56) present in one molecule of haemoglobin is
Answer» 6 1 4 2 Solution :FE PRESENT in 67200 u `=0.33/100 XX 67200` `=222U = 222/56 = 4` atoms.

Discussion

46.	Haemoglobin contains 0.33% of iron by weight . The mol. Wt. of haemoglobin is approx.67200. The no of iron atoms (at.wt. of Fe = 56) present in one molecule of haemoglobin are :
Answer» 1 2 4 6 Answer :C

Discussion

47.	Haemoglobin contains 0.25% iron by mass . The molecular mass of haemoglobin is 89600 then the number of iron atoms per molecule of haemoglobin(Atomic mass of fe = 56)
Answer» ANSWER :4

Discussion

48.	Haemoglobin contains 0.25% iron by mass. The molecular mass of haemoglobin is 89600. calculate the number of iron atoms per molecule of haemoglobin.
Answer» ANSWER :4 ATOMS

Discussion

49.	Haemoglobin contains 0.33% of iron by weight. The molecular mass of haemoglobin is about 67200. The number of iron atoms (at. mass of Fe=56) present in one molecule of haemoglobin is:
Answer» 6 4 2 1 Solution :Mass of 1 molecule of haemoglobin ` = (67200)/(6.022 XX 10^23)` Mass of IRON ` = (67200)/(6.022 xx 10^23) xx (0.33)/(100)` `= 3.682 xx 10^(-22) g ` 56 g of iron contains `6.02 xx 10^23` atoms ` 3.682 xx 10^(-22) g ` of iron contains ` = (6.02 xx 10^23)/(56) xx 3.682 xx 10^(-22)` = 4 atoms

Discussion

50.	Haemoglobin contain 0.33% or iron by weight. The molecular weight of haemoglobin is of haemoglobin is
Answer» 6 1 4 2 Solution :`"Fe PRESENT in 67200 u"=(0.3)/(100)xx67200=222u=(222)/(56)="4 atoms."`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

Half life period of a reaction is directly proportional to initial concentration of the reactant. What is the order of this reaction?

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Half life period of a first-order reaction is 1386 seconds . The specific rate constant of the reaction is

Half-life period of a first order reaction is

Half-life period of a first - order reaction is 1386 seconds. The specific rate constant of the reaction is

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Half - life of polonium is of

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Half-life of a reaction is found to be inversely proportional to the cube of initial concentration. The order of reaction is

Half-life of a radioactive substance is 120 days. After 480 days, 4 g will be reduced to

Half-life of a radioactive element is 100 yr. The time in which it disintegrates to 50% of its mass, will be

Half-life of a radioactive disintegration (A rarr B) having rate constant 231 s^(-1) is

Half-life of a first order reaction is 5xx10^(4)s. What percentage of the initial reactant will react in 2 hours? Calculate.

Half life of a first order reaction completes in 5 minutes. What persent of reactant reacts after 40 minutes ?

Half life of a certain zero order reaction, A rightarrow P is 2 hour when the initial concentration of the reactant, 'A' is 4 mol L^(-1). The time required for its concentration to change from 0.40 to 0.20 mol L^(-1) is

Half-lifeof ""^(210)Pb is 22 years. 2gram of lead is allowed decay for 11 years. (a) How much lead is left?and (b) Whatis the percentage decay ?

Half-life is the time in which 50% of radioactive element disintegrates. Carbon -14 disintegrates 50% in 5770 years. Find the half-life of carbon -14

Half-cell reactions are given below : Mn^(2+)+2e^(-) to Mn,E_(0)=-1.18V 2(Mn^(3+)+e^(-) to Mn^(2+)),E_(0)=+1.51V then what is E_(0) for 3Mn^(2+) to Mn+2Mn^(3+) ?

Half-life for radioactive C^(14) is 5760 years. In how many years 200 mg of C^(14) sample will be reduced to 25 mg

Half life for radioactive ""^(14)C is years. In how many years, 200 mg of ""^(14)C sample will be reduced to 25 mg ?

Hair shampoos belong to which class of synthetic detergent ?

Hair shampoos belong to which class of detergents ?

Hair dyes contain:

Hair dyes contain

Hair cream is ..................

Hair and nail contains

Hair cream is

Haemorrhage disease is caused by deficiency of which vitamin ?

Haemoglobin, the red pigment of blood acts as ______.

Haemophilia is a disease caused by deficiency of:

haemoglobin is a complex of

Human haemoglobin is made up of

Haemoglobinin the example of

Haemoglobin is ……..

Haemoglobin is…………………

Haemoglobin contains 0.334% of Fe by weight. The molecular weight of haemoglobin is approximately 67200. The number of Fe atoms present in one molecule of haemoglobin is

Haemoglobin contains:

Haemoglobin contains 0.33% of iron by weight. The molecular weight of haemoglobin is approximately 67200. The number of iron atoms (at. wt. of Fe = 56) present in one molecule of haemoglobin is

Haemoglobin contains 0.33% of iron by weight . The mol. Wt. of haemoglobin is approx.67200. The no of iron atoms (at.wt. of Fe = 56) present in one molecule of haemoglobin are :

Haemoglobin contains 0.25% iron by mass . The molecular mass of haemoglobin is 89600 then the number of iron atoms per molecule of haemoglobin(Atomic mass of fe = 56)

Haemoglobin contains 0.25% iron by mass. The molecular mass of haemoglobin is 89600. calculate the number of iron atoms per molecule of haemoglobin.

Haemoglobin contains 0.33% of iron by weight. The molecular mass of haemoglobin is about 67200. The number of iron atoms (at. mass of Fe=56) present in one molecule of haemoglobin is:

Haemoglobin contain 0.33% or iron by weight. The molecular weight of haemoglobin is of haemoglobin is