52671 + Interview Questions in Chemistry in Class 11

1.	In a Homogenous catalysis:-
Answer» the catalyst and the REACTANTS should be gases The catalyst and the reactants should FORM a single phase Catalyst and the reactants are all solids The catalyst and the reactions are all liquids. Answer :B

Discussion

2.	In a hydrogen atom , in transition of electron a photon of energy 2.55 eV is emitted , then the change in wavelenghtof the eletron is
Answer» `3.32Å` `6.64Å` `9.97Å` NONE of these SOLUTION :`2pir_(n)=nlambda` `4 to 2` `lambda_(4)=(2pir_(4))/(4)""lambda_(2)=(2pir_(2))/(2)` `lambda_(4)lambda_(2P)=2pi[(0.529(4)^(2))/(1xx4)-(0.529(2)^(2))/(2)]` =`2pi(0.529)(2)` `=4pir_(1)`=`6.64 Å`

Discussion

3.	In a historical experiment to dtermine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for wavelength (lamda) of incident light and the corresponding stopping potential (V_(0)) are given below Given that c=3xx10^(8)ms^(-1) and e=1.6xx10^(-19)C, Planck's constant (in units of J s) found from such an experiment is:
Answer» `6.0xx10^(-34)` `6.4xx10^(-34)` `6.6xx10^(-34)` `6.8xx10^(-34)`. Solution :For photoelectronic emission, `(hc)/(LAMDA)=E_(0)+eV_(0)` . . . .(i) Here, kinetic ENERGY of photoelectron `((1)/(2)mv^(2))=eV_(0)` `(hxx3xx10^(8))/(0.4xx10^(-6))=E_(0)+1.6xx10^(-19)xx2` . . . .(ii) `(hxx3xx10^(8))/(0.4xx10^(-6))=E_(0)+1.6xx10^(-19)xx1` . . . .(III) Substracting eq. (iii) from eq. (ii) we get `h[10^(15)-0.75xx10^(15)]=1.6xx10^(-19)` ltBrgt `hxx0.25xx10^(15)=1.6xx10^(-19)` `h=6.4xx10^(-34)`Js

Discussion

4.	In a hexagonal close packed (hcp) structure of spheres, the fraction of the volume occupied by the sphere is A.In a cubic close packed structure, the fraction is B. The relation for A and B is
Answer» `A=B` `ALTB` `AGTB` A is equal to the fraction in a SIMPLE cubic lattice Solution :Fraction of the SPACE occupied in c.c.p. and h.c.p. arrangement is the same.

Discussion

5.	In a H-atom, the transition takes place from L to K shell. If R = 1.08 xx 10^(7) m^(-1), the wave length the light emitted is nearly
Answer» 4400 Å 1250 Å 1650 Å 1850 Å Answer :B

Discussion

6.	In a group with decrease in electronegativity the metallic character (a)decreases across a period(b)ncreases down the group(c )does not very
Answer» a only B only both (a) and (b) C only Answer :b

Discussion

7.	In a group from top to bottom effective nuclear charge
Answer» INCREASES DECREASES constant can not be PREDICTED ANSWER :C

Discussion

8.	Writeanytwopropertieswhichare decreasedfromtoptobottomin agroup?
Answer» increaes decreases constant can not be predicted Answer :A

Discussion

9.	In a gobar gas plant, gobar gas in obtained by bacterial fermentation of animal refuse. The main combustible gas present in the gobar is found to be methane ( 80% by weight) whose heat of combustion is 809kJ mol^(-1)) . How much gobar gas would have to produced per day for a village of 100 families, if the average consumption of a family is 20,000kJ per dayto meet all its energy requirements .
Answer» Solution :Total energy requirements for 100 families per day `= 20,000 xx 100kJ = 2 xx 10^(6) kJ` Methane `(CH_(4))` undergoes combustion as follows `:` `CH_(4)(g) rarr 2O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(L), Delta H = - 809 kJ MOL^(-1)` For 809 kJ of energy , `CH_(4)`REQUIRED`=16g` ( Molar mass of `CH_(4) = 16 g mol^(-1))` For ` 2 xx 10^(6)` kJ mol energy, `CH_(4)` required `(16)/(809) xx 2 xx 10^(6)g = 39.56kg` As gobar gas contains `80%` by weight of methane , therefore gobar gas required`=(100)/(80) xx 39.56 kg` `= 49.45kg`

Discussion

10.	In a glass tube columns of water and mercury appear as shown. This is best attributed to the differences in their:
Answer» DENSITIES MOLAR masses surface TENSIONS viscosities Answer :C

Discussion

11.	In a given shell the order of screening effect is
Answer» <P>`sgtpgtdgtf` `sgtpgtfgtd` `fgtdgtpgts` `fgtpgtsgtd` SOLUTION :F is GREATER than d is greater than p is greater than s

Discussion

12.	In a gaseous reaction the volume ratio of reactants and product is in agreement with their molar ration Volume of gas in inversely proportional to its mole at definite pressure and temperature .
Answer» <P> Solution :`V PROP` moles at CONSTANT `P,T` .

Discussion

13.	50.0 kg of N_(2)(g) and 10.0 kg of H_(2)(g) are mixed to produce NH_(2)(g). Calculate the NH_(2)(g) formed. Identify the limiting reagent in the production of NH_(3) in this situation.
Answer» Solution :The gasesous phase chemical equation for the production of ammonia is GIVEN as `N_(2)+3H_(2) to 2NH_(3)` 1 mole of `N_(2)="3 moles of "H_(2)` 28kg of `N_(2)=(2 XX 3)" kg of "H_(2)` 46.67 kg of `N_(2)` =10 kg of `H_(2)` 50kg of `N_(2)=10.71"kg of "H_(2)` Hence, HYDROGEN is the limiting reagent. 3 moles of `H_(2)="2 moles of "NH_(3)` 10 kg of `H_(2)=?` Maximum amount of ammonia produced is `(2 xx 17 xx 10)/(3 xx 2)=56.6kg`

Discussion

14.	In a gaseous mixture at 4 atm pressure, 25% of molecules are Nitrogen, 40% of molecules are carbon dioxide and the rest are oxygen. The partial pressure of oxygen in the mixture is
Answer» 1.40 ATM 1.6 atm 1 atm 0.9 atm Solution :`P_(O_2) = (V_(O_L))/(V_("TOTAL")) P IMPLIES (0.35)/(1)(4) = 1.4 atm`.

Discussion

15.	In a gas lighter, mechanical energy is converted into electrical energy by using crystals of barium titanate. Barium titanate is
Answer» piezoelectric but not ferroelectric ferroelectric but not piezoelectric both piezoelectric as WELLAS ferroelectric neither piezoelectric nor ferroelectric Solution :BARIUM TITANATE is both piezoelectric as well a ferroelectric.

Discussion

16.	In a gas phase reaction, 50 kg of nitrogen and 10 g of hydrogen are mixed to produce ammonia. Identify the limiting reagent. Calculate the maximum amount of ammonia produced.
Answer» Solution :The gaseous PHASE CHEMICAL equation for the PRODUCTION of ammonia is given as `N_(2)+3H_(2)rarr2NH_(3)` 1 mole of `N_(2)-=` moles of `H_(2)` `"28 kg of "N_(2)=(2XX3)" kg of "H_(2)` `"46.67 kg of "N_(2)="10 kg of "H_(2)` `"50 kg of "N_(2)="10.71 kg of "H_(2)` Hence, hydrogen is the limiting reagent, `"3 moles of "H_(2)="2 moles of "NH_(3)` `(3xx2)" kg of "H_(2)=(2xx17)" kg of "NH_(3)` `"10 kg of "H_(2)=?` Maximum amount of ammonia produced is `=(2xx17xx10)/(3xx2)=56.6kg`

Discussion

17.	In a fuel cell (device used for producing electricity directly from a chemical reaction )methanol is used as a fuel and oxygen gas is used asan oxidizer. The standard enthalpyof combustion of methanol is -721kJ mol^(-1) . The standard free energiesof formation of CH_(3)OH(l, CO_(2)(g) andH_(2)O(l) are- 166.3 , -394.4 and - 237 .1 kJ mol^(-1) respectively. The standard entropy change accopanying the cell reaction willbe
Answer» `+ 24.11 JK^(-1)` `-24.1JK^(-1)MOL^(-1)` `+19.1JK^(-1) mol^(-10` `-19.1 JK^(-1) mol^(-1)` Solution :`DELTAG^(@) = DeltaH^(@) -T DeltaS^(@)` or `DeltaS^(@) = (DeltaH^(@) - DeltaG^(@))/( T)` ,brgt `= (- 726.0 - ( - 720.3) KJ mol^(-1))/(298K)` ` = - ( 5.7 kJ mol^(-1))/( 298K) = - ( 5700Jmol^(-1))/(298K)` `=- 19.1 JK^(-1) mol^(-1)`

Discussion

18.	In a fuel cell (device used for producing electricity directly from a chemical reaction )methanol is used as a fuel and oxygen gas is used asan oxidizer. The standard enthalpyof combustion of methanol is -721kJ mol^(-1) . The standard free energiesof formation of CH_(3)OH(l, CO_(2)(g) andH_(2)O(l) are- 166.3 , -394.4 and - 237 .1 kJ mol^(-1) respectively. The standard internal energy change of the cell reaction will be
Answer» `-727.24 KJ mol^(-1)` `-724.76kJ mol^(-1)` `-728.48 kJ mol^(-1)` `- 723. 42 kJ mol^(-1)` SOLUTION :`DeltaU^(@) = DeltaH^(@) -Deltan_(G) RT` `= - 726 kJ mol^(-1) - (-(1)/(2))mol` `( 8.314 xx 10^(-3) kJ K^(-1) mol^(-1)) ( 298K)` `= - 726+ 1.24 kJ mol^(-1) = - 724 .76kJ mol^(-1)`

Discussion

19.	In a fuel cell (device used for producing electricity directly from a chemical reaction )methanol is used as a fuel and oxygen gas is used asan oxidizer. The standard enthalpyof combustion of methanol is -721kJ mol^(-1) . The standard free energiesof formation of CH_(3)OH(l, CO_(2)(g) andH_(2)O(l) are- 166.3 , -394.4 and - 237 .1 kJ mol^(-1) respectively. The efficiencyof the fuel cell will be
Answer» `96.7 %` `66.2%` `41.3%` `85.1%` SOLUTION :EFFICIENCY `=(DELTAG)/(DELTAH) = ( - 702.3)/( - 726) = 0.967= 96.7%`

Discussion

20.	In a fuel cell (device used for producing electricity directly from a chemical reaction )methanol is used as a fuel and oxygen gas is used asan oxidizer. The standard enthalpyof combustion of methanol is -721kJ mol^(-1) . The standard free energiesof formation of CH_(3)OH(l, CO_(2)(g) andH_(2)O(l) are- 166.3 , -394.4 and - 237 .1 kJ mol^(-1) respectively. The standard free energy change of the reaction will be
Answer» `-597.8 kJ mol^(-1)` `-298 kJmol^(-1)` `- 465.2 kJ mol^(-1)` `-702.3kJ mol^(-1)` Solution :The reaction is `CH_(3)OH(l) + (3)/(2) O_(2)(g) rarr CO_(2)(g) + 2H_(2)O(l) ` `Delta_(R)G^(@) = SigmaDelta_(f)G^(@)(` PRODUCTS) - `SigmaDelta_(f)G^(@)(`REACTANTS) `= [ Delta_(f)G^(@) ( CO_(2) +2Delta_(f) G^(@) (H_(2)O)] -[Delta_(f)G^(@) (CH_(3)OH)+(3)/(2)Delta_(f)G^(@) (O_(2))]` `= [( - 394.4) + 2 ( - 237.1) ] - [ - 166.3 + (3)/(2) (0)]` `=- 702 . 3 kJ mol^(-1)`

Discussion

21.	In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions . If one atom of B is missing from one of the face centred points, the formula of the compound is :
Answer» `AB_2` `A_2B_3` `A_2B_5` `A_2B` Solution :Contribution by atoms A on the corners 4=`8xx1/8=1` Contribution by atoms B on the face centres =`5xx1/5=5/2` (Total atoms on face centres=6. As ONE ATOM is missing, atoms actually present=5) `therefore` Ratio of A:B=`1:5/2=2:5` HENCE, FORMULA =`A_2B_5`

Discussion

22.	In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atoms of B is missing from one of the face centred points, the formula of the compound is :
Answer» `AB_(2)` `A_(2)B_(3)` `A_(5)B_(5)` `A_(2)B` SOLUTION :Contribution by atoms A on the CORNERS ` 4= 8 xx 1/8 =1 ` Contribution by atoms B on the FACE CENTRES ` 5 xx 1/2 = 5/2 ` ( Total atoms on face centres = 6. As one atoms is missing , atoms actually present = 5) : Ration of A: B = ` 1 : 5/2 = 2: 5` Henece, formula ` = A_(2)B_(5)`

Discussion

23.	In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is
Answer» `A_(2)B` `AB_(2)` `A_(2)B_(3)` `A_(2)B_(5)` ANSWER :D

Discussion

24.	In a face centred cubic lattice, atom A occupies the corner positionsand atom B occupies the face centre positions.If one atomof B is missing from one of the face centred points.
Answer» `A_(2)B` `AB_(2)` `A_(2)B_(3)` `A_(2)B_(5)` Solution :NUMBER of atoms of A per unit cell =`8times1/8=1` Number of B atoms per unit cell =`(6-1)/2=2.5` Formula of the compound =`A_(1)B_(2.5)orA_(2)B_(5)`

Discussion

25.	In a face centredcubic lattice, a cell is shared equally by how many unit cells ?
Answer» 4 2 6 8 Solution :In a face CENTRED cubiclattice, a UNIT cell is SHARED EQUALLY by six unit cells.

Discussion

26.	In a face-centred cubic crystal, the nearest neighbour distance is ___________ times the edge of the crystal
Answer» ANSWER :0.707

Discussion

27.	In a face centred cubic cell, an atom at the face centre is shared by:
Answer» 4 UNIT CELL 2 unit cells 1 unit cells 6 unit cells ANSWER :B

Discussion

28.	In a face centered cubic lattice atoms A are at the corner points and atoms B at the face centered points. If atoms B is missing from one of the face centered points, the formula of the ionic compound is
Answer» `AB_(2)` `A_(5)B_(2)` `A_(2)B_(3)` `A_(2)B_(5)` ANSWER :D

Discussion

29.	In a exothermic reaction, heat isevolved , andsystem losses heat to the surrounding . For such system:
Answer» `q_(q)`WIL be negative `Delta_(r)H` will benegative `q_(r)` will be positive `Delta_(r)H` will bepositive Answer :a,B

Discussion

30.	In a Dumas nitrogen estimation method, 0.30 g of an organic compound gave 50 cm^(3) of N_(2) collected at 300 K and 715 mm Hg pressure. Calculate the percentage composition of nitrogen in the compound. (Vapour pressure of water at 300 K is 15 mm Hg)
Answer» Solution :`P_(1)=715-15=700 mm HG` , `P_(2)=760 mm Hg` `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))` `T_(1)=300 K` , `T_(2)=273 K` , `V1=50 cm^(3)` , `V_(2)=?` `V_(2)=(700 xx 50 xx 273)/(300 xx 760)=41.9 cm^(3)` `"% of N"=(28)/(22400) xx 41.9 xx (100)/(W)=17.46%`

Discussion

31.	In a electrochemical cell
Answer» POTENTIAL energy changes into KINETIC energy Kinetic energy changes into potential energy Chemical energy changes into electrical energy Electrical energy changes into chemical energy SOLUTION :In electrochemical cell chemical energy converted

Discussion

32.	In a cubic vessel of side length 5 cm, a Helium molecule moving with a velocity 10^5 cm/sec in X-direction then rate of change in momentum in X-direction is (in g cm/sec)
Answer» `2.56 XX 10^(-14)` `1.66 xx10^(-14)` `10^10 ` `10^(-4)` ANSWER :B

Discussion

33.	In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present ?
Answer» 3 DOUBLE bonds , 9 SINGLE bonds 6 double bonds , 6 single bonds 3 double bonds , 12 single bonds Zero double bonds , 12 single bonds Answer :A

Discussion

34.	In a crystalline solid, anions B are arranged in a cubic close packing. Cations A are equally distrilbuted between octahedral and tetrahedral voids. If all the octahedral voids are occupied, what is the formula of the solid ?
Answer» Solution :Suppose the number of anions B = n. Then number of OCTAHEDRAL voids = n . Number of tetrahedral voids = 2n. As octahedral and tetradhedral voids are equally occupied by cations A and all the octahedral voids are occupied (given)THEREFORE, n capitan A are PRESENT in octahedral voids and n cationsA are present in tetrahedral voids. In other WORDS, correshponding to n anions B , there are n +n = 2n cations A. Thus, A and anuions B are in the ratio 2n: n =2 : 1 hence, the FORMUAL of the solid will be `A_(2) B` .

Discussion

35.	In a crystalline solid, anions B are arranged in a cubic close packing. Cations A are equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied, what is the formula of the solid ?
Answer» SOLUTION :Suppose the number of anions B=n . Then number of octahedral voids=n Number of TETRAHEDRAL voids =2n As octahedral and tetrahedral voids are equally occupied by cations A and all the octahedral voids are occupied (given ), therefore n cations A are present in octahedral voids and n cations A are present in tetrahedral voids. In other words, CORRESPONDING to n anions B, there are n+n=2n cations A . THUS, cations A and anions B are in the ratio 2n:n=2:1 . Hence , the formula of the solid will be `A_2B`

Discussion

36.	In a crystalline solid, W atoms are at corners, O atoms at edge centres and Na at cube centre, The formula of solid is
Answer» `Na_(4)WO_(3)` `NaWO_(3)` `NaW_(8)O_(3)` `NaWO_(4)` SOLUTION :W (at corner) `=8XX(1)/(2)=1` , O (at edges) `=12xx(1)/(4)=3` Na (at centre )=1 `therefore` FORMULA is `NaWO_(3)`

Discussion

37.	In a crystal of an ionic compound, the ions B form the close packed lattice and the ions A occupy all the tetrahedral voids. The formula of the compound is
Answer» `AB_(2)` `A_(2)B` AB `AB_(3)` SOLUTION :The number of tetrahedral voids is two PER ATOM of the crystal in a close packed structure. Since tetrahedral voids are occupied by A ions, there are two A ions per B ion inthe crystal.The formula of the compound is `A_(2)B`.

Discussion

38.	In a crystal, all the lattice sites are found to be occupied by covalent molecules. To which kind of solid does it belong ?
Answer» Ionic Molecular Covalent None Solution :CONCEPTUAL QUESTION. SEE COMPREHENSIVE REVIEW.

Discussion

39.	In a crystal, Frenkel defect is not shown by alkali metal halides but silver halides show. Why ?
Answer» Solution :Frenkel DEFECT is found in silver HALIDES due to SMALL size of `Ag^+` ION which can enter into interstitial site. It is not found in alkali metal halides, as the alkali metal ions are too big to fit into the interstitial SITES.

Discussion

40.	In a crystal, Frenkel defect is not shown by alkali metal halides buy sliver halides show. Why ?
Answer» Solution :FRENKEL defect is found in sliver HALIDES due to small size of `Ag^(+)` ion which can enter into interstitial SITE. It is not found in ALKALI metal halides, as the alkali metal ions are too BIG to fit into interstitial sites.

Discussion

41.	In a container of constant volume at a particular temparature N_(2) and H_(2) are mixed in the molar ratio of 9:13. The following two equilibria are found o be coexisting in the container N_(2)(g)+3H_(2)(g)harr2NH_(3)(g) N_(2)(g)+2H_(2)(g)harr N_(2)H_(4)(g) The total equiibrium pressure is found to be 305 atm while partial pressure of NH_(3)(g) and H_(2)(g) are 0.5 atm and 1 atm respectivly. Calculate of equilibrium constants of the two reactions given above.
Answer» Solution :Let the initial PARTIAL pressure of `N_(2)` be `9P` and `13P` respectively Total pressure`=P_(N_(2))+P_(H_(2))+P_(NH_(3))+P_(N_(2)H_(4))=3.5` ATM `(9P-x-y)+(13P-3x-2y)+2x+y=3.5`atm .....(1) ` P_(NH_(3))=2x=0.5` atm .....(2) `P_(H_(2))=(13P-3x-2y)= 1` atm ....(3) from (1) `implies(9P-x-y)+1 ` atm `+0.5+y=3.5` `implies(9P-x)=2` atm so `9P=2.25` `P=0.25` atm from (3) equation `2y=1.5` `y=0.75` atm so `P_(N_(2))=9P-x-y=1.25` atm `P_(H_(2))=1` atm `P_(NH_(3))=0.5` atm `P_(N_(2)(H_(4))=0.75` atm So, `K+(P_(1))=(P_(NH_(3))^(2))/P_(H_(2)^(3).P_(N_(2)))=(0.5xx0.5)/(1xx1xx1xx1.25)=0.2atm^(-2)` `K_(P_(2)=P_(N_2H_(4))/P_(N_(2).P_(H_(2))^(2))=(0.75)/(1xx1xx1.25)=0.6 atm^(-2)`

Discussion

42.	In a container m g of a gas is placed. After some time some gas is allowed to escape from container. The pressure of the gas becomes half and its absolute temperature 2//3 rd. The amount of gas escaped is
Answer» `2//3` `1//2` `1//4` `1//6` Solution :We KNOW that `PV = w (RT)/(M)` `:. (P_(1))/(P_(2)) = (w_(1))/(w_(2)) (T_(1))/(T_(2))` `P_(1) = P, P_(2) = (P)/(2), T_(1) = T, T_(2) = (2)/(3) T` `w_(1) = m, w_(2) = m - x` `(P)/(P//2) = (m)/(m x) xx (T)/(2T//3)` `2 = (m)/(m - x) xx (3)/(2)` `4 (m - x) = 3m` `4 m - 4 x = 3m` `m = 4x` or `x = m//4`

Discussion

43.	In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burtn in excess oxygen at 298.0K. The temperaure of the calorimeter was found to increase from 298.0 Kto298.45 K due to the combustion process.Given that the heat energy capacityof the calorimeteris 2.5 kJ K^(-1),the numerical value for the enthalpy of combustion of the gas in kJ mol^(-1)is
Answer» ANSWER :9

Discussion

44.	In a constant volume calorimeter, 3.5 g of a gas with molecular weight 28 was burnt in ress oxygen at 298 K. The temperature of the calorimeter was found to Increase from 298 K to 298.45 K due to the combustion process. Given that the calorimeter constant is 2.5 KJ K^(-1) .Calculate the enthalpy of combustion of the gas in "kJ mol"^(-1).
Answer» Solution :Given `T_i`=298 K `T_f` =298.45 K `k=2.5 kJ K^(-1)` m=3.5 g `M_m`=28 heat EVOLVED=`kDeltaT` `=k(T_f-T_i)` `=2.5 kJ K^(-1)(298.45-298)K` =1.125 kJ `DeltaH_C=1.125/3.5xx28 "kJ MOL"^(-1)` `DeltaH_C=9 "kJ mol"^(-1)`

Discussion

45.	In a constant pressure calorimeter, 224 mL of 0.1 M KOH (aq) solution is reacted with 50ml of 0.1 MH_(2)SO_(4)(aq) solution then increase in temperature of solution will be 9assume heat capacity of calorimeter is negligible): Given : Specific heat of solution =1cal/g-K Density of solution =1g/mL
Answer» `0.5K` `1K` `2K` `4K` ANSWER :a

Discussion

46.	In a conjugate pair of reductant and oxidant, the oxidant has
Answer» HIGHER OX. no LOWER ox.no. same ox.no EITHER of these Solution :Higher O.N

Discussion

47.	In a conjugate pair of reductant and oxidant, the oxidant has "________________"
Answer» HIGHER OXIDATION number lower oxidation number same oxidation number either of these Answer :A

Discussion

48.	In a concentration cell the
Answer» two electrodes are of different elements two electrolytic SOLUTIONS of the same electrolyte but having different CONCENTRATIONS are used electrolyte of ONE strength but elecrtrodes of two different concentrations are used Both (B) and (C) Solution :In CONCENTRATION cell two electrolytic solutions of the same electrolyte but having different concentrations are used. Moreover, electrolyte of one strength but electrodes of two different concentration of `H^(+)` ions.

Discussion

49.	In a compression process, P_("ext") is …………..
Answer» <P>`(P_("INT")+DP)` `(P_("int")-dP)` `(dP-P_("int"))` `(-P_("int")+dP)` ANSWER :A

Discussion

50.	In a compound oxide ions are arranged in cubic close packing arrangement. Cations A occupy one-sixth of the tetrahedral voids and cations B occupy one-third of the octahdral voids. The formual of the compound is A_(x), B_(y), O_(z) then find the value of x + y + z
Answer» ANSWER :`5 = (1 + 1 + 3)`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

In a Homogenous catalysis:-

In a hydrogen atom , in transition of electron a photon of energy 2.55 eV is emitted , then the change in wavelenghtof the eletron is

In a hexagonal close packed (hcp) structure of spheres, the fraction of the volume occupied by the sphere is A.In a cubic close packed structure, the fraction is B. The relation for A and B is

In a H-atom, the transition takes place from L to K shell. If R = 1.08 xx 10^(7) m^(-1), the wave length the light emitted is nearly

In a group with decrease in electronegativity the metallic character (a)decreases across a period(b)ncreases down the group(c )does not very

In a group from top to bottom effective nuclear charge

Writeanytwopropertieswhichare decreasedfromtoptobottomin agroup?

In a glass tube columns of water and mercury appear as shown. This is best attributed to the differences in their:

In a given shell the order of screening effect is

In a gaseous reaction the volume ratio of reactants and product is in agreement with their molar ration Volume of gas in inversely proportional to its mole at definite pressure and temperature .

50.0 kg of N_(2)(g) and 10.0 kg of H_(2)(g) are mixed to produce NH_(2)(g). Calculate the NH_(2)(g) formed. Identify the limiting reagent in the production of NH_(3) in this situation.

In a gaseous mixture at 4 atm pressure, 25% of molecules are Nitrogen, 40% of molecules are carbon dioxide and the rest are oxygen. The partial pressure of oxygen in the mixture is

In a gas lighter, mechanical energy is converted into electrical energy by using crystals of barium titanate. Barium titanate is

In a gas phase reaction, 50 kg of nitrogen and 10 g of hydrogen are mixed to produce ammonia. Identify the limiting reagent. Calculate the maximum amount of ammonia produced.

In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions . If one atom of B is missing from one of the face centred points, the formula of the compound is :

In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atoms of B is missing from one of the face centred points, the formula of the compound is :

In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centred points, the formula of the compound is

In a face centred cubic lattice, atom A occupies the corner positionsand atom B occupies the face centre positions.If one atomof B is missing from one of the face centred points.

In a face centredcubic lattice, a cell is shared equally by how many unit cells ?

In a face-centred cubic crystal, the nearest neighbour distance is ___________ times the edge of the crystal

In a face centred cubic cell, an atom at the face centre is shared by:

In a face centered cubic lattice atoms A are at the corner points and atoms B at the face centered points. If atoms B is missing from one of the face centered points, the formula of the ionic compound is

In a exothermic reaction, heat isevolved , andsystem losses heat to the surrounding . For such system:

In a Dumas nitrogen estimation method, 0.30 g of an organic compound gave 50 cm^(3) of N_(2) collected at 300 K and 715 mm Hg pressure. Calculate the percentage composition of nitrogen in the compound. (Vapour pressure of water at 300 K is 15 mm Hg)

In a electrochemical cell

In a cubic vessel of side length 5 cm, a Helium molecule moving with a velocity 10^5 cm/sec in X-direction then rate of change in momentum in X-direction is (in g cm/sec)

In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present ?

In a crystalline solid, anions B are arranged in a cubic close packing. Cations A are equally distrilbuted between octahedral and tetrahedral voids. If all the octahedral voids are occupied, what is the formula of the solid ?

In a crystalline solid, anions B are arranged in a cubic close packing. Cations A are equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied, what is the formula of the solid ?

In a crystalline solid, W atoms are at corners, O atoms at edge centres and Na at cube centre, The formula of solid is

In a crystal of an ionic compound, the ions B form the close packed lattice and the ions A occupy all the tetrahedral voids. The formula of the compound is

In a crystal, all the lattice sites are found to be occupied by covalent molecules. To which kind of solid does it belong ?

In a crystal, Frenkel defect is not shown by alkali metal halides but silver halides show. Why ?

In a crystal, Frenkel defect is not shown by alkali metal halides buy sliver halides show. Why ?

In a container m g of a gas is placed. After some time some gas is allowed to escape from container. The pressure of the gas becomes half and its absolute temperature 2//3 rd. The amount of gas escaped is

In a constant pressure calorimeter, 224 mL of 0.1 M KOH (aq) solution is reacted with 50ml of 0.1 MH_(2)SO_(4)(aq) solution then increase in temperature of solution will be 9assume heat capacity of calorimeter is negligible): Given : Specific heat of solution =1cal/g-K Density of solution =1g/mL

In a conjugate pair of reductant and oxidant, the oxidant has

In a conjugate pair of reductant and oxidant, the oxidant has "________________"

In a concentration cell the

In a compression process, P_("ext") is …………..

In a compound oxide ions are arranged in cubic close packing arrangement. Cations A occupy one-sixth of the tetrahedral voids and cations B occupy one-third of the octahdral voids. The formual of the compound is A_(x), B_(y), O_(z) then find the value of x + y + z