94420 + Interview Questions in Chemistry in Class 12

1.	An element belonging to chalcogen group is :
Answer» CARBON Phosphorus Chlorine Sulphur. Answer :D

Discussion

2.	An element (atomic mass = 60) having fcc structure has density of 6.23 g cm^(-3). The edge length of its unit cell is :
Answer» 400 PM 276 pm 126 pm 470 pm Answer :A

Discussion

3.	An element (atomic mass = 31) crystallises in a cubic structure. The density of the metal is 5.4g*cm^(-3). The number of unit cells is 3.1g of metal is 6.022xx10^(22). The number of atoms per unit cell is-
Answer» 6 1 4 2 Answer :B

Discussion

4.	An element (atomic mass = 100 // mol) having bcc structure has unit cell edge 400 pm. The density of the element is :
Answer» `10.376g cm^(-3)` `5.1888 g cm^(-3)` `7.289 g cm^(-3)` `2.144 g cm^(-3)` Solution :DENSITY , `d = ( Z xx M)/( a^(3) xx N_(A))` Z = 2 ( bcc), M100 g `mol^(-1)` a = `400 xx 10^(-10) cm = 4 xx 10^(-8) cm`, `N_(A) = 6.02 xx 10^(23)` `:. d = ( 2 xx 100)/( ( 4 xx 10^(-8))^(3) xx ( 6.02 xx 10^(23)))` `= 5.19 g cm^(-3)`

Discussion

5.	An element (atomic mass = 100 g/mole) having b.c.c. structure has unit cell edge 4.00 Å. The density in g/cc of the element is
Answer» 10.376 5.188 7.289 2.144 Answer :B

Discussion

6.	An element (at. mass = 60) having face centred cubic unit cell has a density of 6.23g cm^(-3). What is the edge length of the unit cell? (Avogadro's constant = 6.023 xx 10^(23) mol^(-1))
Answer» SOLUTION :M=60, Z = 4 for fcc unit cell, d=6.23 g `cm^(-3)`, `N_A = 6.023 XX 10^(23) mol^(-1), d = (ZM)/(Ν_Α xx a^3)` `a^3=(4 xx 60)/(6.023 xx 10^(23) xx 6.23) cm^3 = 6.396 xx 10^(-23) cm^3`, `a = 4 xx 10^(-8)` cm = 400 PM

Discussion

7.	An element A occupies group number 17 and period number 2, shows anomalous behaviour . A reacts with water forms a mixture of B,C and acid D. B and C are allotropes. A also reacts with hydrogen violently even in dark to given an acid D. Identify A,B,C and D write the reactions.
Answer» Solution :i) The element A that occupies group number 17 and period number 2 is fluorine ii) Fluorine reacts with water and FORMS a mixture of B and C `2F_2 + 2H_2 O to 4HF + O_2` `3F_2 + 3H_2 O to 6HF + O_3` Therefore B is Oxygen and C is Ozone. iii) Fluorine reacts with hydrogen to give D. `F_2 + H_2 to 2HF` D is hydrofluorine ACID.

Discussion

8.	An element A occupies group number 17 and period number 2, is the most electronegative element. Element A reacts with another element B, Which occupies group number 17 and period number 4, to give a compound C. CompoundC undergoes sp^3 d^2hybridisation and has octahedral structure. Identify the elements A and B and the compound C. Write the reactions.
Answer» Solution :(i) The element of group number 17 and period number 2 is fluorine (A). (ii) The element of group number 17 and period number 4 is BROMINE (B). (III) (A) and (B) react to give an interhalogen COMPOUND (C) bromine penta fluoride. `UNDERSET((A)) (Br_(2))+underset((B))(5F_(2)) rarr underset((C))(2BrF_(5))` (C) undergoes `sp^(3)p^(2)` hybridisation and has octahedral structure.

Discussion

9.	An element A has three isotopes A^(20), A^(21), A^(22). The % abundance of A^(20) is 90% by mole and average atomic mass of the element A is 20.18 then mole % abundance of A^(21) would be -
Answer» `8%` `10%` `2%` `0.5%` Solution :`{:(,A^(20),A^(21),A^(22),M_(avg) = 20.18,),(,darr,darr,darr,,),(% "by moles",90%,(10 - x)%,x%,,):}` `M_(avg) = sum %` of ISOTOPE `xx` atomic mass `20.18 = (90 xx 20 + (10 - x) 21 + x xx 22)/(100)` `2018 = 1800 + 210 - 21 x + 22 x` `2018 = 2018 + x` `x = 8 %` `%` abundance of `.^(21)A = (10 - x) = 2%`

Discussion

10.	An element A forms a chloride which contains 29.34% by weight of chloride and is isomorphous with KCl. The atomic weight of A is
Answer» 85.49 40 23 137.5 Answer :A

Discussion

11.	An element (A) extracted from kernite. A reacts with nitrogen at high temperature gives B. A reacts with alkali to form C. Find out A, B and C. Give the chemical equations.
Answer» SOLUTION :(i) An element (A) extracted form kernite is boron. (ii) Boron reacts with NITROGEN at high TEMPERATURE gives Boron nitride (B) `UNDERSET((A))(2B)+N_(2) overset(Delta)(to)underset((B))(2BN)`

Discussion

12.	An element 'A' exists as a yellow solid in standard state. It forms a volatile hydride 'B' which is a foul smelling gas and is extensively used in qualitative analysis of salts. When treated with oxygen, 'B' forms an oxide 'C' which is a colourless and pungent smelling gas. The gas when passed through acidified KMnO_(4) solution, decolourises it. 'C' gets oxidised to another oxide 'D' in the presence of heterogeneous catalyst. Identify A, B, C, D and also give the chemical equations of reaction of 'C' with acidified KMnO_(4) solution and for conversion of 'C' into 'D'.
Answer» Solution :(i) The data suggests that the yellow solid 'A' is sulphur `(S_(8))`. The volatile hydride 'B' with a punget smell is hydrogen sulphides `(H_(2)S)` gas. Upon treatment with oxygen, `H_(2)S` is converted to sulphur dioxide `(SO_(2))` which is a colourless pungent smelling gas 'C'. (iii) The gas 'C' DECOLOURISES acidified `KMnO_(4)` solution and is also converted another oxide `SO_(3)` 'D' in the presence of heterogeneous catalyst platinised ASBESTOS. The chemical reactions involved are as follow : `underset((A))(S_(8)(s))+8H_(2)(g) overset("heat") to underset((B))(8H_(2)S(g))` `underset((B))(2H_(2)S(g))+3O_(2)(g) overset("heat") to underset((C ))(2SO_(2)(g))+2H_(2)O(g)` `underset((C ))(2SO_(2)(g))+O_(2)(g) overset(Pt)to underset((D))(2SO_(3))(g)` `2KMnO_(4)+3H_(2)SO_(4)to K_(2)SO_(4)+2MnSO_(4)+3H_(2)O+5(O)` `[SO_(2)+2H_(2)O to H_(2)SO_(4)+2H]xx5` `(2H+O to H_(2)O xx5)/(underset(("Violet"))(2KMnO_(4))+5SO_(2)+2H_(2)O to underset(("Colourless"))(K_(2)SO_(4)) +underset(("Colourless"))(2MnSO_(4))+2H_(2)SO_(4)`

Discussion

13.	An element 'A' exists as a yellow solid in standard state. It forms a volatile hydride 'B' which is a foul smelling gas and is extensively used in qualitative analysis of salts. When teated with oxygen, 'B' forms an oxide 'C' which is a colourless, pungent smelling gas. This gas when passed through acidified KMnO_(4) solution, decolourises it. 'C' gets oxidised to another oxide 'D' in the presence of a heterogeous catalyst. Identify A, B, C, D and also give the chemical equation of reaction of 'C' with acidified KMnO_(4) solution and for conversion of 'C' to 'D'.
Answer» Solution :(i) SINCE element 'A' is a yellow solid in the standard state and forms a foul smelling gas 'B' which is extensively USED in qualitative analysis of salts, therefore, the element 'A' must be sulphure `(H_(2)S)`. `UNDERSET(("Yellow solid, A"))underset("Sulphur")(S) + H_(2) overset("Heat")rarr underset(("Foul smelling gas, B"))underset("Hydrogen sulphide")(H_(2)S)` (ii) Since the volatile gas 'B' reacts with OXYGEN to form a colourless pungent smelling gas 'C' which decolourises acidified `KMnO_(4)` solution, the gas 'C' must be sulphur dioxide `(SO_(2))`. `underset("Hydrogen suphide (B)")(2H_(2)S) + 3O_(2) rarr 2 H_(2)O + underset(("Colourless pungent smelling gas C"))underset("Sulphur dioxide")(2SO_(2))` The gas C being a reducing agent reduces acidified `KMnO_(4)` solution. `{:(""2KMnO_(4)+3H_(2)SO_(4) rarr K_(2)SO_(4)+2MnSO_(4)+3H_(2)O+5[O]),(""SO_(2)+H_(2)O+[O] rarr [H_(2)SO_(4)] xx 5),(bar(underset(("Pink"))(2KMnO_(4))+5SO_(2)+2H_(2)rarr K_(2)SO_(4)+underset(("Colourless"))(2MnSO_(4))+2H_(2)SO_(4))):}` (iii) Since the colourless pungent smelling gas 'C', i.e., `SO_(2)`, can be oxidised to another oxide 'D' in the presence of a heterogeneous CATALYST (i.e., Pt or `V_(2)O_(5)`), therefore, the oxide 'D' must be sulphur trioxide `(SO_(3))`. `underset("Sulphur dioxide (C)")(2SO_(2)) + O_(2) rarr underset("Sulphur trioxide (D)")(2SO_(3))` Thus, 'A' is sulphur (S), 'B' is hydrogen sulphide, `(H_(2)S)`. 'C' is sulphur dioxide `(SO_(2))` and 'D' is sulphur trioxide `(SO_(3))`.

Discussion

14.	An element A exist in two isotopic forms A^15 and A^16 . If the average atomic mass of A was found to be 15.24, then the % relative abundance of A^15 will be
Answer» 0.1 0.2 0.4 0.76 Answer :D

Discussion

15.	An element A emits an alpha particle and form B. A and B are
Answer» isotopes isobars ISOTONES nuclides SOLUTION :`""_(Z)^(A)A OVERSET(-ALPHA)(rarr) ""_(Z-2)^(A-4)B.` A and B are simply nuclides.

Discussion

16.	An element Adissolves both in acid and alkali. It is an example of
Answer» Allotropic NATURE of A Dimorphic nature of A Amorphous nature of A AMPHOTERIC nature of A Solution :Amphoteric SUBSTANCE can REACT with both acid and base .

Discussion

17.	An element A dissolves both in acid and alkali. It is an example of :
Answer» ALLOTROPIC NATURE of A Dimorphic nature of A Amorphous nature of A Amophoteric nature of A Answer :D

Discussion

18.	An element A belongs to 14th group and occupies period number 6. A reacts with conc. HCl to give B an acid. A is used to prepare C which is used as an antiknock in automobilies. Identify the element A and the compounds B and C write the reactions
Answer» Solution :1. As per the position in the periodic table, the element A is lead. 2. Lead with Conc. HCl GIVES B `PB + 4HCl to H_2 PbCL_4 + H_2 ` `therefore ` Compound B is chloroplumbic ACID. 3. Compound C is tetraethyl lead.

Discussion

19.	An element A (at.wt.=75) and B (at.wt. =25) combine to form a compound . The compound contains 75% A by weight. The formula of the compound will be :
Answer» `A_2B` `A_3B` `AB_3` AB Answer :D

Discussion

20.	An element A and B constiture bcc type crystallinestructure. Element A occupies body centre positionand Bis at thecorners of cube. What is the formula of thecompound ? Whatarethecoordinationnumbers of A and B?
Answer» SOLUTION :FORMULA = AB, coordinationnumberof A= 8, coordinationnumber= of B= 8

Discussion

21.	An element ._(96)X^(227) emits 4alpha and 5 beta particles to form new element Y. Then atomic number and mass number of Y are
Answer» 93, 211 211,93 212,88 88,212 Solution :`._(96)X^(227) rarr Y + 4 alpha + 5 BETA` On equations mass number `227 = y + 4 xx 4 + 0, y = 211` On EQUATING atomic number `96 + y + 2 xx 4 - 5, y = 93`

Discussion

22.	An element ‘X’ (At mass = 40 g mol–1) having f.c.c. structore, has unit celledge length of 400 pm.Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’ (N_A = 6.022 × 10^23 mol^–1)
Answer» SOLUTION :`d=(zM)/(a^3N_A)` =`(4 times 40)/((4 times 10^-8) times 6.022 times 10^-23)` `=4.15g//cm^3` No of UNIT cells=total no of atoms 4 =`[4/40 times 6.022 times 10^23]//4` `=1.5 times 10^22`

Discussion

23.	An element ‘A’ exist as a yellow solid in standard stae. It forms a voilet hydride ‘B’ which is a foul smelling gas and is extensively used in qualitative analysis of salts. When reated with oxygen. ‘B’ forms an oxide ‘C’ which is a colourless and pungent smelling gas. The gas when passed through acidified kMnO_(4) solution, decolourises it, ‘C’ gets oxidised to another oxide ‘D’ in the presence of heterogenous catalyst. Identifier A, B, C, D and also give the chemical equation of reaction ‘C’ with acidified KmnO_(4) solution and for conversion of ‘C’ into ‘D’.
Answer» Solution :`underset(("A"))(S_(8)(s))+8H_(2)(g)overset("heat")rarrunderset(("B"))(8H_(2)S(g))` `underset(("B"))(2H_(2)S(g))+3O_(2)(g)overset(Delta)rarrunderset(("C"))(2SO_(2))(g)+2H_(2)O(g)` `underset(("C"))(2SO_(2))(g)+O_(2)(g)overset("Pt.")rarrunderset(("D"))(2SO_(3))(g)` OVERALL : `underset(("voilet"))(2KMnO_(4))+5SO_(2)+2H_(2)Orarrunderset(("colourless"))(K_(2)SO_(4))+underset(("colourless"))(2MnSO_(4))+2H_(2)SO_(4)`

Discussion

24.	An electroytic cell contains a solution of Ag_2SO_4 and platinum electrodes. A current is passeduntil 1.6 g of O_2 has been liberated at anode. The amount of Ag deposited at cathode would be :
Answer» `1.6g` `0.8g` `21.6g` `107.88g` ANSWER :C

Discussion

25.	An electrophilic reagent is :
Answer» Electron DEFICIENT SPECIES Electron RICH species NEGATIVELY CHARGED species A Lewis base. Answer :A

Discussion

26.	An electronic vacuum tube wassealed off during an experiment at a pressure of 8.2xx10^(-10) atm at 27^(@)C The volume of the tube was 30dm^(3). The number of gas molecules remaining in the tube are
Answer» `6.02xx10^(14)` `49.4xx10^(23)` `24.6xx10^(6)` `8.2xx30xx6.02xx10^(23)` Solution :`PV=nRT,n=(PV)/(RT)=(8.2xx10^(-10)xx30)/(0.082xx300)` `=10^(-9)` mol`=6.02xx10^(14)` MOLECULES

Discussion

27.	An electronic vacuum tube was sealed off during an experiment at a pressure of 8.2 xx 10^(-10) atm at 27^(@)C. The volume of the tube was 30 dm^(3). The number of gas molecules remaining in the tube are :
Answer» `6.023 XX 10^(14)` `8.2 xx 6.02 xx 10^(23)` `24.6 xx 10^(6)` `8.2 xx 30 xx 6.02 xx 10^(23)` Solution :Moles of gas remaining,n`= ( pV)/( RT)` `n = ( 8 .2 xx 10^(-10)xx 30)/( 0.082 xx 300) = 1 xx 10^(-9)` Molecules of gas `= 1 xx 10^(-9) xx 6.02 xx 10^(23)` `= 6.02 xx 10^(14)`

Discussion

28.	An electron will have the highest energy in the set:
Answer» 3, 2, 1, 1/2 4, 2, -1, 1/2 4, 1, 0, -1/2 5, 0, 0, 1/2 Answer :B

Discussion

29.	An electron will have highest energy with which one of the followng sets of four quantum numbers?
Answer» `{:(n""l""m""s),(3""2""1""+1/2):}` `{:(n""l""m""s),(4""2""-1""+1/2):}` `{:(n""l""m""s),(4""2""0""-1/2):}` `{:(n""l""m""s),(5""0""0""-1/2):}` Solution :`A(3d),B(4d),C(4P),D(5s)`.The SEQUENCE of energies is 3d, 4p, 5s, 4d.

Discussion

30.	An electron, practically at rest, is initially accelerated through a potential difference of 100 volts. It then has a de Broglie wavelength =lambda_(1) Å. It then get retarded through 19 volts and then has a wavelength lambda_(2) Å. A further retardation through 32 volts changes the wavelength to lambda_(3)Å, What is (lambda_(3)-lambda_(2))/(lambda_(1))?
Answer» ANSWER :`20/63`

Discussion

31.	An electron practically at rest, is initially accelerated through a potential difference of 100 volts. It then has a de Broglie wavlength =lambda_(1)Å. It then get retorted through 19 volts and then has a wavelength lambda_(2)Å . A further retardation through 32 volts changes the wavelength to lambda_(3). What is the value of (lambda_(3)-lambda_(2))/(lambda_(1))?
Answer» `(20)/(41)` `(10)/(63)` `(20)/(63)` `(10)/(41)` ANSWER :C

Discussion

32.	An electron jumps from n^(th) level to 1^(st) level, the fact ( s) which is // are correct of H - atoms is // are
Answer» Number of spectral lines `= ( N ( n - 1))/( 2)` number of spectral lines `= Sigma ( n - 1)` If n = 4, the number of spectral lines = 6 Number of spectral lines `= n ( n - 1) ` Answer :A,B,C

Discussion

33.	An electron of a velocity x is found to have a certain wavelength.The velocity to be possessed by the neuron to have half the de Broglie wavelength possessed by electron is:
Answer» `x//1840` `x//920` `3680x` `x//3680` SOLUTION :No. of unpaired ELECTRONS =3 TOTAL spin `=3xx1/2=3/2`

Discussion

34.	An electron is allowed to move freely in a closed cubic box of length 10 cm. The maximum uncertainty in its velocity will be observed as
Answer» `4xx10^(-3)m//s` `5xx10^(-4)m//s` `4xx10^(-5)m//s` `4xx10^(-6)m//s` Answer :B

Discussion

35.	An electron is a hydrogen atom is its ground state absorbs 1.50 times as much energy as the minimum required for its escape from the atom. What is the wavelength of the emitted electron?
Answer» `4.70Å` 4.70 NM `9.4 Å` 9.40 nm Answer :A

Discussion

36.	An electron in the ground state of hydrogen was excited to a higher energy level using monochromatic radiations of wave length (lambda )975 Å . The longest wave length that appears in the resulting spectrum is due to transition from:
Answer» `n_(4) to n_(1)` `n_(4) to n_(3)` `n_(5) to n_(4)` `n_(5) to n_(1)` Solution :The energy assoicated with radiation of WAVE length `975 Å`is given by `E = (12400)/(975)eV = 12.72 eV` [For a radiaton of WAVELENGHT x `Å` we know `E = (12400)/(x) eV`] `E = 13.6 [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]eV` ` or12.72 = 13.6 [1 - (1)/(n_(2)^(2))] ""[becauseE = 12.72 eV]` or ` (1)/(n_(2)^(2))k = (13.6-12.72)/(13.6) = (0.88)/(13.6) = (1)/(16)` or `n_(2)^(2) = 16 , n_(2) ~= 4` The transition `n_(4) to n_(3)`willgivethe longestwave lenght.

Discussion

37.	An electron in an isolataed atom, may be described by four quantum number n,l,m and s. The value of m for electron most easily removed from a gaseous atom of n-alkaline earth metal is
Answer» same as the maximum VALUE of N for the element any NUMBER from `-(n-1)` to `+(n-1)` an positive numberfrom 1 to (n-1) zero Answer :C

Discussion

38.	An electron in a hydrogne atom in its ground state absorbs 1.50 times as much energy as the minium required for its escape from the atom. What is the wavelength of the emitted electron ?
Answer» `4.7 Å` `4.7 NM` `9.4 Å` 9.40 nm ANSWER :A

Discussion

39.	An electron has spin quantum number +1//2 and a magnetic quantum number -1. It cannot be present in
Answer» s-orbital p-orbital d-orbital f-orbital Answer :A

Discussion

40.	An electron has a total energy of 2 MeV. Calculate the effective mass of the electron in kg and its speed. Assume rest mass of electron 0.511 MeV.
Answer» ANSWER :`2.9xx10^(10)CM"SEC"^(-1)`

Discussion

41.	an electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength is equal to 1.6 angestrom ?
Answer» `58.90`V `85.75V` `45.35`V `105.31` V Answer :A

Discussion

42.	An electrolytic cell is constructed for preparing hydrogen. For the average current of 1 ampere inthe circuit, the time required for producing 112 ml of H_2 at STP is approximately
Answer» 500 SEC 800 sec 1930 sec 965 sec Answer :D

Discussion

43.	An electrolytic cell contains a solution of AgNO_(3) and Pt electrodes. A current is passed until 1.6 gm of O_(2) has been liberated at anode. The amount of Ag deposited has liberated at cathode would be
Answer» 1.6 G 0.8 g 21.6 g 10.788 g Solution :1.6 g of `O_2-=0.2F `of ELECTRICITY Now 1F `-=` 107.8 g of AG. `THEREFORE 0.2F -=21.6` g of Ag.

Discussion

44.	An electrolytic cell contains a solution of Ag_(2)SO_(4)and platinum electrodes. Current is passed until 1.6 g. of O_(2)has been liberated at anode. The amonut of Ag deposited at cathode would be
Answer» 0.8g 1.6g 21.6g 107.88g Answer :C

Discussion

45.	An electrolytic cell contains a solution of Ag_(2)SO_(4) and platinum electrodes. A current is passed until 1.6 gof O_(2) has been liberated at anode. The amount of silver deposited at cathode would be
Answer» `107.88g` `1.6g` `0.8g` `21.60g` SOLUTION :`(W_(A))/(E_(A))=(W_(B))/(E_(B)),(16)/(8)=(WT. of Ag)/(108)` `therefore`of `Ag =21.6` g

Discussion

46.	An electrolytec cell contains a solution of AgNO_3 andhave platinum electrodes. A current is passed untill 1.6 g ofO_2 has been liberated at anode. The amount of silver peposited at cathode would be :
Answer» `107.88`G `1.6`g `0.8`g `21.60 g` ANSWER :D

Discussion

47.	An electrolyte
Answer» FORMS complex ions in solution gives ions only when electricity is passed possesses ions even in soid state gives ions ony when DISSOLVED in water. Solution :ELECTROLYTES posses ions even in SOLID state.

Discussion

48.	An electrode is prepared by dipping an Ag strip into a solution saturated withsilver thiocyanate, AgSCN, and containing 0.10 M SCN^- . The emf of the voltaic cell constructed by connecting this electrode as the cathode to the standard hydrogen half cell as the anode is 0.45 V. What is K_(sp)of AgSCN?
Answer» SOLUTION :`1 XX 10^(-2)`

Discussion

49.	An electrode is prepared by dipping a silver strip into a solution saturateed with AgACN and contanining 0.10 M SCN^(-). The e.m.f. of voltaic cell constructed by connecting this, as the cathode, to the standard hydrogen half-cell as anode was found to be 0.45 V. What is the solubility product of AgSCN. (Given E_(Ag^(+)//Ag)^(@) = 0.80 V)
Answer» ANSWER :`1.169 XX 10^(-7) ;`

Discussion

50.	An electrochemical cell is set up as follows : Pt (H_(2), atm ) \| 0.1 M HCl \|0.1 M acetic acid \| (H_(2) , 1 atm ) Pt EMF of this cell will not be zero because
Answer» the temperature is constant the pH of 0.1 M HCl and 0.1 M acetic ACID is not the same acids used in the two COMPARTMENTS are different EMF of a CELL depends on molarities of the acids used. Solution :It is a concentration cell . EMF `ne 0` because 0.1 MHCl and 0.1 M `CH_(3)COOH` have different `[H^(+)]` , i.e., their pH are not equal .

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

An element belonging to chalcogen group is :

An element (atomic mass = 60) having fcc structure has density of 6.23 g cm^(-3). The edge length of its unit cell is :

An element (atomic mass = 31) crystallises in a cubic structure. The density of the metal is 5.4g*cm^(-3). The number of unit cells is 3.1g of metal is 6.022xx10^(22). The number of atoms per unit cell is-

An element (atomic mass = 100 // mol) having bcc structure has unit cell edge 400 pm. The density of the element is :

An element (atomic mass = 100 g/mole) having b.c.c. structure has unit cell edge 4.00 Å. The density in g/cc of the element is

An element (at. mass = 60) having face centred cubic unit cell has a density of 6.23g cm^(-3). What is the edge length of the unit cell? (Avogadro's constant = 6.023 xx 10^(23) mol^(-1))

An element A occupies group number 17 and period number 2, shows anomalous behaviour . A reacts with water forms a mixture of B,C and acid D. B and C are allotropes. A also reacts with hydrogen violently even in dark to given an acid D. Identify A,B,C and D write the reactions.

An element A has three isotopes A^(20), A^(21), A^(22). The % abundance of A^(20) is 90% by mole and average atomic mass of the element A is 20.18 then mole % abundance of A^(21) would be -

An element A forms a chloride which contains 29.34% by weight of chloride and is isomorphous with KCl. The atomic weight of A is

An element (A) extracted from kernite. A reacts with nitrogen at high temperature gives B. A reacts with alkali to form C. Find out A, B and C. Give the chemical equations.

An element A exist in two isotopic forms A^15 and A^16 . If the average atomic mass of A was found to be 15.24, then the % relative abundance of A^15 will be

An element A emits an alpha particle and form B. A and B are

An element Adissolves both in acid and alkali. It is an example of

An element A dissolves both in acid and alkali. It is an example of :

An element A belongs to 14th group and occupies period number 6. A reacts with conc. HCl to give B an acid. A is used to prepare C which is used as an antiknock in automobilies. Identify the element A and the compounds B and C write the reactions

An element A (at.wt.=75) and B (at.wt. =25) combine to form a compound . The compound contains 75% A by weight. The formula of the compound will be :

An element A and B constiture bcc type crystallinestructure. Element A occupies body centre positionand Bis at thecorners of cube. What is the formula of thecompound ? Whatarethecoordinationnumbers of A and B?

An element ._(96)X^(227) emits 4alpha and 5 beta particles to form new element Y. Then atomic number and mass number of Y are

An element ‘X’ (At mass = 40 g mol–1) having f.c.c. structore, has unit celledge length of 400 pm.Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’ (N_A = 6.022 × 10^23 mol^–1)

An electroytic cell contains a solution of Ag_2SO_4 and platinum electrodes. A current is passeduntil 1.6 g of O_2 has been liberated at anode. The amount of Ag deposited at cathode would be :

An electrophilic reagent is :

An electronic vacuum tube wassealed off during an experiment at a pressure of 8.2xx10^(-10) atm at 27^(@)C The volume of the tube was 30dm^(3). The number of gas molecules remaining in the tube are

An electronic vacuum tube was sealed off during an experiment at a pressure of 8.2 xx 10^(-10) atm at 27^(@)C. The volume of the tube was 30 dm^(3). The number of gas molecules remaining in the tube are :

An electron will have the highest energy in the set:

An electron will have highest energy with which one of the followng sets of four quantum numbers?

An electron jumps from n^(th) level to 1^(st) level, the fact ( s) which is // are correct of H - atoms is // are

An electron of a velocity x is found to have a certain wavelength.The velocity to be possessed by the neuron to have half the de Broglie wavelength possessed by electron is:

An electron is allowed to move freely in a closed cubic box of length 10 cm. The maximum uncertainty in its velocity will be observed as

An electron is a hydrogen atom is its ground state absorbs 1.50 times as much energy as the minimum required for its escape from the atom. What is the wavelength of the emitted electron?

An electron in the ground state of hydrogen was excited to a higher energy level using monochromatic radiations of wave length (lambda )975 Å . The longest wave length that appears in the resulting spectrum is due to transition from:

An electron in an isolataed atom, may be described by four quantum number n,l,m and s. The value of m for electron most easily removed from a gaseous atom of n-alkaline earth metal is

An electron in a hydrogne atom in its ground state absorbs 1.50 times as much energy as the minium required for its escape from the atom. What is the wavelength of the emitted electron ?

An electron has spin quantum number +1//2 and a magnetic quantum number -1. It cannot be present in

An electron has a total energy of 2 MeV. Calculate the effective mass of the electron in kg and its speed. Assume rest mass of electron 0.511 MeV.

an electron beam can undergo diffraction by crystals. Through what potential should a beam of electrons be accelerated so that its wavelength is equal to 1.6 angestrom ?

An electrolytic cell is constructed for preparing hydrogen. For the average current of 1 ampere inthe circuit, the time required for producing 112 ml of H_2 at STP is approximately

An electrolytic cell contains a solution of AgNO_(3) and Pt electrodes. A current is passed until 1.6 gm of O_(2) has been liberated at anode. The amount of Ag deposited has liberated at cathode would be

An electrolytic cell contains a solution of Ag_(2)SO_(4)and platinum electrodes. Current is passed until 1.6 g. of O_(2)has been liberated at anode. The amonut of Ag deposited at cathode would be

An electrolytic cell contains a solution of Ag_(2)SO_(4) and platinum electrodes. A current is passed until 1.6 gof O_(2) has been liberated at anode. The amount of silver deposited at cathode would be

An electrolytec cell contains a solution of AgNO_3 andhave platinum electrodes. A current is passed untill 1.6 g ofO_2 has been liberated at anode. The amount of silver peposited at cathode would be :

An electrolyte

An electrochemical cell is set up as follows : Pt (H_(2), atm ) | 0.1 M HCl |0.1 M acetic acid | (H_(2) , 1 atm ) Pt EMF of this cell will not be zero because