94420 + Interview Questions in Chemistry in Class 12 Page 501 InterviewSolution

25001.	The radioactive isotope ""_(27)^(60)Co which is used in the teratment of cancer can be made by (n,p) reaction. For this reaction, the target nucleus is :
Answer» `""_(27)^(59)Co` `""_(28)^(60)Ni` `""_(27)^(60)Co` `""_(28)^(59)Ni` ANSWER :B

Discussion

25002.	The radioactive decay._(83)^(211)Birarr_(81)^(207)Ti , takes place in 100L closed vessel at 27^(@)C Starting with 2 mols of _(83)^(211)Bi(t_(1//2)=130sec) , the presuure development in the vessel after 520 sec will be:
Answer» `1.875` ATM `0.2155atm` `0.4618` atm `4.618` atm ANSWER :C

Discussion

25003.	The radioactive decay of ""_(35)^(88)X by a beta-emission produces an unstable nucleus which spontaneously emits a neutron. The final product is :
Answer» `""_(37)^(88)X` `""_(35)^(89)Y` `""_(34)^(88)Z` `""_(36)^(87)W` Answer :D

Discussion

25004.	The radio-nuclide Th_90^234undergoes two successive beta- decays followed by one alpha- decay. The atomic number and the mass number respectively of the resulting radionuclide are:
Answer» 92 and 234 94 and 230 90 and 230 92 and 230 ANSWER :C

Discussion

25005.	The radioactive decay follows :
Answer» ZERO oder FIRST ORDER second order order more than three. Answer :B

Discussion

25006.	The radio isotope, tritium (""_(1)^(3)H) has a half life period of 12-3 years. If the initial amount of tritium is 32 mg, how many milligrams of it would remain after 49-2 years?
Answer» 1mg 2 mg 4 mg 8 mg Solution :AMOUNT LEFT after N half lives `=([A]_(0))/(2^(n))=(32)/(2^(4))=2MG`

Discussion

25007.	The radio action substance decompose after 100 min and its concentration becomes part of 8 to original concentration.Calculate data constant and half life time.
Answer» SOLUTION :`k=2.08xx10^(-2) MIN^(-1),t_((t)/(2))=33.3 min^(-1)`

Discussion

25008.	The radii of P^(+) and Q^(-) are 0.95 overset()A and 1.81 overset()A respectively. Predict whether the coordination number of P^(+) is 6 or 4.
Answer» SOLUTION :The co-ordination NUMBER of PT is 6.

Discussion

25009.	The radii of Ar is greater than the radii of chlorine
Answer» Solution :In chlorine, the radii means the atomic or covalent radii which are actually HALF the inter-nuclear distance between 2 atoms whereas in ARGON the radii means the van der Waals radii as Argon is not a diatomic molecule. van der Waals radii are actually half the distance between adjacent molecules. So van der Waals radil being LARGER than atomic radil, Argon has GOT a larger radii than chlorine

Discussion

25010.	The radical can be identified by Borax Bead test is
Answer» `MG^(2+)` `Ca^(2+)` `CU^(+)` `Cu^(2+)` SOLUTION :Borax bead test is performed only with coloured salts. `Cu^(2+)` is coloured ion having unpaired d-electrons.

Discussion

25011.	The radiations having high penetrating power and not affected by electrical and magnetic field are
Answer» ALPHA rays Beta rays GAMMA rays Neutrons Solution :`gamma`-rays have MAXIMUM PENETRATING power

Discussion

25012.	The radiations from a naturally ocuurring radio element, as seen after deflection in a magnetic field in one direction, are
Answer» DEFINITELY `ALPHA`-rays Definitely `BETA-` rays Both `alpha and beta`-rays Either `alpha or beta-` rays Solution :DUE to it's NATURE

Discussion

25013.	The radiation having specific biological effect but unable to cause ionisation is-
Answer» UV-radiation `beta-RAY` `y- ray` `X-ray` ANSWER :A

Discussion

25014.	The radiant energy of the sun is due to:
Answer» Disintegration Nuclear fission Nuclear FUSION COMBUSTION. ANSWER :C

Discussion

25015.	The radial wave equation for hydrogen of radial nodes from nucleus are: Psi_(1s)=(1)/(16sqrt(4))(1/a_(0))^(3//2) [("x"-1)("x"^(2)-8"x"+12)]e^(-x//2) where, x=2r//a_(0),a_(0)= radius of firstBohr orbitThe minimum and maximum position of radial nodes from nucleus are:
Answer» `a_(o), 3a_(o)` `0.5a_(o), 3a_(o)` `0.5a_(o), a_(o)` `0.5a_(o), 4a_(o)` ANSWER :B

Discussion

25016.	The radial probability distribution curve for an orbitals comprises of 3 maxima. If the orbital has 3 angular nodes as well, then the orbital can be:
Answer» 5f 7d 6f 7f Answer :C

Discussion

25017.	The ractions of ethylene glycol with PI_(3) gives
Answer» `ICH_(2)CH_(2)I` `CH_(2)=CH_(2)` `CH_(2)=CHI` ICH=CHI Answer :B

Discussion

25018.	Theradiactive isotope of hydrogen is
Answer» HYDROGEN PROTIUM DEUTERIUM TRITIUM ANSWER :D

Discussion

25019.	the raction of hlaogen aicds on an ether, has the following order of ractivity:
Answer» HCI GT HBr gt HI HI gt HCI gt BHR HI gt HBr gt HCI HCI gt HI gt HBr Solution :NA

Discussion

25020.	The racemic mixture of Alanine (CH_(3)-underset(NH_(2))underset(\|)(C )H-COOH) can be resolved by using, (1) (+) -2-Butanol(2) (l)-2-Chlorobutanoic acid (3) (+-) -2 - Butanol(4) (d l mix)-2-Chlorobutanoic acid
Answer» 1 & 2 only 1 & 3 only 2 & 4 only 3 & 4 only Answer :A

Discussion

25021.	The R/S designation for the following stereoisomers of 1,3-Dibromo-2-methylbutane is :
Answer» 2R, 3R 2R, 3S 2S, 3R 2S, 3S Answer :A

Discussion

25022.	The racemic mixture in liquid/gaseous state will have
Answer» Same BOILING point as that of its pure ENANTIOMER Same refractive index as that of its pure enantiomer. Same DENSITY as that of its pure enantiomer All of the above Answer :D

Discussion

25023.	The (R ) - (S) - enantiomers of an optically active compound differ in ……
Answer» their reactivity with ACHIRAL ragents. their optical rotation of plane polarized light. their melting points. their solubility in achiral reagents. Solution :(R ) - and (S) - ENATIOMERS differ only in their BEHAVIOUR towards plane polarized light.

Discussion

25024.	The R-O-R bond angle in ether is
Answer» `180^(@)` `120^(@)` `111.7^(@)` `104^(@)` ANSWER :C

Discussion

25025.	The (R-) and (S-) enantiomers of an optically active compound differ in
Answer» their reactivity with achiral reagents their opticalrotation of plane POLARIZED light their MELTING points None of these Answer :B

Discussion

25026.	The questions consist of two atatements each, printed as Assertion and Reason. While answering these questions you are required to choose any one of the following four responses : The conductance measurement of an electrolytic solution is made by using two condutivity cells having defferent cell constants gives different values of conductance. The conductivity of solution obtained will be different but conductivity of solution would be same ,
Answer» If both the assertion and reason are TRUE but the reason is ont the CORRECT EXPLANATION of assertion If both the assertion and reason are true but the reason is not the correct explanation of assertion. If the assertion is true but reason is false. If assertion is false but reason is rue Solution :Both are correct. The `K` is same in both cases because number of ions per c.c remains same.

Discussion

25027.	The questions consist of two atatements each, printed as Assertion and Reason. While answering these questions you are required to choose any one of the following four responses : Copper is dissolved at anode and deposited at cathode when Cu electrodes are used and electrolyted is1M CuSO_4 (aq) solution . SOP of Cu is less than SOP of waer and SRP of Cu is greate than SRP of water .
Answer» If both the ASSERTION and reason are TRUE but the reason is ont the correct EXPLANATION of assertion If both the assertion and reason are true but the reason is not the correct explanation of assertion. If the assertion is true but reason is false. If assertion is false but reason is rue Solution :SOP FI Cu is greater than SOP of water `(H_2 O rarr O_2)` and `SRP` of Cu si greater than SRP of water `(H_2O rarrH_2)`.

Discussion

25028.	The questions consist of two atatements each, printed as Assertion and Reason. While answering these questions you are required to choose any one of the following four responses : Electrolyss of an aqueous solution of KI gives I_2 at the anode but that ofKF givesO_2 at the anode and not f_2 F_2 is more reactive than I_2
Answer» If both the assertion and reason are true but the reason is ont the correct explanation of assertion If both the assertion and reason are true but the reason is not the correct explanation of assertion. If the assertion is true but reason is FALSE. If assertion is false but reason is rue Solution :Correct explanation. `I^-` ions have much higher OXIDATION POTENTIAL than of qater while `F^-` ions have much LOWER oxidation potential than that of water.

Discussion

25029.	The question consist of two statements, Statement I and Statement II. You are to examine these two statements carefully and select the answer to these items using the codes given below. Statement I : Conversion of blue copper sulphate to black cupric oxide on heating is a physical change. Statement II : A change in which chemical composition does not change is called physical change.
Answer» Both the statements individually TRUE and Statement II is the correct explanation of Statement I. Both the statements are individually true but Statement II is not correct explanation of Statement I. Statement I is true but Statement II is false. Statement I is false but Statement II is true. Solution :On heating BLUE copper sulphate TURNS to a WHITE residue which is called anhydrous copper sulphate because the water of crystallization disappears on heating.

Discussion

25030.	The quantum numbers for the valence electron of an atom are n=3,l=0,m=0, s=+1//2 The element is
Answer» Calcium SODIUM Lithium Potassium Solution :This CORRESPONDS to `3s^(1)` ELECTRON and HENCE it is of sodium.

Discussion

25031.	The quantum numbers for the last electron in an atom are n = 3, l = 1 and m = -1. The atom is:
Answer» Al Si Mg C Answer :A

Discussion

25032.	The quantum numbers for the designation of 3d are:
Answer» `n=3,m=-3` `n=3,l=3` `n=4,l=1` `n=3,l=2` ANSWER :D

Discussion

25033.	The quantumnumber + 1//2 and -1//2 for the electronspin represent
Answer» ROTATION of the electron in CLOCKWISE and anticlockwise directions respectively rotation of the electron in anticlockwise and clockwise directions respectively magnetic moment of the electron pointing up and down respectively two quantum mechanical spin STATES which have no CLASSICAL analogue Answer :D

Discussion

25034.	The quantum numbers for the 19^(th) electron of Cr (Z=24) are :
Answer» ANSWER :C

Discussion

25035.	The quantum numbers +1/2 and -1/2 for the electron spin represent :
Answer» rotation of the ELECTRON in clockwise and anticlockwise direction respectively. rotation of the electron in ANTI clockwise and anticlockwise direction respectively. Magnetic MOMENT of the electron POINTING up and down respectively two quantum mechanical spin STATES which have no classical analogue Answer :D

Discussion

25036.	The quantum number not obtained from Schrodinger wave equation is
Answer» n l m s Solution :SPIN QUANTUM number was not obtained from Schrodinger WAVE EQUATION.

Discussion

25037.	The quantity remaining constant on dilution is
Answer» NUMBER of MOLES of solute Molarity of the solution Mole FRACTION of solute Mass of the solution Answer :A

Discussion

25038.	The quantity of K in a rate of expression
Answer» is independent of CONCENTRATION of REACTANTS Is CALLED Arrhenius constant Is dimesionless Is independent of temperaturee Answer :A

Discussion

25039.	The quantity of heat measured for a reaction in a bomb calorimeter is equal to
Answer» `DeltaG` `DeltaH` is negative `pDeltaV` `DeltaE` Solution :Heat of combustion is USUALLY measured by placing a known mass of a COMPOUND in a steel container at CONTANT volume called BOMB calorimeter. `q_("reaction")("heat of reaction")=C_(V)Deltat=DeltaE`

Discussion

25040.	The quantity of electricity required to liberate 112cm^(3)of hyrogen at STP from acidified water is
Answer» 965 C 1 Faraday 0.1F 96500 C. Answer :A

Discussion

25041.	The quantity of electricity required to electrolyse separately 1 M aqueous solutions of ZnSO_(4), AlCl_(3) "and" AgNO_(3) completely in the ratio of -
Answer» `2:1:1:` `2:1:3` `2:2:1` `2:3:1` ANSWER :D

Discussion

25042.	The quantity of electricity needed to separately electrolyse 1 M solution of ZnSO_(4), AlCl_(3) and AgNO_(3) completely is in the ratio of
Answer» `2:3:1` `2:1:1` `2:1:3` `2:2:1` Solution :`Zn^(2+) (1 M) + 2E^(-e) RARR Zn , 2 ` Faraday `Al^(3+)(1 M) + 3 e^(-) rarr AI , 3` Faraday `Ag^(+)(1 M) + e^(-) rarr Ag ,1` Farady Ration of quantity of electricity required `=2:3:1`

Discussion

25043.	The quantity of charge requried to obtain one mole of aluminium from Al_(2)O_(3) is
Answer» 1F 6F 3F 2F Solution :`Al_(2)O_(3) to 2AL,` i.e., `2Al^(3+)+6e^(-)to2Al` or `Al^(3+)+3e^(-)toAl`. Hence to OBTAIN one mole of Al. CHARGE requried=3F.

Discussion

25044.	The quantity of electricity needed to liberate 0.5 gram equivalent of an element is ________.
Answer» 48250 Faradays 48250 Coulombs 193000 Faradays 193000 Coulombs Answer :B

Discussion

25045.	The quantity of electricity in Faradays required to reduce 1.23 gm of nitro benzene to aniline is
Answer» 0.12 0.03 0.06 0.6 Answer :C

Discussion

25046.	The quantity of charge required to obtain one mole of aluminium from Al_(2)O_(3) is_______
Answer» 1F 6F 3F 2F Solution :* Reaction (Ionization : `Al_(2)O_(3) to AL^(3+) +3O^(2-)` Reaction forming Al reduction : `Al^(3+) + 3e^(-) to Al` * So 3 Mole `e^(-) to `1 mole So, 3F current PRODUCE 1 mole of Al.

Discussion

25047.	The quantity of charge required to obtain 1 mole of aluminium from Al_(2)O_(3) is _________.
Answer» 2F 3F 6F 12F Answer :B

Discussion

25048.	The quantity of .^(14)C as well as that of .^(14)CO_(2) present in the atmosphere remains constant. The concentration of .^(14)C in all living organisms remains almost constant during their life time. After their death, .^(14)C is not taken up by them but the content of .^(14)C assimilated begins to decay by emitting beta- particles, with half-life period of 5568 years. The decay rate at the time of death of plant is 16.1counts per minute per gram of carbon. Hence, by measuring the decay rate of the dead matter, the age of matter can be calculated , e.g. if decay rate of sample of wood is found to be N distengrations per minute per gram of carbon after t years, then N=N_(0)e^(-lambda t) where lambda= disntegration constant and N_(0)= number of disntegrations per minute per gram when the plant had just died. The basis for carbon - 14 dating method is that the
Answer» `C-14` FRACTION is same in all objects `C-14` is highly UNSTABLE and is readily lost from the objects ratio of `.^(14)C` to `.^(12)C` in our atmosphere has ALWAYS been constant living tissue will not absorb `C-14` but will absorb `C-12` from their sources of carbon Answer :C

Discussion

25049.	The quantity of .^(14)C as well as that of .^(14)CO_(2) present in the atmosphere remains constant. The concentration of .^(14)C in all living organisms remains almost constant during their life time. After their death, .^(14)C is not taken up by them but the content of .^(14)C assimilated begins to decay by emitting beta- particles, with half-life period of 5568 years. The decay rate at the time of death of plant is 16.1counts per minute per gram of carbon. Hence, by measuring the decay rate of the dead matter, the age of matter can be calculated , e.g. if decay rate of sample of wood is found to be N distengrations per minute per gram of carbon after t years, then N=N_(0)e^(-lambda t) where lambda= disntegration constant and N_(0)= number of disntegrations per minute per gram when the plant had just died. .^(14)C is present in environmental because of
Answer» artifical transmutation cosmic neutron bombardment of nitrogen it being a part of radioactive SERIES happening naturally `.^(12)C` when GETS bombarded with neutrons, TRANSMUTES to `.^(14)C` in environment Answer :B

Discussion

25050.	The quantity of .^(14)C as well as that of .^(14)CO_(2) present in the atmosphere remains constant. The concentration of .^(14)C in all living organisms remains almost constant during their life time. After their death, .^(14)C is not taken up by them but the content of .^(14)C assimilated begins to decay by emitting beta- particles, with half-life period of 5568 years. The decay rate at the time of death of plant is 16.1counts per minute per gram of carbon. Hence, by measuring the decay rate of the dead matter, the age of matter can be calculated , e.g. if decay rate of sample of wood is found to be N distengrations per minute per gram of carbon after t years, then N=N_(0)e^(-lambda t) where lambda= disntegration constant and N_(0)= number of disntegrations per minute per gram when the plant had just died. .^(14)C is
Answer» an ARTIFICIAL RADIOACTIVE isotope a NATURAL radioactive isotope a natural non-radioactive isotope an ARTIFICAL non-radioactive isotope Answer :A

Discussion

Explore topic-wise InterviewSolutions in .

The radioactive isotope ""_(27)^(60)Co which is used in the teratment of cancer can be made by (n,p) reaction. For this reaction, the target nucleus is :

The radioactive decay._(83)^(211)Birarr_(81)^(207)Ti , takes place in 100L closed vessel at 27^(@)C Starting with 2 mols of _(83)^(211)Bi(t_(1//2)=130sec) , the presuure development in the vessel after 520 sec will be:

The radioactive decay of ""_(35)^(88)X by a beta-emission produces an unstable nucleus which spontaneously emits a neutron. The final product is :

The radio-nuclide Th_90^234undergoes two successive beta- decays followed by one alpha- decay. The atomic number and the mass number respectively of the resulting radionuclide are:

The radioactive decay follows :

The radio isotope, tritium (""_(1)^(3)H) has a half life period of 12-3 years. If the initial amount of tritium is 32 mg, how many milligrams of it would remain after 49-2 years?

The radio action substance decompose after 100 min and its concentration becomes part of 8 to original concentration.Calculate data constant and half life time.

The radii of P^(+) and Q^(-) are 0.95 overset(*)A and 1.81 overset(*)A respectively. Predict whether the coordination number of P^(+) is 6 or 4.

The radii of Ar is greater than the radii of chlorine

The radical can be identified by Borax Bead test is

The radiations having high penetrating power and not affected by electrical and magnetic field are

The radiations from a naturally ocuurring radio element, as seen after deflection in a magnetic field in one direction, are

The radiation having specific biological effect but unable to cause ionisation is-

The radiant energy of the sun is due to:

The radial wave equation for hydrogen of radial nodes from nucleus are: Psi_(1s)=(1)/(16sqrt(4))(1/a_(0))^(3//2) [("x"-1)("x"^(2)-8"x"+12)]e^(-x//2) where, x=2r//a_(0),a_(0)= radius of firstBohr orbitThe minimum and maximum position of radial nodes from nucleus are:

The radial probability distribution curve for an orbitals comprises of 3 maxima. If the orbital has 3 angular nodes as well, then the orbital can be:

The ractions of ethylene glycol with PI_(3) gives

Theradiactive isotope of hydrogen is

the raction of hlaogen aicds on an ether, has the following order of ractivity:

The racemic mixture of Alanine (CH_(3)-underset(NH_(2))underset(|)(C )H-COOH) can be resolved by using, (1) (+) -2-Butanol(2) (l)-2-Chlorobutanoic acid (3) (+-) -2 - Butanol(4) (d l mix)-2-Chlorobutanoic acid

The R/S designation for the following stereoisomers of 1,3-Dibromo-2-methylbutane is :

The racemic mixture in liquid/gaseous state will have

The (R ) - (S) - enantiomers of an optically active compound differ in ……

The R-O-R bond angle in ether is

The (R-) and (S-) enantiomers of an optically active compound differ in

The quantum numbers for the valence electron of an atom are n=3,l=0,m=0, s=+1//2 The element is

The quantum numbers for the last electron in an atom are n = 3, l = 1 and m = -1. The atom is:

The quantum numbers for the designation of 3d are:

The quantumnumber + 1//2 and -1//2 for the electronspin represent

The quantum numbers for the 19^(th) electron of Cr (Z=24) are :

The quantum numbers +1/2 and -1/2 for the electron spin represent :

The quantum number not obtained from Schrodinger wave equation is

The quantity remaining constant on dilution is

The quantity of K in a rate of expression

The quantity of heat measured for a reaction in a bomb calorimeter is equal to

The quantity of electricity required to liberate 112cm^(3)of hyrogen at STP from acidified water is

The quantity of electricity required to electrolyse separately 1 M aqueous solutions of ZnSO_(4), AlCl_(3) "and" AgNO_(3) completely in the ratio of -

The quantity of electricity needed to separately electrolyse 1 M solution of ZnSO_(4), AlCl_(3) and AgNO_(3) completely is in the ratio of

The quantity of charge requried to obtain one mole of aluminium from Al_(2)O_(3) is

The quantity of electricity needed to liberate 0.5 gram equivalent of an element is ________.

The quantity of electricity in Faradays required to reduce 1.23 gm of nitro benzene to aniline is

The quantity of charge required to obtain one mole of aluminium from Al_(2)O_(3) is_______

The quantity of charge required to obtain 1 mole of aluminium from Al_(2)O_(3) is _________.

The radii of P^(+) and Q^(-) are 0.95 overset()A and 1.81 overset()A respectively. Predict whether the coordination number of P^(+) is 6 or 4.