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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
An isotope of Potassium `._(19)^(40) K ` has a half life of `1.4 xx 10^(9)`year and decays to Argon `._(18)^(40)Ar`which is stable. (i) Write down the nuclear reaction representing this decay. (ii) A sample of rock taken from the moon contains both potassium and argon in the ratio `1//3`. find age of rock |
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Answer» Correct Answer - (i) `._(19)^(40) K rarr . _(18)^(40) Ar + ._(+1) beta ^(0) +v` (i) `._(19)^(40)Krarr._(18)^(40)Ar+.._(+1)" "^(0)e` (positron) (ii) `t=(1)/(lambda) ln .(1)/((1//4))" " (K)/(Ar)=(1)/(3)` (Given ratio) `=(1.4 xx10^(9))/(ln2) ln4 " " k=((1)/(4))` `=2.8 xx 10^(9)` years |
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| 152. |
Two reactions having their energy of activation `E_(1)` and `E_(2)` temperature coefficients `T_(c_(1))` and `T_(c_(2))` respectively within the temperature 300 and `310K`. The ratio of their temperature coefficient is:A. `e^(E_(1)//E_(2))`B. `e^((E_(1)-E_(2))xx10^(-4)//R)`C. `10^(E_(1)//E_(2))`D. `e^((E_(1) - E_(2))//4)` |
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Answer» Correct Answer - `(b)` `2.303 log t_(C_(1)) = (E_(1))/(R) [(T_(2) - T_(1))/(T_(1) T_(2))]` `2.303 log T_(c_(2)) = (E_(2))/(R) [(T_(2) - T_(1))/(T_(1) T_(2))]` `2.303 [log T_(C_(1)) - log T_(C_(2))]` `= ((E_(1) - E_(2)))/(R) xx [(310 - 300)/(300xx310)]` `:. 2.303 log ((T_(C_(1)))/(T_(C_(2)))) = [(E_(1) - E_(2))/(R)] xx 1.07xx10^(-4)` or In `(T_(c_(1)))/(T_(C_(2))) = [(E_(1) - E_(2))/R] xx 10^(-4)` `:. (T_(c_(1)))/(T_(c_(2))) = e^([(E_(1) - E_(2))/(R)]xx10^(-4))` |
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| 153. |
The activity of a radioactive substance is `R_(0)` at `t = 0`, `R_(1)` at `t = t` and `R_(2)` at `t = 2t`. The decay constant of this species is/are given by:A. `(log_(e) R_(1) - log_(e) R_(2))/(t)`B. `(log_(e) R_(0) - log_(e) R_(1))/(t)`C. `(log_(e) R_(2) - log_(e) R_(1))/(2t)`D. `(log_(e) R_(0) - log_(e) R_(1))/(2t_(2))` |
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Answer» Correct Answer - `(a,b,d)` `R_(1) = R_(0) e^(-lambda t)` `R_(2) = R_(0) e^(-lambda 2t)` `:. (R_(1))/(R_(2)) = e^(-lambda t)` or `lambda = (log_(e) R_(1) - log_(e) R_(2))/(t)` Similarly, `lambda = (log_(e) R_(0) - log_(e) R_(1))/(t)` `lambda =(log_(e) R_(0) - log_(e) R_(2))/(2t)` |
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| 154. |
The rate of decay of a radioactive species is given by `R_(1)` at time `t_(1)` and `R_(2)` at later time `t_(2)`. The mean life of this radioactive species is:A. `(t_(2) - t_(1))/("In" R_(1) - "In" R_(2))`B. `(t_(2) - t_(1))/("In" R_(2) - "In" R_(1))`C. `(t_(2) + t_(1))/("In" R_(2)+ "In" R_(1))`D. `(t_(2) - t_(1))/("In" R_(2)+ "In" R_(1))` |
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Answer» Correct Answer - `(a)` `R_(1) = lambdaN_(1), R_(2) = lambda N_(2)` `:. (R_(1))/(R_(2)) = (N_(1))/(N_(2))` Also `t_(1) = (1)/(lambda) log_(e) ((N_(0))/(N_(1)))` `t_(2) = (1)/(lambda) log_(e) ((N_(0))/(N_(2)))` `:. T_(1) - t_(2) = (1)/(lambda) [log_(e) ((N_(0))/(N_(1))) - log_(e) ((N_(0))/(N_(2)))]` `= (1)/(lambda) log_(e) ((N_(2))/(N_(1)))` `= (1)/(lambda) log_(e) ((R_(2))/(R_(1)))` `T = (1)/(lambda) = (t_(1) - t_(2))/(log_(e) R_(2) - log_(e) R_(1))` |
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| 155. |
The rate of decay of a radioactive species is given by `R_(1)` at time `t_(1)` and `R_(2)` at later time `t_(2)`. The mean life of this radioactive species is:A. `T = ((t_(1) - t_(2)))/("In" (R_(2)//R_(1)))`B. `T = ((t_(2) - t_(1)))/("In" (R_(2)//R_(1)))`C. `T = ((t_(2) - t_(1)))/("In" (R_(1)//R_(2)))`D. `T = ("In" (R_(2))//R_(1))/((t_(2)- t_(1)))` |
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Answer» Correct Answer - `(a)` `R_(1) = R_(0) e^(-lambda t_(1))` and `R_(2) = R_(0) e^(-lambda t_(2))` Now `(R_(2))/(R_(1)) = e^(lambda(t_(1)-t_(2)))` or In `((R_(2))/(R_(1))) = lambda (t_(1) - t_(2))` `(1)/(lambda) = ((t_(1) - t_(2)))/("In"(R_(2)//R_(1)))` or `T = (t_(1) - t_(2))/("In" (R_(2)//R_(1)))` |
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| 156. |
A radioactive species undergoes decay for time `t` where `t = 4t_(1//2)` . The average life `(T)` of species can therefore be given by:A. `2t` in `2 = T`B. `4t` in `2 = T`C. `T = (t)/(4 "In" 2)`D. `T = (t)/(2 "In" 2)` |
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Answer» Correct Answer - `(c)` `T = (1)/(lambda) = (t_(1//2))/(0.693) = (t)/(4xx0.693) = (t)/(4 "in" 2)` |
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| 157. |
The half line of a radioactive element is `2n` year. The fraction decayed in `n` year.A. `0.10`B. `0.29`C. `1.414`D. `0.414` |
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Answer» Correct Answer - `(b)` `N = (N_(0))/((2)^(1//2)) = (N_(0))/(sqrt(2))` `:.` Fraction decayed `((N_(0) - N))/(N_(0)) = (sqrt(2) - 1)/(sqrt(2)) = 0.29` |
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| 158. |
The half of a radioactive sample is `2n` year. The fraction left undecyed after `n` year is:A. `(1)/(2)`B. `(1)/(sqrt(2))`C. `(1)/(sqrt(3))`D. `2` |
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Answer» Correct Answer - `(b)` `T = t_(1//2) xx p` `n = 2n xx p :. P = (1)/(2)` `:. N = (N_(0))/(sqrt(2))` |
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| 159. |
Calculate the no. of `alpha-` and `beta-` particles given during the change: `._(93)^(237)Np rarr ._(83)^(209)Bi` Also report the nature and name of this radioactive series. |
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Answer» `._(93)Np^(237) rarr ._(83)Bi^(209) + a_(2) He^(4) + b _(-1)e^(0)` Where `a` amd `b` are `alpha` and `beta-` particles given out during change. Equating mass number: `237 = 209 + 4a + 0 xx b` `a = 7` Equating atomic number, keeping `a = 7` `93 = 83 + 2xx7 + b xx (-1)` `b = 4` Thus `7 alpha`, and `4beta-` are given out during the change also `237//4 = 59` with remainder 1 and thus it is `(4n + 1)` series, the artificial radioactive sereis also called as napunium series with `n = 59`. |
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| 160. |
A radioactive element gets spilled over the floor of a room. Its half-life period is 30 days. If the initial activity is ten times the permissible value, after how many days will it be safe to enter the room ?A. 50 daysB. 30 daysC. 10 daysD. 100 days |
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Answer» Correct Answer - D Let the permissible value of value of activity =A Initial activity =10A Given `t_(1//2)` =30 days `" " :.k=(ln2)/(30)` Let after time t activity reduces from 10A to permissible value (A) `t=(l)/(k) ln ((10A)/(A))=(30)/(ln(2))ln(10)=(30 xx 2.303)/(0.693)=100 ` days |
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| 161. |
An acid solution of `KReO_(4)` sample containing `26.83mg` of combined rhenium was reduced by passage through a column of granulated zinc. The effulent solution including the washings from the column, was then titrated with `0.5N KMnO_(4)`. `11.45mL` of the standard `KMnO_(4)` was required for the reoxidation of all the rhendium of all the rehniuym to the perryhentate ion `Re_O_(4)^(-)` . Assuming that rhenium was the only element reduced, what is the oxisation state to which rhenium was reduced by the zinc column? |
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Answer» Correct Answer - `+3` |
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| 162. |
A sample of pithchbelen is found to contain `50%` uranium and `2.425%` lead. Of this lead only `93%` was `Pb^(206)` isotope. If the disintegration constant is `1.52xx10^(-10)yr` how old could be the pithblende deposits? |
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Answer» Correct Answer - `3.3xx10^(8)` year |
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| 163. |
The number of neutrons emitted when `._(92)^(235)U` undergoes controlled nuclear fission to `._(54)^(142)Xe` and `._(38)^(90)Sr ` is |
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Answer» Correct Answer - 4 `._(92)^(235)U+._(0)n^(1) rarr ._(54)^(142) Xe+._(38)^(142)Xe+._(38)^(90)Sr+x ._(0)n^(1)` Applying mass number balance `235+1=142+90+x` `x=4` |
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| 164. |
One mole of `N_(2) H_(4)` loses `10` mole electrons to form a new compound `Y`. Assuming that all the `N_(2)` appears in new compound, what is oxidation state of `N` in `Y` ? There is no charge in oxidant state of `H`. |
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Answer» Correct Answer - `+3` |
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| 165. |
A `10g` mixture of `Cu_(2)S` and `CuS` was treated with `200 mL` of `075M MnO_(4)^(-)` in acid solution producing `SO_(2), CU^(2+)` and `Mn^(2+)`. The `SO_(2)` was boiled off and the excess of `MnO_(4)^(-)`. The `SO_(2)` was boiled off and the excess of `MnO_(4)^(-)` was titrated with `175 mL` of `1M Fe^(2+)` solution. Caculaate `%` fo `CuS` in original mixture. |
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Answer» Correct Answer - `57.94%` |
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| 166. |
(a) `CusO_(4)` reacts with `KI` in acidic medium to liberate `I_(2)`: `2CuSO_(4) + 4KI rarr Cu_(2)I_(2) + 2K_(2) SO_(4) + I_(2)` (b) Merrcuric per isodate `Hg_(5)(IO_(6))_(2)` reacts with a mixture of `KI` and `HCI` follwing the equaction: `Hg_(5)(10_(6))_*(2) + 34KI + 24HCI rarr 5K_(2) HgI_(4) + 8I_(2) + 24KCI + 12H_(2)O` (c) The liberted iodine s titrated against `Na_(2)S_(2)O_(3)` solution. One mL of whcih is equivalent to `0.0499g` of `CuSO_(4).5H_(2)O. 5H_(2)O`. What volume in `mL` of `Na_(2)S_(2)O_(3)` solution will be required to react with `I_(2)` liberated from `0.7245g` of `Hg_(5) (10_(6))_(2)`? `M. wt`, of `Hg_(5) (10_(6))_(2) = 1448.5` and `M.wt` of `CuSO_(4). 5H_(2)O = 249.5` |
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Answer» Correct Answer - `40mL` |
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| 167. |
Natural nitrogen atoms has found to exist in two isotopic forms, `._(7)N^(14)` with mass `14.0031` and `._(7)N^(15)` with mas `15.0001` amu. Which isotope is more stable? Assume mass of `n` and `p` to eb `1.00893` and `1.00757` amu. |
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Answer» Correct Answer - `N^(15)` |
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| 168. |
Consider an `alpha-` particle just in contact with a `._(92)U^(238)` nucleus. Calculate the coulombic repulsion energy (i.e., the height fo coulombic barrier between `U^(238)` and `alpha-` particle.)Assume that the distance between them is equal to the sum of their radii. |
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Answer» Correct Answer - `26.14 MeV` |
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| 169. |
In the transormation fo `._(992)^(238)U` to `._(92)^(234)U`, if one emission is an `alpha-` particle, what should be the other emission(s)?A. Two `beta^(-)`B. Two `beta^(-)` and one `beta^(+)`C. One `beta^(-)` and one `gamma`D. One `beta^(+)` and one `beta^(-)` |
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Answer» Correct Answer - `(a)` `._(92)^(238)U rarr ._(92)^(234)U + ._(2)^(4)He + 2 ._(-1)^(0)e` |
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| 170. |
`._(92)U^(238)` is a neutral `alpha-` emitter. After `alpha-` emission, the residual nucleus callled `UX_(1)` in turns emits `a beta^(-1)` particle to produce another nucleuis `UX_(2)` Find out the atomic and mass numbers of `UX_(1)` and `UX_(2)`. Also if uranium belongs to `III gp` to which group `UX_(1)` and `UX_(2)` belong. |
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Answer» Correct Answer - `90,234,91,234,III gp` |
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| 171. |
The nucler reacion, `._(29)^(63)Cu + ._(2)^(4)He rarr ._(17)^(37)Cl + 14 ._(1)^(1)H + 16 ._(0)^(1)n` represents:A. Artofoca, radopactivityB. Induced radioactivityC. Nuclear fissionD. Spallation reaction |
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Answer» Correct Answer - `(d)` A large mp. Of sub-atomic particles are emitted out during spallation reaction. |
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| 172. |
What mass of `K_(2)Cr_(2) O_(7)` is required to produce `5.0` litre `CO_(2)` at `75^(@)C` and `1.07` atm pressure from excess of oxlaic acid? Also report the volume of `0.1N NaOH` required to neutralise the `CO_(2)` evolved. |
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Answer» Correct Answer - `K_(2)Cr_(2)O_(7) = 9.18g, V = 3.746`litre |
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| 173. |
`KMnO_(4)` oxidizes `X^(n+)` ion to `XO_(3)^(-)` itself changing to `Mn^(2+)` in acid solution. `2.68xx10^(-3)` moel of `K_(2)Cr_(2)O_(7)` was required `1.61xx10^(-3)` mole of `MnO_(4)^(-)`. What is the value of `n` ? Also calcualte the atomic mass of `X`, if the weight of `1g` equivalent of `XCl_(n)` is 56 |
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Answer» Correct Answer - `2,97` |
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| 174. |
A steel sample is to be analysed fo r`Cr` and `Mn` simultaneously. By suitable treatment the `Cr` is oxidised to `Cr_(2)O_(7)^(2-)` and the Mn to `MnO_(4)^(-)` . A 10.00g sample of steel is used to produce `250.0mL` of a solution containing `Cr_(2)O_(7)^(2-)` and `MnO_(4)^(-).10 mL` of this solution is added to `BaCl_(2)` solution and by prooper adjument of the `pH`, the chromium is completely precipitate as `BaCrO_(4) (0.0549g)`. The second `10 mL` solution protion requires exactly `15.95 mL` of `0.0750 M` standard for complete reduction of this solution. Calculate `%` of `Mn` and `Cr` in this sample. |
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Answer» Correct Answer - `Cr = 2.823%, Mn = 3.29%` |
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| 175. |
A steel sample is to be analysed fo r`Cr` and `Mn` simultaneously. By suitable treatment the `Cr` is oxidised to `Cr_(2)O_(7)^(2-)` and the Mn to `MnO_(4)^(-)` . A 10.00g sample of steel is used to produce `250.0mL` of a solution containing` Cr_(2)O_(7)^(2-)` and `MnO_(4)^(-)` (a) A `10.00mL` portion of this solution is added to a `BaCl_(2)` solution and by proper adjument of the acidity, the chromimum is completeley precipitated as `0.0549g BaCrO_(4)`. (b) A second `10.00mL` portion of this soltuion requires exactily `15.95mL` of `0.0750M` standard `Fe^(2+)` solution for its titaration (in acid solution). Calculate the `%` of `Cr` in the steel sample. `(Cr = 52, Mn = 55, Ba = 137)` |
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Answer» Correct Answer - `%` of `Cr = 2.821%` of `Mn = 1.496%` |
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| 176. |
Two solutions of `0.1M cr_(2) O_(7)^(2-)(aq.)` and `0.1M MnI_(4)^(-) (aq.)` are to be used to titatre (titrating solution) be required for a given solution of `fe^(2+)(aq.)` (b) If a given titration requires `24.50 mL` of `0.100M Cr_(2)O_(7)^(2-) (aq.)`,how many `mL` of `0.100M MnO_(4)^(-) (aq.)` would have been required if it had been used instead? |
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Answer» Correct Answer - (a) Vol. Of `MnO_(4)^(-)` required will be `1.2` times fo vol. of `Cr_(2)O_(7)^(2-)`, (b) `29.4mL` |
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| 177. |
`0.84g` iron are containing `X` per cent of iron was taken in a solution containg all the iron in ferrous state. The solution required `X ml` of a porassium dichromate soltuion for oxidation of iron content to ferric state. Calculate the strength of potassium dichromate solution. |
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Answer» Correct Answer - `7.35 g//"litre"` |
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| 178. |
`30mL` of a soltuion containg `9.15g//"litre"` of an oxalte `K_(X) H_(Y) (C_(2)O_(4))_(Z).nH_(2)O` are required for titrating `27mL` of `0.12N NaOH` and `36mL` of `0.12N KMnO_(4)` separalty. Calculalte `X,Y,Z` are in the simple ratio of `g` atoms. |
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Answer» Correct Answer - `X = 1, Y = 3, Z = 2, n = 2` |
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| 179. |
`0.2828g` of iron wire was dissolved in excess dilute `H_(2)SO_(4)` and the solution was made up to `100 mL`. `20mL` of this solution required `30mL` of `N//30 K_(2)Cr_(2)O_(7)` doluyion for exact oxidation. Calculate `%` purity of `Fe` in wire. |
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Answer» Correct Answer - `99%` |
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| 180. |
A sample of `._(131)^(53) I` is iodide ion was admistered to a patient in a carrier consisting of `0.10 mg` of stable iiodide ion. After 4 day `67.7%` of initial radioactivity was detected in the throid gland of the patient. What mass of stable iodide ion had migrated to thyroid glad? (`t_(1//2)` for iodide ion = 8 day) |
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Answer» If `t = 4`day, `lambda = (0.693)/(8)`, then since `r_(0) prop N_(0)` and `r prop N` `:. (r_(0))/(r) = (N_(0))/(N)` `:. t = (2.303)/(lambda) log (r_(0))/(r)` `4 = (2.303xx8)/(0.693) log (1)/(0.693) log (r_(0))/(r)` `r = 0.707 r_(0)` Thus iodide ion left is `0.707` part of intially injected sample, however the rate decreases only `67.7%` or `0.688` in 4 days, thus If `0.707` is left then iodidie ion migrated to thryorid `=1` Thus `0.677` is left than iodide ion migrated to thyroid `= (1xx0677)/(0.707) = 0.958` or `95.8%` of the iodide ion is migrated to gland. |
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| 181. |
For estimating ozone in the air, a ceratin volume of air is passed through an acidified or neutral `KI` solution when oxygen is evolved and iodide is oxidised to give iodine. When such a solution is acidified, free iodine is evolved which can be be titrated with standard `Na_(2)S_(2)O_(3)` solution. In an experiment `10` litre of air at1 atm and `27^(@)C` were passed through an alkaline `KI` solution, at the end, the iodine entrapped in a solution on titration as above required `1.5mL` of `0.01N Na_(2) S_(2) O_(3)` solution. Calculate volume `%` fo `O_(3)` in sample. |
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Answer» Correct Answer - `1.847x10^(-3)%` |
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| 182. |
A `200 mL` sample of a citrus fruit drinks containing ascorbic acid (vitamin `C, mol. We 176.13`) was acidified with `H_(2)SO_(4)` and `10mL` of `0.250M I_(2)` was added. Some of the iodine was reduced by the ascorbic acid to `I^(-)`. The excess of `I_(2)` required `4.6mL` of `0.01M Na_(2)S_(2)O_(3)` for reduction. What was the vitamin `C` content of the drink in `mg` vitamin per `mL` drink? The reactions are: `C_(6) H_(8) O_(6) + I_(2) rarr C_(6) H_(6) O_(6) + 2HI` `5H_(2)O + S_(2)O_(3)^(2-) + 4I_(2) rarr 2SO_(4)^(2-) + 8I^(-) + 10H^(-)` |
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Answer» Correct Answer - `0.058mg//mL` |
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| 183. |
A mixture containg `As_(2)O_(3)` and `As_(2)O_(5)` and required `20.10mL` of `0.05N` iodine for titration. The resulting solution is then acidified and excess fo `KI` was added. The liberated iondine required `1.1113g` hypo `(Na_(2)S_(2)O_(3) 5H_(2)O)` for complete reaction. Calculate mass of mixture. The reactions are: `AsO_(3) + 3l_(2) + 2H_(2)O rarr As_(2)O_(5) + 4H^(+) + 4I^(-)` `As_(2) O_(5) + 4H^(+) + 4l^(-) rarr As_(2) O_(3) + 2l_(2) + 2H_(2)O` |
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Answer» Correct Answer - `0.2496g` |
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| 184. |
A `2.5g` sample containing `As_(2) O_(5) Na_(2) HAsO_(3)` and inert substance is dissolved in water and the `pH` is adjusted to neutral with excess of `NaHCO_(3)`. The solution is titrated with `0.15M I_(2)` solution, requiring `11.3 mL` to just reach the end point, then the solution is acidified with `HCl, KI` is added and the liberated iodine requires `41.2 mL` fo `0.015M Na_(2)S_(2)O_(3)` under basic conditions where it converts to `SO_(4)^(2-)`. Calculate per cent compositon of mixture. |
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Answer» Correct Answer - `Na_(2) HAsO_(3) = 11.53%, As_(2)O_(5) = 3.57%` inert material `= 84.9%` |
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| 185. |
Hydroxylamine reduces iron `III` according to the equaction, `4Fe^(3+) + 2NH_(2)OH rarr N_(2) O + H_(2)O + 4 Fe^(2+) + 4H^(+)`. Orpm `II` thus produced is estimatedf by tiration with standard `KMnO_(4)` solution. The reaction is `MnO_(4)^(-) + 5Fe^(2+) + 8H^(+) rarr Mn^(2+) + 5Fe^(3+) + 4H_(2)O`. A `10 mL` of hydroxyamine solution was diluted to one litere. `50 mL` of this diluted soltuion was boiled with an excess of `Fe^(3+)` solution. The resulting solution required `12mL` of `0.02M KMnO_(4)` solution for complete oxidation of `Fe^(2+)`. Calculate the weight of `NH_(2)OH` in one litre of orignal solution. |
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Answer» Correct Answer - `39.6 g//"litre"` |
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| 186. |
A gaseous substance AB has a radioactive element A with a half life of 2 days. The gaseous substance decomposes to `A_((g))` and `B_((g))` in a closed container and an equilibrium gets established rapidly within a fraction of second . The substance to which A decays into is non gaseous and does not effect volume of the container. Which of the statement are correct regarding shifting of equilibria due to radioactive decay if data in question 1 is true.A. Due to radioactive decay, equilibria will shift rightwards.B. Equilibria will shift leftC. No shift in equilibria due to radioactive decay will occur.D. Equilibria may shift in any direction. |
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Answer» Correct Answer - C `K_(P)=(P_(A) xx P_(B))/(P_(AB))` As A and AB both are ratio active due to decay there mole ratio will not changes as both have same half life so NO shift in equilibria due to decay. |
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| 187. |
Radioactive decay follows first order kinetics and the rate constant is often termed as decay constant. Certain radioactive substances may undero sequential decays in order to convert into a stable nucleus. The series comprising all such elements is termed as radioactive disintegration series. A substance A undergoes sequential decay as shown `A overset(lambda_(1))rarr B overset (lambda_(2))rarr C`. If the decay constant `lambda_(1)` and `lambda_(2)` are `4 xx10^(-2) "min"^(-1)` and `16 xx 10^(5) "min" ^(-1)` respectively then the molar ratio of B to A after a very long time will be :A. `2.5 xx 10^(-8)`B. `4 xx 10^(-2)`C. `(1)/(16 )xx 10^(-5)`D. `4 xx 10^(7)` |
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Answer» Correct Answer - A `[._(z)X^(A) underset (-alpha)rarr ._(z-2)X^(A-4)underset(-beta)rarr ._(z-1)X^(A-4)underset(-beta)rarr ._(z)X^(A-4)` `:.` In this series there will be three will be three different element with atomic no A, A-2, Z-1] |
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| 188. |
Radioactive decay follows first order kinetics and the rate constant is often termed as decay constant. Certain radioactive substances may undero sequential decays in order to convert into a stable nucleus. The series comprising all such elements is termed as radioactive disintegration series. A radioactive series is formed such that after each ` alpha` decay there are two consecutive `beta` decay and the cycle repeats. How many different elements this series can have if there are 12 members in the series ?A. 12B. 4C. 3D. 6 |
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Answer» Correct Answer - C After a very long time `: lambda_(1)N_(A)=lambda_(2)N_(B)` `(N_(A))/(N_(B))=(lambda_(1))/(lambda_(2))=(4 xx 10^(-2))/(16 xx10^(-5))=(1)/(4)=10^(-7)=2.5 xx 10^(-8) ` |
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| 189. |
A drug is given intravenously and drug concentrations in blood measured at 1 and 4 hour are 26 and `18mu g//mL`. What is the half-line of drug and at what time will the level decreases to `10mu g//mL` |
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Answer» `t = (2.303)/(lambda) log (N_(0))/(N)` `t = 1 hr, N = 26 mu g//mL` `:. 1 = (2.303)/(lambda) log (N_(0))/(26)` ...(i) `t = 4hr, N = 18 mug//mL` `:. 4 = (2.303)/(lambda) log (N_(0))/(18)` ....(ii) `:.` By eqn. (i) and (ii) `lambda = 0.123 hr^(-1)` and `N_(0) = 29.40 mug//mL` `:. t_(1//2) = (0.693)/(lambda) = (0.693)/(0.123) = 5.63 hr` Now `t = (2.303)/(0.123) log (29.40)/(10) = 8.77 hr` |
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| 190. |
A sample of wooden aircrafts is found to undergo `9 dp m g^(-1)` of `.^(14)C`. What is appoximate age of aircraffts? The half line of `._(6)^(14)C` is `5730` year and rate of disintergration of wood recently cut down is `15 dp m g^(-1)` do `._(6)^(14)C`? |
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Answer» Given, `r_(0) = 15` dpm, `r = 9` dpm, `t_(1//2) = 5730 yr` `t = (2.303)/(lambda) log (r_(0))/(r)` `= (2.303xx5730)/(0.693) log (15)/(9)` `= 4224.47 yr` |
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| 191. |
Tritium, `._(1)T^(3)` (an isotope of `H`) combine wityh flurine to form a weak acid `TF` whch ionises to give `T^(+)` prepared dilute aquios solution of `TF` has a `pt` (euivalent of `pH`) fo 1.7 and frezze at `- 0.372^(@)C`. If `600mL` of freshly prapared soltuion were allowed to stand for `24.8` years, calculate: (i() Ionisation constation of `TE` (ii) Charge carried by `beta-` particles emitted by tritium in faraday. Given: `K_(f)` for `H_(2)O = 1.86, t_(1//2) (T) = 12.4yrs`. |
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Answer» Correct Answer - (i) `2.5xx10^(-3)`, (ii) `0.054` faraday |
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| 192. |
What is the charge in mass when 2 mole of hydroghen atoms combine to form 1 mole fo `H_(2)` molecule if: `2H rarr H_(2), Delta E = -436 kJ`? Comment on the result. |
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Answer» Let total energy is obtained as a result of mass decay. `Delta m = (Delta E)/(c^(2)) = (436xx10^(3))/((3.0xx10^(8))^(2))` `= 4.84xx10^(-12)kg = 4.84xx10^(-9)g` Thus loss is mass accompanying formation of 1 mole of `H_(2)` form its constrituent atoms is `4.84xx10^(9)g`. It is too small to be deterctable by any balance currently available. Thus the energy difference can be given due to attactions among `H-` atom to form `H-H` bond. |
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| 193. |
Statement: parent element of `(4n + 1)` series is Plutonim `-241` Explanation: It decays to give `8alpha` and `5beta` particles.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - `c` parent element of `(4n + 1)` series is `.^(241)Pu`. The series gives `8alpha` and `5beta-` particlles to give finally stable element `.^(209)Bi`. |
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| 194. |
The parent nucleus of `(4n + 3)` seires is:A. `.^(228)Ac`B. `.^(235)U`C. `.^(238)U`D. `.^(237)Th` |
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Answer» Correct Answer - `(b)` The `(4n + 3)` series or actinium series has parent element `.^(235)U` |
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| 195. |
True or False Statements : In a radioactive element the fraction of initial amount remaining after its mean life time is l/e. |
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Answer» Correct Answer - [T] `(c_(1))/(c_(0))=e^(-lambdax(1)/(lambda))=e^(-1)=(1)/(e)` |
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| 196. |
What is the value of `n` for the present element of `(4n + 3)` series?A. `59`B. `58`C. `57`D. `60` |
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Answer» Correct Answer - `(b)` parent element of `(4n + 3)` series is `._(92)^(235)U`, Thus `(235)/(4) = [58xx4+3]` |
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| 197. |
If proton/neutron ratio of an isotope is less than one then which of the following emission will be shown by the isotopeA. `beta`-particleB. Positron emissionC. K electron captureD. Either (B) or (C) |
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Answer» Correct Answer - A [If `(P)/(n) lt 1` Then `._(0)n^(1) rarr._(1)P^(1) +._(-1)beta^(@)` Hence there will be `beta` emission |
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| 198. |
Statement: Neutron decay result in `beta-` emission and emission of neutrino. Explanation: Higher values of `n//p` ratio give rise to neutron decay.A. `S` is correct but `E` is wrong.B. `S` is wrong but `E` is correct.C. Both `S` and `E` are correct and `E` is correct explanation of `S`.D. Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - `b` Neutron decay occurs due to high `n//p` ratio as: `._(0)^(1)n rarr ._(+1)^(1)p + ._(-1)^(0)e overline(v)` (antineutrino) |
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| 199. |
Select the types of redox reaction form the following on the basis of type of tedox changes: (a) intermolecular redox, (b) intramolecular redox (c) auto redox. If none, write none. (i) `C_(6)H_(5) CHO overset(NaOH)(rarr) C_(6)H_(5)CH_(2) OH + C_(6) H_(5) COONa` (iI) `Cr_(2)O_(7)^(2-) + 2OH^(-) rarr 2CrO_(4)^(2-) + H_(2)O` (iii) `2Mn_(2)O_(7) rarr 4MnO_(2) + 3O_(2)` (iv) `NO_(3)^(-) + H_(2)S + H_(2)O + H^(+) rarr NH_(4)^(+) + HSO_(4)^(+-)` (v) `Fe + N_(2) H_(4) rarr NH_(3) + Fe(OH)_(2)` (vi) `2KOH + Br_92) rarr KBr + KBrO` (vii) `2Cu^(+) rarr Cu + Cu^(2+)` (viii) `Ag(NH_(3))_(2)^(+) overset(2H^(+))(rarr) Ag^(+) + 2NH_(4)^(+)` (ix) `5KI + KIO_(3) + 6HCl rarr 3I_(2) + 6KCI + 3H_(2)O` |
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Answer» Correct Answer - (a) `(iv),(v),(ix)`, (b) `(iii)`, (c) `(i),(vi),(vii)` |
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