94420 + Interview Questions in Chemistry in Class 12

1.	An endothermic reactio AtoB ha s an activation energy as kKJ"mol"^(-1) of A. If energy change of the reaction is yKJ, the activation energy of the reverse reaction is
Answer» `-X` `x-y` `x+y` `y-x` ANSWER :B

Discussion

2.	An endothermic reaction AtoB has an activatio energy 15 kaca/mole and the heat of reaction is 5kcal/mole. The activation energy of the reaction BtoA is
Answer» `20` kcal/mole `15` kcal/mole `10` kcal/mole zero Answer :C

Discussion

3.	An endothermic reaction A rarr B have an activation energy 15 kcal//mol and the heat of the reaction is 5 kcal//mol. The activation energy of the reaction B rarr A is :
Answer» `20kcal//mol` `15kcal//mol` `10kcal//mol` `ZERO` ANSWER :C

Discussion

4.	An enantiomerically pure acid is treated with racemic mixture of and alcohol having one chiral carbon. The ester formed will be:
Answer» optically ACTIVE MIXTURE PURE enantiomer meso compound RACEMIC mixture. Answer :A

Discussion

5.	An enantiomerically pure acid is treated with racemic mixture of an alcohol having one chiral carbon. The ester formed will be
Answer» Optically ACTIVE MIXTURE Pure enantiomer Meso compound Racemic mixture ANSWER :A

Discussion

6.	An enantiomerically pure acid is treated with racemic mixture of an alcohol having one chiral carbon. The ester formed will be :
Answer» OPTICALLY ACTIVE MIXTURE PURE enantiomer meso compound racemic mixture Solution :N//A

Discussion

7.	An enantiomerically pure acid is treated with a racemic mixture of an alcohol having one chiral carbon. The ester formed will be
Answer» OPTICALLY active mixture Pure enantiomer Meso compound Racemicmixture Solution :The optically active acid will react with d and l forms of alcohol present in the RACEMIC mixture at DIFFERENT rates to form two diastereomers in UNEQUAL ammmts leading to OPTICAL activity of the product.

Discussion

8.	An emulsion is a colloidal system consisting of:
Answer» TWO SOLIDS two liquids ONE GAS and one solid one gas and one liquid. Answer :B

Discussion

9.	An emulsion cnnot be broken by……….and……….
Answer» heating adding more amount of dispersion medium FREEZING adding emulsifying agent Solution :An EMULSION can be broken by heating or COOLING, i.e., (a) and (C ) and not by (B0 and (d).

Discussion

10.	An emulsifier is an agent which:
Answer» STABILIZES EMULSION Homogenises an emulsion Accelerates the dispersion Aids the FLOCCULATION of an emulsion Solution :Stabilizes emulsion

Discussion

11.	An emulsion cannot be broken by ............ and
Answer» HEATING Adding more AMOUNT of dispersion medium FREEZING Adding emulsifying agent SOLUTION :An emulsion is broken by heating or freezing.

Discussion

12.	An elements is in M^(3+) form. Its electronic configuration is [Ar]3d^(1) the ion is
Answer» `Ti^(3+)` `Ti^(4+)` `CA^(2+)` `Sc^(+)` Solution :`(Ar)3D^(1)+3=Ti`, it means `M^(3+)` form `Ti^(3+)` ION.

Discussion

13.	An elementary step is characterised by its ...........
Answer» SOLUTION :MOLECULARITY

Discussion

14.	An elementary reaction occurs in a closed vessel. 2CO_((g))+O_(2(g))O_(2(g))to2CO_(2(g)) If the volume of the reaction vessel is made one third of its original volume at constant temperature,the order of the reaction….of its original rate.
Answer» becomes TWENTY seven times becomes NINE times becomes three times becomes eighteen times Answer :A

Discussion

15.	An elementary reaction is given as 2P + Qtoproducts. If concentration of Q is kept constant and concentration of P is doubled then rate of reaction is
Answer» doubled halved QUADRUPLED remains same SOLUTION :`r_1 = k [P]^2[Q]` `r_2 = k [2P]^2[Q]` `:.r_2/r_1 = 4`

Discussion

16.	An elemental crystal has density of 8570 kg m^(-3). The packing efficiency is 0.67. If the closest distance between neighbouring atoms is 2.86 Å. The mass of one atom is (1 amu = 1.66 xx 10^(-27))kg)
Answer» ANSWER :93

Discussion

17.	An element ._(Z)M^(A) undergoes an alpha- emission followed by two successive beta- emissions. The element formed is ……………………….. .
Answer» Solution :`._(Z)M^(A)overset(-alpha)RARR._(z-2)M^(Z-4)``overset(-BETA)rarr._(Z-1)M^(A-4)overset(-beta)rarr._(Z)M^(A-4)`

Discussion

18.	An element Xloses one and twoßparticles in three successive stages. The resulting element will be:
Answer» an ISOBAR of X an ISOTOPE of X an isotone of X X itself. Answer :B

Discussion

19.	An element X with the electronic configuration 1s^(2),2s^(2)2p^(6), 3s^(2) would be expected to form the chloride with the formula
Answer» `XCl_(3)` `XCl_(2)` XCl `X_(2)CL` SOLUTION :This X element is a second GROUP element so its CHLORIDE will be `XCl_(2)`

Discussion

20.	An element (X) which is the most abundant metal in the earth's crustand the third most abudant element, is extracted by the electrolysis of its fused oxide in melted cryolite and fluorspar. XCl_(3) exists as (XCl_(3))_(n) in crystalline state and is only dimeric (X_(2)Cl_(6)) in fused state X+3HCl+6H_(2)O to XCl_(3).6H_(2)O(s)+(3)/(2)H_(2) Anhydrous XCl_(3) fumes in moist air and is very hygroscopic When XCl_(3).6H_(2)O(s) is heated strongly, hte products formed are:
Answer» `XCl_(3) and H_(2)O` `X_(2)O_(3),HCL and H)_(2)O` `X(OH)_(3) and HCl` no effect Answer :B

Discussion

21.	An element X with an atomic mass of 60 g/mol has density of 6.23 g cm^(-3). If the edge length of its cubic unit cell is 400 pm, identify the type of cubic unit cell. Calculate the radius of an atom of this element.
Answer» Solution :`M="60 g mol"^(-1), "a = 400 pm "=400xx10^(-10)cm, d="6.23 g cm"^(-3)," z needs to be CALCULATED."` `z=(d XXA^(3)xxN_(A))/(M)` Substituting the values, we have `z=(6.23xx400^(3)xx10^(-30)xx6.023xx10^(23))/(60)=(6.23xx64xx10^(-1)xx6.023)/(60)=4` In has face - CENTRED cubic (fcc) structure. `4r=sqrt2a or r=(sqrt2a)/(4)=(1.414xx4xx10^(-8))/(4)=1.414xx10^(-8)cm=141.4" pm"`.

Discussion

22.	An element (X) which is the most abundant metal in the earth's crustand the third most abudant element, is extracted by the electrolysis of its fused oxide in melted cryolite and fluorspar. XCl_(3) exists as (XCl_(3))_(n) in crystalline state and is only dimeric (X_(2)Cl_(6)) in fused state X+3HCl+6H_(2)O to XCl_(3).6H_(2)O(s)+(3)/(2)H_(2) Anhydrous XCl_(3) fumes in moist air and is very hygroscopic Consider of the following is correct? Which of the following is correct?
Answer» `alpha gt BETA and b gt a` `beta gt alpha and b LT a` `alpha gt beta and a gt b` `alpha lt beta and b lt a` ANSWER :A

Discussion

23.	An element 'X' which occurs in the first short period has an outer electronic structure s^(2)p^(1). What is the formula and acid-base character of its oxides
Answer» `XO_(3)`, BASIC `X_(2)O_3)`, basic `X_(2)O_(3)`, acidic `XO_(2)` acidic Answer :C

Discussion

24.	An element 'X' present in its ground state, the value of principal annd azimuthal quantum number for last electron of element 'X' is n=3 and l=1 and spin multiplicity for given element is 4. then according to given information correct statement(s) regarding givenn element 'X' is/are:
Answer» Element 'X' is 3rd acid period and 15th GROUP element In valence SHELL of element 'X' electron density is SYMMETRICALLY distributed Element 'X' has full filled valence shell. none of these Answer :A::B

Discussion

25.	An element X of atomic mass 25.0 exists as X)_(4) in benzene to the extent of 100%. When 10.30g of saturated solution of X in benzene is added to 20.0 g of benzene, the depression in freezing point of the resulting solution is 0.51 K. If K_(f) for benzene is "5.1 K kg mol"^(-1), the solubility of X in 100 g of benzene will be
Answer» 3.0 g 2.7 g 0.30 g 0.27 g Solution :Suppose saturated solution of X in BENZENE contains w g of X (present as `X_(4)`). Hence, benzene present `=(10.30-w)g` `therefore"Total benzene present"=20+(10.30-w)` `=(30.30-w)g` `DeltaT_(F)=(1000K_(f)w_(2))/(w_(1)xxM_(2))therefore0.51=(1000xx5.1xxw)/((30.30-w)XX100) ` `""(M_(2)or X_(4)=25xx4=100)` `or 51(30.3-w)=5100 w or 30.3-w=100 w` Thus, 10.30 g of saturated solution contains 0.3 g of X and 10.0 g of benzene. `therefore` Solubility of X in 100 g of benzene = 3.0 g

Discussion

26.	An element X occurs in short period having configuration ns^2np^1 . The formula and nature of its oxide is :
Answer» `XO_3` ,BASIC `XO_3` , ACIDIC `X_2O_3` , AMPHOTERIC `X_2O_3` , basic Answer :C

Discussion

27.	An element X ( molar mass = 80 g / mol ) having fcc structure , calculate number of unit cells in 8 g of X.
Answer» `0.4 * N_A` `0.1 * N_A` `4 * N_A` `0.025 * N_A` ANSWER :D

Discussion

28.	An element (X) having same cheical properties as that of hydrogen atom. The diatomic "mole"cule X_(2) is gaseous in nature. The energy required to remove an electron from the outermost shell of (X) atom is 13.5eV. It is also found that the time taken for diffusion of equal volume of gaseous X_(2) and O_(2) at the same pressure is in the ratio sqrt(3):4. the normal freezing point of pure X_(2)O is 0^(@)C. However, on adding 0.02 "mole"s of a non-electrolyte solute to 0.8 "mole"s of X_(2)O, the freezing point of solution is found to be-1.25^(@)C. X-atom also obey Bohr's model. What will be the energy required to excite 5 (X) atoms to second exicted state?
Answer» 30eV 60eV 120eV 15eV Answer :B

Discussion

29.	An element (X) having same cheical properties as that of hydrogen atom. The diatomic "mole"cule X_(2) is gaseous in nature. The energy required to remove an electron from the outermost shell of (X) atom is 13.5eV. It is also found that the time taken for diffusion of equal volume of gaseous X_(2) and O_(2) at the same pressure is in the ratio sqrt(3):4. the normal freezing point of pure X_(2)O is 0^(@)C. However, on adding 0.02 "mole"s of a non-electrolyte solute to 0.8 "mole"s of X_(2)O, the freezing point of solution is found to be-1.25^(@)C. X-atom also obey Bohr's model. What amount of electricity is required to produce 3.3 gm of x_(2) gas by electrolysis of X_(2)O?
Answer» `0.1F` `1.1F` `0.3F` `0.4F` ANSWER :D

Discussion

30.	An element (X) having same cheical properties as that of hydrogen atom. The diatomic "mole"cule X_(2) is gaseous in nature. The energy required to remove an electron from the outermost shell of (X) atom is 13.5eV. It is also found that the time taken for diffusion of equal volume of gaseous X_(2) and O_(2) at the same pressure is in the ratio sqrt(3):4. the normal freezing point of pure X_(2)O is 0^(@)C. However, on adding 0.02 "mole"s of a non-electrolyte solute to 0.8 "mole"s of X_(2)O, the freezing point of solution is found to be-1.25^(@)C. X-atom also obey Bohr's model. When 3 gm of a weak monoprotic acid (M_(w)=60) is added to 0.8 "mole"s of X_(2)O, the resulting solution freeze at -3.75^(@) C. What is the degree of dissociation for weak acid?
Answer» `0.10` `0.20` `0.30` `0.40` ANSWER :B

Discussion

31.	An element X has the following isotopic composition.""^(200)X=90%, ""^(199)X=8.0%, ""^(202)X=2.0%Its average atomic mass is
Answer» 199amu 200amu 201 AMU 202amu Solution :Average ATOMIC MASS of X `=0.9xx200+0.08xx199+0.02xx202` `=180+15.92+404=199.96` `=200` amu

Discussion

32.	An element X has the following isotopic composition, .^(200)X:90% .^(199)X:8% .^(2002)X:2% the weighted average atomic mass of the naturally-occurring element 'X' is closed to:
Answer» 201amu 202amu 199amu 200amu Solution :AVERAGE ATOMIC MASS of `X=[(99)/(100)xx200]+[(8)/(100)xx199]+[(2)/(100)xx202]` `=199.96` amu =200amu.

Discussion

33.	An element 'X' has its electronic configuration of 'K' shell is (n-5)s^(2) and it has total number of electrons in its outermost, penultimate and antipenultimate shell are 2,8 and 25 respectively then find out total number of unpaired electrons in element 'X' in their ground state.
Answer» ANSWER :7

Discussion

34.	An element X decays first by positron emission and then two alpha - particles are emitted in successive radioactive decay. If the product nucleus has a mass number 229 and atomic number 89, the mass number and the atomic number of element X are
Answer» 273, 92 237, 94 238, 93 273, 93 Answer :B

Discussion

35.	An element X belongs to group I or 2 or 15. Its oxide react with water to produce highly acidic solution the elements belong to which group ?
Answer» SOLUTION :The ELEMENT BELONGS to GROUP 15.

Discussion

36.	An element X (Atomic mass = 25) exists as X4 is benzene. 51g of saturated solution of X in benzene was added to 50.0 g of pure benzene. The resulting solution showed a depression of freezing point of 0.55 K. Find the solubility of X per 100 g of benzene. (Kf for benzene = 5.5 K kg "mol"^(-1)
Answer» Solution :LET X g be the mass of element in 51.0 g of saturated solution. Mass of benzene in 51.0 g of saturated solution = 51.0 -x g Total mass of benzene containing x g of solution `=50+51 -x=(101-x)g` `DeltaT_(f) = (1000 K_(f)W_(B))/(M_(B)W_(A)) = (1000 xx 5.5 xx x)/(4 xx 25 xx (101 -x))` =0.55 (given) `rArr x=1.0 g` Hence, solubility `=(W_(B) xx 100)/W_(A) =1/(51-1) xx 100 = 2.0 g`

Discussion

37.	An element with molar mass 2.7 xx 10^(-2)" kg mol"^(-1) forms a cubic unit cell with edge length 405 pm. If the density is 2.7 xx 10^(3)" kg m"^(-3), what is the nature of the cubice unit cell?
Answer» Solution :`"Density, d"=(zxxM)/(a^(3)xxN_(A))""(rho)=(nM)/(a^(3)N_(A))` `therefore""n=(rhoxxa^(3)xxN_(A))/(M)=((2.7xx10^(3)"kg m"^(-3))(4.05xx10^(-10)m)^(3)(6.022xx10^(23)mol^(-1)))/(2.7xx10^(-2)" kg mol"^(-1))` `=3.99=4` Thus, there are 4 atoms of ELEMENTS present per unit cell, hence, the CUBIC unit cell must be FACE - centred or cubic CLOSE - packed (ccp).

Discussion

38.	An element with molar mass 2.7xx10^(-2) kg mol^(-1) forms a cubic unit cell with edge length 405 pm. If its density is 2.7xx10^3 kg m^(-3), what is the nature of the cubic unit cell?
Answer» Solution :`a=405pm = 405xx10^(-12)m` `d= FRAC{zM}{a^3N} therefore z= frac {da^3N}{M}=4` Since there are 4 atoms present per UNIT CELL, it is FACE centred cubic unit cell or cep structure.

Discussion

39.	An element with molar mass 2.7 xx 10^(-2) "kg mol"^(-1) forms a cubic unit cell with edge length 405 pm. If its density is 2.7 xx 10^3 "kg m"^(-3), what is the nature of the cubic unit cell ?
Answer» Solution :Molar Mass (M) `= 2.7 xx 10^(-2) "kg mol"^(-1)` Edge length of a UNIT cell = 405 pm ` = 4.05 xx 10^(-10) m` Density of unit cell = `2.7 xx 10^3 "kg m"^(-3)` Using formula `d = (Z xx M)/(a^3 xx N_A)` we get , `Z = (2.7 xx 10^(3) xx 6.022 xx 10^(23) xx (4.05 xx 10^(-10))^3)/(2.7 xx 10^(-2))` Z = 4. Hence, unit cell is face-centred CUBIC.

Discussion

40.	An element with molar mass 2.7 xx 10^(-2) kg mol^(-1) forms a cubic unit cell with edge length 405 pm. If its density is 2.7 xx 10^(3) kg m^(-3), what is the nature of the cubic unit cell ?
Answer» SOLUTION :Applying the FOLLOWING relation, we have `"Density "=(zxxM)/(a^(3)N_(A)) or Z=("Density"xxa^(3)xxN_(A))/(M)"….(i)"` `M=2.7xx10^(-2)" kg mol"^(-1), a="405 pm "=405xx10^(-12)m=4.05xx10^(-10)m` `"Density "=2.7xx10^(3)" kg m"^(-3)` Thus, all the values of are in SI units. Substituting the values in equation (i) above, we have `z=(2.7xx10^(3)xx4.05^(3)xx10^(-30)xx6.022xx10^(23))/(2.7xx10^(-2)) or z=4` As there is 4 atoms in a unit cell, it is a face-centred CUBIC.

Discussion

41.	An element with molar mass 27 g mol^(-1) forms a cubic unit cell with edge length 4.05 xx 10^(-8)cm. If its density is 2.7 g cm^(-3), what is the nature of cubic unit cell ?
Answer» Solution :We shell apply the relation : `d=(zM)/(a^(3)N_(A)) or z=(DA^(3)N_(A))/(M)` Here, `d=2.7gcm^(-3), a=4.05xx10^(-8)cm, M="27 G mol"^(-1) and N_(A)=6.022xx10^(23)" atoms mol"^(-1)` Substituting the VALUES in the equation, we have `z=("2.7 g cm"^(-3)xx(4.05xx10^(-8))^(3)cm^(3)xx6.022xx10^(23)" atoms mol"^(-1))/("27 g mol"^(-1))` `=(2.7xx(4.05)^(3)xx10^(-24)xx6.022xx10^(23))/(27)" atoms = 4 atoms"` The element has a face - centred cubic cell.

Discussion

42.	An element with molar mass 27 g mol^(-1) forms a cubic unit cell with edge length 4.05 xx 10^(-8) cm. If its density is 2.7g cm^(-3), what is the nature of the cubic unit cell?
Answer» Solution :Density, `d=(Z xx M)/(a^3 xx N_A)` or `Z=(d xx a^3 xx N_A)/(M)` ...(i) According to the given data, `M = 27 g MOL^(-1), a=4.05 xx 10^(-8)` CM, `d = 2.7 g cm^(-3), N_A = 6.022 xx 10^(23) mol^(-1)` Substituting these values in expression (i), we get `Z=(2.7 g cm^(-3) xx (4.05 xx 10^(-8) cm)^3 xx 6.022 xx 10^(23) mol^(-1))/(27 g mol^(-1))` = 3.99 = 4 Since, there are 4 ATOMS of the element present per unit CELL, the cubic unit cell must be face centred.

Discussion

43.	An element with density 2.8 g cm^(-3) forms a fcc unit cell with edge length 4 xx 10^(-8) cm. Calculate the molar mass of the element. (Given : N_(A) = 6.022 xx 10^(23) mol^(-1))
Answer» Solution :d = 2.8 g `CM^(-3)`, z = 4 (for FCC), a = 4 `XX 10^(-8)` cm, `N_(A) = 6.022 xx 10^(23) mol^(-1)` `d = (Z xx M)/(N_(A) xx a^(3))` `M = (d xx a^(3) xx N_(A))/(Z)` `= (2.8 g cm^(-3) (4 xx 10^(-8) cm)^(3) xx 6.022 xx 10^(23))/(4)` `M = 2.8 xx 16 xx 20^(-1) xx 6.022` = 26.97 g `mol^(-1)`

Discussion

44.	An element with density 11.2 g cm^(-3) forms a fcc lattice with edge length of 4 xx 10^(-8) cm. Calculate the atomic mass of the element. (Given : N_(A) = 6.022 xx 10^(23) mol^(-1))
Answer» Solution :d = 11.2 `g//cm^(3)`, Z = 4, a = 4 `xx 10^(-8)` cm `d = (Z xx M)/(N_(A) xx a^(3))` 11.2 `= (4 xx M)/(6.022 xx 10^(23) xx (4 xx 10^(-8))^(3))` `M = (11.2 xx 6.022 xx 10^(23) xx 4 xx 10^(-8) xx 4 xx 10^(-8) xx 4 xx 10^(-8))/(4)` `M = 11.2 xx 6.022 xx 16 xx 10^(-1)` M = 107.9 g `mol^(-1)` or 107.9 u

Discussion

45.	An element with density 11.2 g cm^(-3) forms a fcc lattice with edge length of 4 xx 10^(-8) cm. Calculate the atomic mass of the element. Give N_(A) = 6.022 xx 10^(23) atoms mol^(-1).
Answer» ANSWER :107.9u

Discussion

46.	An element with density 11.2 g cm^(-3) forms a fcc lattice with edge length of 4 xx 10^(-8) cm. Calculate the atomic mass of the element. [Given : N_(A) = 6.022 xx 10^(23)"atoms" mol^(-1)]
Answer» SOLUTION :Use the relation `d=(zM)/(a^(3)N_(A)) or M=(da^(3)N_(A))/(z)` Here, z = 4 (for fcc CRYSTAL), `d=11.2" G cm"^(-3) and a=4XX10^(-8)cm` Substituting the VALUES in the above equation, we have `M=(11.2xx(4xx10^(-8))^(3)xx6.022xx10^(23))/(4)=(11.2xx64xx10^(-24)xx6.022xx10^(-23))/(4)` `=(11.2xx64xx6.022)/(4xx10)=107.9u`

Discussion

47.	An element with atomic number 51 belongs to group
Answer» 11 14 16 16 Solution :The configuration of the element with atomic number 51 is : `[Kr]^(36) 5s^(2) 4d^10 5p^3.` Since the OUTER SHELL's configuration has FIVE electrons `(5s^2 5p^3)`, therefore, it belongs to group

Discussion

48.	An element with atomic number 84 and mass number 218 loses one alpha-particle and two beta-particles in three successive stages, the resulting element will have
Answer» At. No. 84 and mass number 214 At. No. 82 and mass number 214 At. No. 84 and mass number 218 At. No. 82 and mass number 218 Solution :`._(84)A^(218) rarr ._(84)B^(214) + ._(2)He^(4) + 2._(-1)e^(0)`

Discussion

49.	An element with atomic number 21 is a
Answer» TRANSITION element alkali metal halogen representative element Solution :The electronic configuration of `._(21)X`is The last ELECTRON enters to d orbital which means it belongs to 'd' block elements or transition elements.

Discussion

50.	An element with atomic number 20 is placed in which period of the periodic table?
Answer» ANSWER :4

Discussion

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An element with molar mass 2.7xx10^(-2) kg mol^(-1) forms a cubic unit cell with edge length 405 pm. If its density is 2.7xx10^3 kg m^(-3), what is the nature of the cubic unit cell?

An element with molar mass 2.7 xx 10^(-2) "kg mol"^(-1) forms a cubic unit cell with edge length 405 pm. If its density is 2.7 xx 10^3 "kg m"^(-3), what is the nature of the cubic unit cell ?

An element with molar mass 2.7 xx 10^(-2) kg mol^(-1) forms a cubic unit cell with edge length 405 pm. If its density is 2.7 xx 10^(3) kg m^(-3), what is the nature of the cubic unit cell ?

An element with molar mass 27 g mol^(-1) forms a cubic unit cell with edge length 4.05 xx 10^(-8)cm. If its density is 2.7 g cm^(-3), what is the nature of cubic unit cell ?

An element with molar mass 27 g mol^(-1) forms a cubic unit cell with edge length 4.05 xx 10^(-8) cm. If its density is 2.7g cm^(-3), what is the nature of the cubic unit cell?

An element with density 2.8 g cm^(-3) forms a fcc unit cell with edge length 4 xx 10^(-8) cm. Calculate the molar mass of the element. (Given : N_(A) = 6.022 xx 10^(23) mol^(-1))

An element with density 11.2 g cm^(-3) forms a fcc lattice with edge length of 4 xx 10^(-8) cm. Calculate the atomic mass of the element. (Given : N_(A) = 6.022 xx 10^(23) mol^(-1))

An element with density 11.2 g cm^(-3) forms a fcc lattice with edge length of 4 xx 10^(-8) cm. Calculate the atomic mass of the element. Give N_(A) = 6.022 xx 10^(23) atoms mol^(-1).

An element with density 11.2 g cm^(-3) forms a fcc lattice with edge length of 4 xx 10^(-8) cm. Calculate the atomic mass of the element. [Given : N_(A) = 6.022 xx 10^(23)"atoms" mol^(-1)]

An element with atomic number 51 belongs to group

An element with atomic number 84 and mass number 218 loses one alpha-particle and two beta-particles in three successive stages, the resulting element will have

An element with atomic number 21 is a

An element with atomic number 20 is placed in which period of the periodic table?