94420 + Interview Questions in Chemistry in Class 12 Page 607 InterviewSolution

30301.	The half life period of a substance is 50 minutes at a certain initial concentration. When the concentration is reduced to one half of its initial concentration, the half life period is found to be 25 minutes. Calculate the order of reaction.
Answer» Solution :Suppose theintial concentration in the first case is `a" mol L"^(-1)`. Then `[A_(0)]_(1)=a,(t_(1//2))_(1)=50" minutes"` `[A_(0)]_(2)=(a)/(2),(t_(1//2))_(2)=25" minutes"` We know that for a reaction of nth order, `t_(1//2)prop(1)/([A_(0)]^(n-1)):.((t_(1//2))_(1))/((t_(1//2))_(2))=([A_(0)]_(2)^(n-1))/([A_(0)]_(1)^(n-1))={([A_(0)]_(2))/([A_(0)]_(1))}^(n-1)` Substituting the VALUES, we get `(50)/(25)=((a//2)/(a))^(n-1)" or "(2)/(1)=((1)/(2))^(n-1)=((2)/(1))^(1-n)` or `""1-n=1" or "n=0.` Hence, the reaction is of zero-order. (b) Graphical method. GRAPHICALLY, half-life method can be used to test the orderof reaction as follows : `t_(1//2)prop(1)/([A_(0)]^(n-1))" or "t_(1//2)=K(1)/([A_(0)]^(n-1))` where K is a constant of proportionality (not the rate constant.) Thus, for zero order, `t_(1//2)=K[A_(0)].` Hence, plot of `t_(1 //2)"vs"[A_(0)]" will be linear passing through the origin and having SLOPE = K*"`.

Discussion

30302.	The half life period of a reaction is halved when the initial concentration is doubled. Calculate the order of reaction.
Answer» SOLUTION :`t_(1//2) propto 1/([A_(0)]^(n-1)` `(t_(1//2))_(1)/(t_(1//2))_(2)= ([A_(0)]_(2)^(n-1))/([A_(0)]_(1)^(n-1))` `(t)/(t_(1//2)) = ((2A)/(a))^(n-1)` `(2)^(1) = (2)^(n-1)` `n-1=1` or n=2

Discussion

30303.	The half life period of a reaction is 100 min . In 400 min the intitial concentration of 2.0 g willbecome :
Answer» `0.25 g ` `0.75g` `0.125g` `0.1g` Solution :(C ) `t_(1//2) = 100 ` mm 400 min =4 half LIFE PERIODS Amount of substance LEFT = `(A_(0))/(2^(4))` `(2.0)/(16) = 0.125 g `

Discussion

30304.	The half-life period of a radioactive substance is 8 years. After 16 years, the mass of the substance will reduce from starting 16.0 g to
Answer» 8.0 g 6.0 g 4.0 g 2.0 g Solution :`N = (16)/(8) = 2, N = (N_(0))/(2^(n)) = (16.0)/(2^(2)) = (16.0)/(4) = 4.0 g`

Discussion

30305.	The half life period of a radioactive substance is 5.27 years (decay constant =2.5xx10^(-7)min""^(-1)). The decay activity of 2.0 g of the sample is about :
Answer» `5XX10^(10)` dpm `7.5xx10^(15)` dpm `5xx10^(5)` dpm `7.5xx10^(20)` dpm. Solution :Decay ACTIVITY, `-(dN)/(DT)=lamdaxxN` `=2.5xx10^(-7)xx(2.0xx6.02xx10^(23))/(60)` `=5.0xx10^(15)` dpm

Discussion

30306.	The half life period of a radioactive substance is 15 minutes. What per cent of radioactivity of that material will remain after 45 minutes ?
Answer» 1 12.5 1.5 7.5 Answer :B

Discussion

30307.	The half-life period of a radioactive substance is 140 days. After 560 days, one gram of the element will be reduced to:
Answer» `(1)/(2)G` `(1)/(4)g` `(1)/(8)g` `(1)/(16)g` ANSWER :D

Discussion

30308.	The half life period of a radioactive substance having radioactive disintegration constant 231sec^(-1)is :
Answer» `3.0xx10^(-2)SEC` `3.0xx10^(-3)sec` `3.3xx10^(-2)` sec `3.3xx10^(-3)` sec. Answer :B

Discussion

30309.	The half-life period of a radioactive nuclide is 3 hours. In 9 hours its activity will be reduced by a factor of
Answer» `1//9` `1//8` `1//27` `1//6` SOLUTION :CAL. `N//N^(0)`

Discussion

30310.	The half life period of a radioactive nuclide is 3 hour . In 9 hour its activity will be reduced by
Answer» `1//9` `7//8` `1//27` `1//6` SOLUTION :

Discussion

30311.	The Half life period of a radioactive element P is same as the mean life time another radioactive element Q. Initially both of them have the same number of atoms.
Answer» <P>`P` and `Q` DECAY at the same rate always `P` will decay at a FASTER rate than `Q` `Q` will decay at a faster rate than `P` `P` and `Q` have the same decay rate initially Answer :B

Discussion

30312.	The half life period of a radioactive nucleide is 1 hour. In three hours its activity will be reduced by a factor of:
Answer» `(1)/(9)` `(1)/(6)` `(1)/(27)` `(1)/(8)` ANSWER :D

Discussion

30313.	The half life period of a radioactive isotope of X is 15 hours. How long will it take for its activity to be reduced to 1/16 of its original value ?
Answer» 30 hours 45 hours 60 hours 120 hours Answer :C

Discussion

30314.	The half-life period of a radioactive element is 30 minutes. One sixteenth of the original quantity of the element will remain unchanged after
Answer» 60 minutes 120 minutes 70 minutes 75 minutes Solution :`(1)/(16) =(1)/(2^(N)) or (1)/(2^(4)) = (1)/(2^(n)) or n = 4` `:.` REQUIRED times `= 4xx t_(1//2) = 120` MIN

Discussion

30315.	The half -life period of a radioactive element is 140 days. After 560 days, one gram of the element will reduced to
Answer» `(1)/(2) g` `(1)/(4) g` `(1)/(8) g` `(1)/(16) g` Solution :No. of half-lifes `= (560)/(140) = 4`, If 'n' half LIVES, the element will reduce of `((1)/(2))^(n) xx " initial weight " = (1)/(16) gm`

Discussion

30316.	The half-life period of a radioactive element is 140 days. After 560 days, one gram of the element will reduce to
Answer» `(1//2) G` `(1//4) g` `(1//8) g` `(1//6) g` Solution :FRACTION REMAINING `= ((1)/(2))^(N)` `= ((1)/(2))^(560//140) = ((1)/(2))^(4) = (1)/(16)`

Discussion

30317.	The half life period of a radioactive element is 140 days . After 560 days , 1 g of element will be reduced to ........
Answer» `(1/2)G` `(1/4)g` `(1/8)g` `(1/16)g` Solution :in 140 days `implies` initial CONCENTRATION reduced to `(1/2)g` in 280 days `implies` initial concentration reduced to `(1/4)g` in 420 days `implies` initial concentration reduced to `(1/8)g` in 560 days `implies` initial concentration reduced to `(1/16)g`

Discussion

30318.	The half life period of a radioactive element is 140 days . After 560 days , 1 g of the element will reduce to
Answer» 0.5 g 0.25 g 1/8 g 1/16 g Solution :`1 g m overset(140) (to) (1)/(2) overset(140) (to) (1)/(4) overset(140)(to) (1)/(8) overset(140) (to) (1)/(16) s`

Discussion

30319.	The half life period of a radioactive element is 140 days. After 560 days, 1 g of element will be reduced to
Answer» `((1)/(2))g` `((1)/(4))g` `((1)/(8))g` `((1)/(16))g` Answer :D

Discussion

30320.	The halflife periodof a radioactive element is 140days.After280days 1g of elementwill bereduced to whichamountof the following ?
Answer» `1/4` `1/16` `1/8` `1/2` SOLUTION :`1/16`

Discussion

30321.	Half-life periodof a radioactive element is 100 seconds. Calculate the disintegration constant and average life. How much time will it take to lose its activityby 90%?
Answer» ANSWER :`0.00693 s^(-1), 144.3 s, 332.3 s`

Discussion

30322.	The half life period of a radioactive element is 120 days. Starting with 1 g, the amount of element decayed in 600 days will be :
Answer» `(1)/(16)G` `(15)/(16)g` `(1)/(32)g` `(31)/(32)g` Answer :D

Discussion

30323.	The half-life period of a order reaction is 15 minutes. The amount of substance left after one hour will be:
Answer» `(1)/(4)` of the ORIGINAL amount `(1)/(8)` of the original amount `(1)/(16)` of the original amount `(1)/(32)` of the original amount Solution :Given `t_((1)/(2)=15`minutes Total time (T)=1hr=60min From `T=ntimest_((1)/(2))` `n=(60)/(15)=4` Now from the formula `(N)/(N_(0))=((1)/(2))^(n)=((1)/(2))^(4)=(1)/(16)` Where `N_(0)=` INITIAL amount N=amount LEFT after times t hence the amount of substance left after 1 hour will be `(1)/(16)`

Discussion

30324.	The half life period of a particular isotope is 10 years. Its decay constant is:
Answer» `6.932"YEAR"^(-1)` `0.6932"year"^(-1)` `0.06932"year"^(-1)` `0.006932"year"^(-1)` ANSWER :C

Discussion

30325.	The half-life period of a first order reaction is given by
Answer» `t_(1//2)= 0.693/K` `t_(1//2)= k/0.693` `t_(1//2)= 2.303` In 2 `t_(1//2) = 2.303 (a/2)` Solution :For a 1ST order REACTION, `t_(1//2) = (0.693)/(k)` .

Discussion

30326.	The half-life period of a first order reaction is 15 minutes. The amount of substance left after one hour will be
Answer» one half one FOURTH one eighth one sixteenth Solution :Half-lives `=n=t/t_1= (1 HR)/(15 MIN)= 60/50=4` AMOUNT lift `=[A]/[A]_0= 1/2^n= 1/2^4=1/16= [A]=[A]_0/16`

Discussion

30327.	The half-life period of a first order reaction is 60 seconds. Calculate the rate constant.
Answer» SOLUTION :`K=0.693/(t_1/2)=0.693/60=0.0115sec^-1`

Discussion

30328.	The half life period of a first order reaction is 60 min. What percentage will be left after 240 min.
Answer» SOLUTION :`6.25% `

Discussion

30329.	The half-life period of a first order reaction is
Answer» directly proportional to the INITIAL CONCENTRATION a inversely proportional to a independent of a independent of the rate constant of the REACTION Solution :For a 1ST order `t_(1//2)` is independent of 'a' . (Initial concentration) .

Discussion

30330.	The half-life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be
Answer» 230.3 minutes 23.03 minutes 46.06 minutes 460.6 minutes Answer :C

Discussion

30331.	The half life period of a first order reactions is 10 mins. What percentage of the reactant will remain after one hour ?
Answer» SOLUTION :`1.563%`

Discussion

30332.	The half life period of a first order reaction is 10 mins, what percentage of the reactant will remain after one hour?
Answer» SOLUTION :`1.563%`

Discussion

30333.	The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log2=0.301)
Answer» 23.03 minutes 46.06 minutes 460.6 minutes 230.3 minutes Answer :B

Discussion

30334.	The half-life period of a certain reaction is directly proportional to initial concentration of the reactant. Predict the order of the reaction and white the expression to calculate the half-life period of the reaction.
Answer» SOLUTION :Zero order `t_(½) = ([R_(0)])/(2K)`

Discussion

30335.	The half life period of a first order chemical reaction is 6.93 minutes . The time required for the completion of 99% of the chemical reaction will be ( log 2 = 0.301 ) :
Answer» 230.3 MINUTES 23.03 minutes 46.06 minutes 460.6 minutes Solution :(C ) `k=(0.693)/(t_(1//2))` `= (0.693)/(6.93)=0.1 "min"^(-1)` `t= (2.303)/(k) "LOG" ([A]_(0))/([A])` `= (2.303)/(0.1) "log" (100)/(1) ` `=(2.303xx2)/(0.1) = 46 .06 min`

Discussion

30336.	The half life period of ._(58)^(141)Ce is 13.11 days. It is a beta-particle emitter and the average energy of the beta-particle emitted is 0.442 MeV. What is the total energy emitted per second in watts by 10 mg of ._(58)^(141)Ce?
Answer» SOLUTION :Rate of disintegration per sec =` LAMBDA xx` No. atoms `= (0.693)/(13.11 xx 24 xx 60 xx 60) xx (6.023 xx 10^(23))/(141) xx 0.01` Total `beta`- particles emitted ` 2.61 xx 10^(23)` Total energy emittd `= 2.61 xx 10^(13) xx 0.422 = 1.1536 xx 10^(12) MEV` Energy in erg `= (1.1536 xx 10^(13))(1.6 xx 10^(-6))` Energy in watt `= (1.536 xx 10^(13) xx 1.6 xx 10^(-6))/(10^(7)) = 1.84` watt

Discussion

30337.	The half life period of a 1st order reaction is 30 seconds. Calculate its rate constant.
Answer» SOLUTION :`t_(1//2)=0.6931/K` `K=0.6931/30=0.0231sce^(-1)`

Discussion

30338.	The half-life period of ._(53)I^(125) is 60 days. What percent of radioactivity would be present after 180 days
Answer» 0.25 `12.5%` 0.5 0.75 Solution :`N = (N_(0))/(2^(n)) n = (180)/(60) = 3` `(N)/(N_(0)) = (1)/(2^(3)) rArr (N)/(N_(0)) xx 100 = (1)/(8) xx 100 = 12.5%`

Discussion

30339.	The half-life period of a 1^(st) order reaction is 60 minutes. What percentage will be left over after 240 minutes ?
Answer» 0.0625 0.0425 0.05 0.06 Solution : `:t_(1//2)=(0.693)/(K)Rightarrow(0.693)/(t_(1//2))=Krightarrow(0.693)/(60)=K` `K=0.01155min^(-1)` `K= (2.303)/(t)LOG ((a)/(a-x))` LET the INITIAL amount (a) 100 `0.01155min^(-1)= (2.303)/(240min)log((100)/(a-x))` `(0.01155min^(-1)xx40min)/(2.303)=log ((100)/(a-x))` `1.204=log100-(a-x)` `1.204=2-log(a-x)` `log(a-x)=2-1.204` log(a-x)=0.796 (a-x)=6.25%

Discussion

30340.	The half life period for the first order reaction XY_(2) rarr X+Y_(2) is 10 minutes . In what period would the concentration of XY_(2) be reduced to 10% of the original concentration ?
Answer» 33.2 min 90 min 120 min 26.3 min SOLUTION :(A ) `K = (0.693)/(t_(1//2)) ` `= (0.693)/(10) = 0.0693 min^(-1)` `t = (2.303 )/(k) "LOG" ([A]_(0))/(0.1[A]_(0))` `= (2.303)/(0.0693)xx1= 33.2 `

Discussion

30341.	The half-lifeperiodfor afirstorderreactionis 69.3 S . Itsrateconstant is
Answer» `10^(-2) s^(-1)` ` 10^(-4)s^(-1)` `10s^(-1)` `10^(2)s^(-1)` SOLUTION :RATECONSTANT `=(0.693)/(t_(1//2)) =(0.693)/(69.3)=10^(-2) S^(-1)`

Discussion

30342.	The half life period for a zero order reaction is equal to
Answer» `0.693/K` `a/(2K)` `1/(KA)` `a/K` ANSWER :C

Discussion

30343.	The half life period for a reaction at initial concentration of 0.2 and 0.4 mol L^(-1) are 100 s and 200 s respectively. The order of the reaction is
Answer» 0 1 2 3 Answer :A

Discussion

30344.	The half life period for a radioactive substance is 15 minutes.How many gms of this substance after one hour?
Answer» 25 46.875 43.75 37.5 Solution :Method-I:`Ao overset(15)tooverset(Ao)underset(2)-overset(30)tooverset(Ao)underset(4)-overset(45)tounderset(8)overset(Ao)tooverset(60)tounderset(16)overset(Ao)-` After 60 MIN ,`(50)/(16)`=3.125 GRAM substance REMAINS.`therefore` 50-3.125=46.875 gram substance decay. Method -II radioactive are ALWAYS first order reaction. `therefore t_((1)/(2))=(0.693)/(K)` `therefore K=(0.693)/(t_((1)/(2)))=(0.693)/(15)=0.0462"minute"^(-1)` Now K=`(2.303)/(t)` log `([R]_(0))/([R]_(t))` 0.0462=`(2.303)/(60)` log `(50)/([R]_(t))` `therefore` log50-log `[R]_(t)` =1.2036 log `[R]_(t)`=0.495 `[R]_(t)`=antilog 0.495`~~3.12` `therefore` 50-3.12`~~`46.875 gram substance decay.

Discussion

30345.	The half life period for a first order reaction is
Answer» PROPORTIONAL to CONCENTRATION INDEPENDENT of concentration inversely proportional to concentration inversely proportional to the SQUARE of the concentration Answer :B

Discussion

30346.	The half-life period for a first order reaction is:
Answer» Proportional to CONCENTRATION Independent of concentration Inversely proportional to concentration Inversely proportional to the SQUARE of the concentration Solution :`t_(1//2) = (0.693)/(K)`, i.e, half LIFE is independent of concentration

Discussion

30347.	The half-life of the ratio element ._(83)Bi^(210) is 5 days. Starting with 20 g of this isotope, the amount remaining after 15 days is
Answer» 10 G 5 g 2.5 g 6.66 g Solution :`n = (15)/(5) = 3, N = (N_(0))/(2^(n)) = (20)/(2^(3)) = (20)/(8) = 2.5 g`

Discussion

30348.	The half life of the nucleide .^(220)Rn is 54.5s. What mass of radon is equivalent to 1 millicurie (mci) ?
Answer» ANSWER :`1.06xx10^(-15)KG`

Discussion

30349.	The half life period for a 1st order reaction is 6.93 sec., its rate constant is
Answer» 10 `sec^-1` 100 `sec^-1` 0.1 `sec^-1` 1 `sec^-1` ANSWER :C

Discussion

30350.	The half life period for a 1st order reaction is 10 min. The initial amount of the reactant was 0.08 M and concentration at any instant is 0.01 M. The time taken for the reaction is
Answer» 10 min 20 min 30 min 40 min Answer :C

Discussion

Explore topic-wise InterviewSolutions in .

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The halflife periodof a radioactive element is 140days.After280days 1g of elementwill bereduced to whichamountof the following ?

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The half life period of a first order reaction is 60 min. What percentage will be left after 240 min.

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The half-life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be

The half life period of a first order reactions is 10 mins. What percentage of the reactant will remain after one hour ?

The half life period of a first order reaction is 10 mins, what percentage of the reactant will remain after one hour?

The half life period of a first order chemical reaction is 6.93 minutes. The time required for the completion of 99% of the chemical reaction will be (log2=0.301)

The half-life period of a certain reaction is directly proportional to initial concentration of the reactant. Predict the order of the reaction and white the expression to calculate the half-life period of the reaction.

The half life period of a first order chemical reaction is 6.93 minutes . The time required for the completion of 99% of the chemical reaction will be ( log 2 = 0.301 ) :

The half life period of ._(58)^(141)Ce is 13.11 days. It is a beta-particle emitter and the average energy of the beta-particle emitted is 0.442 MeV. What is the total energy emitted per second in watts by 10 mg of ._(58)^(141)Ce?

The half life period of a 1st order reaction is 30 seconds. Calculate its rate constant.

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